Mathematics 10: Grade 10 Mathematics Worktext/Workbook 9781983254550, 198325455X

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MATHEMATICS 10

CEDRIC A. BORRES

PATTERNS AND ALGEBRA GEOMETRY PROBABILITY AND STATISTICS

LORETA L. BORRES

AAL CEDBOR ENTERPRISES

Published by:

AAL CEDBOR ENTERPRISES

©Philippine Copyright, 2015 by:

Cedric A. Borres Loreta L. Borres

All Rights Reserved. No part of this workbook maybe reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior permission from the authors.

ISBN 978-198-3254-55-0

Preface This workbook in Mathematics Grade 10 seeks to develop mastery mathematics skills. The major goal of the authors in writing this workbook is to present, develop critical thinking and problem solving skills, implement discovery and inquiry-based learning, improve cooperative learning, constructivism, impose reflective learning, experiential and situated learning of the grade 10 junior high School Students (in k to 12 curriculum) in a simple and understandable language. Exercises are based on the topics listed in the official curriculum guide of all Grade 10 Junior High School. These are: 1. 2. 3. 4. 5. 6.

Sequences. Polynomial Functions Circles. Measures of Positions. Basic Combinational Concepts. Probability.

It is expected that through this workbook in Grade 10 “Mathematics 10”, mathematics education will be made fruitful for the Grade 10, Junior High School. Cedric A. Borres Loreta L. Borres

Table of Contents Chapter 1.

Sequences 1.1 1.2 1.3 1.4 1.5

2.

3.

Definitions and Types of Sequences Arithmetic Sequence Sum of Terms of an Arithmetic Sequence Geometric Sequence Finite and Infinite Geometric Sequence

1 2 9 16 22 32

Polynomial Functions

36

2.1 2.2 2.3 2.4 2.5 2.6

37 40 44 53 57 62

Linear Functions Quadratic Functions Polynomial Functions on Higher Degree Synthetic Division Remainder Theorem and Factor Theorem Zeros and Graphs of Polynomial Functions

Circles

67

3.1 3.2 3.3 3.4

68 71 77 84

Parts of a Circle Relation Among the Parts of a Circle Theorems Relating the Parts of a Circle Theorems on Secant and Tangent Lines and Segments

4.

Geometric Figures on the Rectangular Coordinate Plane 4.1 Distance Formula 4.2 Rule of Distance formula on Geometric Properties 4.3 Equation of a Circle 4.4 Problems Involving Geometric Figures in the Coordinate Plane

5

6

7

90 91 101 107 116

Measures of Position

125

5.1 Measures of Positions 5.1.1 Quartiles 5.1.2 Deciles 5.1.3 Percentiles 5.2 Problems Involving Measures of Position 5.3 Constructs Box Plot

126 126 137 149 157

Basic Combinational Concepts

175

6.1 Count Occurrences of an Event 6.2 Fundamental Counting Principle 6.3 Permutations and Combinations

161 190 204

Probability of Compound Events

224

7.1 7.2 7.3 7.4

225 231 243 254

Cardinality of a Union of Two Sets Union and Intersection of Events Probability of a Union of Two Sets Probabilities of a Union and Intersection of Events

167

This workbook is dedicated

To our God Almighty

for shading light and guidance

in writing the workbook,

1.0

sequence

1

Chapter 1 SEQUENCES 1.1 1.2 1.3 1.4 1.5 1.6

DEFINITIONS AND TYPES OF SEQUENCES ARITHMETIC SEQUENCE SUM OF TERMS OF AN ARITHMETIC SEQUENCE GEOMETRIC SEQUENCE FINITE AND INFINITE SEQUENCE PROBLEMS INVOLVING QUADRATIC FUNCTIONS AND EQUATIONS

Specific Objectives: at the end of the chapter the students should be able to;          

Generalize a pattern, Illustrate a sequence and some types of sequences, Graph arithmetic sequence, Determine the terms of an arithmetic sequence including the general nth term of the sequence, Define a geometric sequence, Graph a geometric sequence, Differentiate finite from infinite geometric sequence, Compare arithmetic from geometric sequence.. Find the sum of terms of a given geometric sequence , both finite and infinite, Solve problems involving sequences and their sum.

Cartan, Elie (Joseph) [kah(r)tã](1869--1951) Mathematician, born in Dolomieu, SE France. He held posts at Montpellier, Lyon, and (1912--40) the Sorbonne, becoming one of the most original mathematicians of his time. He worked on Lie groups and differential geometry, and founded the subject of analysis on differentiable manifolds, which is essential to modern fundamental physical theories. Among his discoveries are the theory of spinors, the method of moving frames and the exterior differential calculus. His son, Henri (Paul) Cartan (1904-- ) also became a mathematician, known for his work in the theory of analytic functions.

2

1.1.

Chapter 1

sequences

DEFINITIONS AND TYPES OF SEQUENCES

A "sequence" (or "progression", in British English) is an ordered list of numbers; the numbers in this ordered list are called "elements" or "terms". A "series" is the value you get when you add up all the terms of a sequence; this value is called the "sum". Example 1. "1, 2, 3, 4" is a sequence, with terms "1", "2", "3", and "4"; The corresponding series is the sum "1 + 2 + 3 + 4", and the value of the series is 10. A sequence may be named or referred to as "A" or "An". The terms of a sequence are usually named something like "Ai" or "An", with the subscripted letter "i" or "n" being the "index" or counter. So the second term of a sequence might be named "A2" (pronounced "Ay-sub-two"), and "A12" would designate the twelfth term. Note: Sometimes sequences start with an index of n = 0, so the first term is actually A0. Then the second term would be A1. The first listed term in such a case would be called the "zero-eth" term. This method of numbering the terms is used. Don't assume that every sequence and series will start with an index of n = 1 Different Types of Sequences We can divide sequences as per the number of elements which are present in any given sequence as Finite Sequence Infinite Sequence Fibonacci Sequence But, sometimes, we can divide sequences according to the definition of it or we can say the expression of it which are present in the given problem. These are Arithmetic Sequence Geometric Sequence Harmonic Sequence

1.1

definitions and types of sequences

3

Finite Sequence A sequence in which the number of elements which are present in it is finite or countable is called the finite sequence. The total number of the elements which are present in the finite sequence is called the length of the finite sequence. Example1. 1, 2, 3, 4, ..........90 is the finite sequence as this is the list of some finite numbers. Infinite Sequence A sequence is usually meant to be a progression of numbers with a clear starting point. Some sequences continue with no end. This type of sequence is called an infinite sequence. The system of natural numbers is an example of an infinite sequence. Example 1. the sequence of successive quotients mentioned above is an infinite sequence. Infinite means "it never ends". Often, it is possible to express the rule, which yields the various terms of a sequence in terms of algebraic formula. Consider for instance, the sequence of even natural numbers 2, 4, 6 … Here, a1 = 2 = 2 × 1, a2 = 4 = 2 × 2, a3 = 6 = 2 × 3, a4 = 8 = 2 × 4 … a24 = 48 = 2 × 24, and so on. Fibonacci Sequence In some cases, an arrangement of numbers such as 1, 1, 2, 3, 5, 8,.. has no visible pattern. But the sequence is generated by the recurrence relation and is given by a1 = a2 = 1 a3 = a1 + a2 an = an - 2 + an - 1, n > 2

4

Chapter 1

sequences

Arithmetic Sequence If we have a sequence as x1, x2 ,x3,......., and the sequence follows the rule as xj xi = d for all values of i and j, then the sequence is called the arithmetic sequence. This type of sequence where any term except the first term is obtained by adding a fixed number to the previous term. Arithmetic Sequence Formula: If we have x1, x2 , x3 ,.....,xn,..... be the any arithmetic sequence, then the sum of this finite series, where n is any positive integer is given by x1+x2+..............+xn=∑ni=1xi If a is the first term, d is the common difference, Tn is the nth term and Sn is the sum of any arithmetic sequence, then nth term given by Tn = a + (n - 1) d Sum of the first n terms of the sequence is Sn = n2 [2a + (n - 1)d] = n2 (a + L), where l is the last (nth) term of the sequence. If a, b and c are in the form of arithmetic sequence, then b is known as the arithmetic mean and expressed as b = a+c2 Example 1. consider 4, 7, 10.... In this sequence, first term is 4 and we get the second term as 7 = 4 + 3. Second term is 7 and we get third term as 10 = 7 + 3. So, it is clear that the difference between the two successive terms is 3 and it is the same for all terms or in the complete sequence. So, this is the example for arithmetic sequence. Consider the sequence 2, 7, 12, 17, 22,....... The nth term of any arithmetic sequence is given by Tn = a + (n - 1)d The 6th term(taking n = 6), T6 = 2 + ( 6 -1) 5 = 27 The 8th term(taking n = 8), T8 = 2 + ( 8 -1) 5 = 37

1.1

definitions and types of sequences

5

Geometric Sequence A geometric sequence or progression is a set of terms each of which is formed by multiplying to the preceding a fixed quantity called the common ratio. The common ratio may be integral or fractional, negative or positive and it can be found by dividing any term of the term that precedes it. Geometric Sequence Formula If a is the first term, r is the common ratio, Tn is the nth term and Sn is the sum of any finite geometric sequence, then nth term is given by Tn = arn – 1 Sum of first n terms is given by the relation Sn=a(1−rn)1−r, where r ≠ 1. Example 1. 3, 9, 27, 81, … is a geometric sequence, 3 being multiplied to any term to get to the next term. 3 therefore is the common ratio To get the missing term of a geometric sequence, we use the formula: an = a1 rn-1 where an = last term a1 = first term n = number of terms r = common ratio Example 2. Consider the sequence 10, 20, 40, 80 Here, a = 10, r = 2 5th term is T5 = 10 (2) 5 - 1 = 10(2) 4 = 160 7th term is T7 = 10(2) 7 - 1 = 10(2) 6 = 640

6

Chapter 1

sequences

Example 3. In the geometric sequence 7, 21, 63, … find the 7th term. Solution:

n =7 a1 = 7 r =3 an = a1 rn-1 = 7 (3)7-1 = 7 (3)6 = 5103 Therefore the 7th term is 5103.

Harmonic Sequence Given: 1, ½, 1/3, ¼… Observe the denominators of each term in the sequence. Try to get the reciprocals. Do these form a sequence? If so, what type of sequence? A sequence whose reciprocals form an arithmetic sequence is called a harmonic sequence. Example1. 1, ½, 1/3, ¼… The reciprocal form an arithmetic sequence the harmonic sequence 2/3, ½, 2/5, 1/3, 2/7… We can say that ½ is the harmonic means between 2/3 and 2/5; ½, 2/5 and 1/3 are the harmonic between 2/3 and 2/7. 2/3+1/2+2/5+… is a harmonic series. Example 2. Find the 12th term of the harmonic sequence 1/9, 1/12, 1/15... Note that the reciprocal forms an arithmetic sequence so we may first find the 12th term of the harmonic sequence which is 1/42

1.1

definitions and types of sequences

7

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.1 1. Find the common difference in the following arithmetic finite sequence: 2, 7, 12, 17, ?, 47

2. Find the common ratio in the following geometric finite sequence: 7, 21, 63, 189, 567

3. What term is 128 in the geometric sequence 1/8, -1/4, -1, 2, … 128

8

Chapter 1

sequences

4. If a1 = 2, a2 = 3 and an + (an – 1)(an – 2) what is a5

5. Find the tenth term and the n-th term of the following sequence: 1/2, 1, 2, 4, 8,...

6. Find the n-th and the 26th terms of the geometric sequence with a5 = 5/4 and a12 = 160.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.1 1. Find the common difference in the following arithmetic finite sequence: 2, 7, 12, 17, ?, 47 The first four terms of the sequence show that the common difference is 5. In other words, we can add 5 to any term in the sequence to get the next term in the sequence. 2. Find the common ratio in the following geometric finite sequence: 7, 21, 63, 189, 567 The sequence has a common ratio of 3 because 21/7 = 63/21 = 189/63 = 567/189 = 3. In other words, we can multiply any term in the sequence by 3 to get the next term in the sequence. 3. What term is 128 in the geometric sequence 1/8, -1/4, -1, 2, … 128 Solution: a1 = 1/8 r = -2 an = 128 n=? an = a1 rn-1 128 = 1/8 (-2)n-1 128(8) = (-2)n-1 1024 = (-2)n-1 Since 1024 = (-2)10 then, (-2)10 = (-2)n-1 therefore, 10 = n-1 n = 11 Thus, 128 128 is the 11th term

4. If a1 = 2, a2 = 3 and an + (an – 1)(an – 2) what is a5 Solution an = (an – 1)(an – 2 a5 = (a4)(a3 a4 = (a3)(a2) a3 = (a2)(a1) By substitution: a3 = (3)(2) = 6 a5 = (18)(6) = 108 a5 = (6)(2) = 18 5. Find the tenth term and the n-th term of the following sequence: 1/2, 1, 2, 4, 8,... The differences don't match: 2 – 1 = 1, but 4 – 2 = 2. So this isn't an arithmetic sequence. On the other hand, the ratios are the same: 2 ÷ 1 = 2, 4 ÷ 2 = 2, 8 ÷ 4 = 2. So this is a geometric sequence with common ratio r = 2 and a = 1/2. To find the tenth and n-th terms, I can just plug into the formula an = ar(n – 1): an = (1/2) 2n–1 a10 = (1/2) 210–1 = (1/2) 29 = (1/2)(512) = 256 6. Find the n-th and the 26th terms of the geometric sequence with a5 = 5/4 and a12 = 160. These two terms are 12 – 5 = 7 places apart, so, from the definition of a geometric sequence, I know that a12 = ( a5 )( r7 ). I can use this to solve for the value of the common ratio r: 160 = (5/4)(r7) 128 = r7 2=r Since a5 = ar4, then I can solve for the value of the first term a: 5/4 = a(24) = 16a 5/64 = a Once I have the value of the first term and the value of the common ratio, I can plug each into the formulas, and find my answers: an = (5/64)2(n – 1) a26 = (5/64)(225) = 2 621 440

1.2

arithmetic sequence

9

1.2 ARITHMETIC SEQUENCE Arithmetic sequence is a sequence in which a constant number can be added to the preceding term to get the next term. Sometimes it is called an arithmetic progression. The constant number that is added is called the common difference Example 1: 1, 5, 9, 13, 17, 21,... is an arithmetic sequence, 4 being added to any term to get the next term. The nth term of an arithmetic sequence is defined as follows: an = a1 + (n – 1) d where: an = nth term of the sequence a1 = the first term of the sequence n = the number of terms d = common difference Example 2. Find the 10th term of the arithmetic sequence 7, 0, -7, . . . an = a1 + (n – 1) d a10 = 7 + 9 (-7) a10 = -56

a1 = 7

n = 10

d = -7

Example 3. In the arithmetic sequence 1, 3/2, 2, . . . which term has a value of 11? an = a1 + (n – 1) d 11 = 1 + (n – 1) ½ 10 (2) = n – 1 n = 20 + 1 n = 21

a1 = 1

d = 1/2

Therefore, 11 is 21st term in the sequence

an = 11

10

Chapter 1

sequences

Example 4. The forth term in the arithmetic sequence is 18 and the ninth term is 43. Write the sequence t4 = 18 and t9 = 43 an = a1 + (n– 1) d a4 = a1 + (4 – 1) d 18 = a1 + 3d à Equation 1 and when n = 9 a9 = a1 + (9 +1) d 43 = a1 + 8d à Equation 2 Subtracting equations 1 and 2 18 = a1 + 3d (-) (-) (-) 43 = a1 + 8d -25 = -5d d=5 Substituting: 18 = a1 + 3d 18 = a1 + 3(5) 18 – 15 = a1 a1 = 3 Therefore the sequence is 3, 8, 13, . . . Continue adding 5 to the sequence until you have 9 terms. Example 5. What is the first term of an arithmetic sequence in which the 78th term is 176 and the common difference is 2? an = a1 + (n– 1) d 176 = a1 + (78 – 1) (2) 176 = a1 + (77) (2) 176 = a1 + 154 a1 = 176 – 154 a1 = 22

1.2

arithmetic sequence

11

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.2 1. Find the common difference and the next term of the following sequence: 3, 11, 19, 27, 35,...

2. Find the common ratio and the seventh term of the following sequence: 2/9, 2/3, 2, 6, 18,...

12

Chapter 1

sequences

3. Find the n-th term and the first three terms of the arithmetic sequence having a6 = 5 and d= 3/2.

4. Find the n-th term and the first three terms of the arithmetic sequence having a4 = 93 anda8 = 65.

1.2

arithmetic sequence

13

5. The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the n th term and the value of the 50 th term

6. The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20 th term

7. An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term.

14

Chapter 1

sequences

8. An arithmetic sequence has a its 5 th term equal to 22 and its 15 th term equal to 62. Find its 100 th term.

9. Find the sum of all the integers from 1 to 1000.

1.2

arithmetic sequence

15

10. Find the sum of the first 50 even positive integers.

11. We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms s50 = 50 (2 + 100) / 2 = 2550. Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.2 1. Find the common difference and the next term of the following sequence: 3, 11, 19, 27, 35,... To find the common difference, I have to subtract a pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other: 11 – 3 = 8 19 – 11 = 8 27 – 19 = 8 35 – 27 = 8 The difference is always 8, so d = 8. Then the next term is 35 + 8 = 43. 2. Find the common ratio and the seventh term of the following sequence: 2/9, 2/3, 2, 6, 18,... To find the common ratio, I have to divide a pair of terms. It doesn't matter which pair I pick, as long as they're right next to each other:

The ratio is always 3, so r = 3. Then the sixth term is (18)(3) = 54 and the seventh term is (54)(3) = 162.

3. Find the n-th term and the first three terms of the arithmetic sequence having a6 = 5 and d= 3/2. The n-th term of an arithmetic sequence is of the form an = a + (n – 1)d. In this case, that formula gives me a6 = a + (6 – 1)(3/2) = 5. Solving this formula for the value of the first term of the sequence, I get a = –5/2. Then: a1 = –5/2, a2 = –5/2 + 3/2 = –1, a3 = –1 + 3/2 = 1/2, and an = –5/2 + (n – 1)(3/2) 4. Find the n-th term and the first three terms of the arithmetic sequence having a4 = 93 anda8 = 65. Since a4 and a8 are four places apart, then I know from the definition of an arithmetic sequence that a8 = a4 + 4d. Using this, I can then solve for the common difference d: 65 = 93 + 4d –28 = 4d –7 = d Also, I know that a4 = a + (4 – 1)d, so, using the value I just found for d, I can find the value of the first term a: 93 = a + 3(–7) 93 + 21 = a 114 = a Once I have the value of the first term and the value of the common difference, I can plug-n-chug to find the values of the first three terms and the general form of the n-th term: a1 = 114, a2 = 114 – 7 = 107, a3 = 107 – 7 = 100 an = 114 + (n – 1)(–7)

5. The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the n th term and the value of the 50 th term Solution: Use the value of the common difference d = 3 and the first term a1 = 6 in the formula for the n th term given above an = a1 + (n - 1 )d = 6 + 3 (n - 1) =3n+3 The 50 th term is found by setting n = 50 in the above formula. a50 = 3 (50) + 3 = 153 6. The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20 th term Solution: Use the value of the common difference d = -10 and the first term a1 = 200 in the formula for the n th term given above and then apply it to the 20 th term a20 = 200 + (-10) (20 - 1 ) = 10 7. An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term. Solution: We use the n th term formula for the 6 th term, which is known, to write a6 = 52 = a1 + 10 (6 - 1 ) The above equation allows us to calculate a1. a1 = 2 Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows. a15 = 2 + 10 (15 - 1) = 142

8. An arithmetic sequence has a its 5 th term equal to 22 and its 15 th term equal to 62. Find its 100 th term. Solution: We use the n th term formula for the 5 th and 15 th terms to write a5 = a1 + (5 - 1 ) d = 22 a15 = a1 + (15 - 1 ) d = 62 We obtain a system of 2 linear equations where the unknown are a1 and d. Subtract the right and left term of the two equations to obtain 62 - 22 = 14 d - 4 d Solve for d. d=4 Now use the value of d in one of the equations to find a1. a1 + (5 - 1 ) 4 = 22 Solve for a1 to obtain. a1 = 6 Now that we have calculated a1 and d we use them in the n th term formula to find the 100 th formula. a100 = 6 + 4 (100 - 1 )= 402 9. Find the sum of all the integers from 1 to 1000. Solution: The sequence of integers starting from 1 to 1000 is given by 1 , 2 , 3 , 4 , ... , 1000 The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above). s1000 = 1000 (1 + 1000) / 2 = 500500

10. Find the sum of the first 50 even positive integers. Solution: 

The sequence of the first 50 even positive integers is given by 2 , 4 , 6 , ...



The above sequence has a first term equal to 2 and a common difference d = 2. We use the n th term formula to find the 50 th term a50 = 2 + 2 (50 - 1) = 100

11. We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms s50 = 50 (2 + 100) / 2 = 2550 Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5. Solution: The first few terms of a sequence of positive integers divisible by 5 is given by 5 , 10 , 15 , ... The above sequence has a first term equal to 5 and a common difference d = 5. We need to know the rank of the term 1555. We use the formula for the n th term as follows 1555 = a1 + (n - 1 )d Substitute a1 and d by their values 1555 = 5 + 5(n - 1 ) Solve for n to obtain n = 311 We now know that 1555 is the 311 th term, we can use the formula for the sum as follows s311 = 311 (5 + 1555) / 2 = 242580

16

Chapter 1

sequences

1.3 Sum of Terms of an Arithmetic Sequence To sum up the terms of this arithmetic sequence: a + (a + d) + (a + 2d) + (a + 3d) + ... use this formula:

It is called Sigma Notation (called Sigma) means "sum up" And below and above it are shown the starting and ending values:

It says "Sum up n where n goes from 1 to 4. Answer=10 Example1. Add up the first 10 terms of the arithmetic sequence: { 1, 4, 7, 10, 13, ... } The values of a, d and n are: a = 1 (the first term) d = 3 (the "common difference" between terms) n = 10 (how many terms to add up) So:

Becomes:

= 5(2+9·3) = 5(29) = 145

1.3

the sum of terms of an arithmetic sequence

First, we will call the whole sum "S": S = a + (a + d) + ... + (a + (n-2)d) + (a + (n-1)d) Next, rewrite S in reverse order: S = (a + (n-1)d) + (a + (n-2)d) + ... + (a + d) + a Now add those two, term by term: S = a + (a+d) + ... + (a + (n-2)d) S

= (a + (n-1)d)

+ (a + (n-2)d)

+ ... + (a + d)

17

+ (a + (n-1)d) + a

2S = (2a + (n-1)d) + (2a + (n-1)d) + ... + (2a + (n-1)d) + (2a + (n-1)d) Each term is the same! And there are "n" of them so ... 2S = n × (2a + (n-1)d) Now, just divide by 2 and we get: S = (n/2) × (2a + (n-1)d) Which is our formula:

Example 1. Find the sum of 1 + 5 + 9 + ... + 49 + 53. Checking the terms, I can see that this is indeed an arithmetic series: 5 – 1 = 4, 9 – 5 = 4, 53 – 49 = 4. I've got the first and last terms, but how many terms are there in total? I have the n-th term formula, "an = a1 + (n – 1)d", and I have a1 = 1 and d = 4. Plugging these into the formula, I can figure out how many terms there are: an = a1 + (n – 1)d 53 = 1 + (n – 1)(4) 53 = 1 + 4n – 4 53 = 4n – 3 56 = 4n 14 = n So there are 14 terms in this series. Now I have all the information I need: 1 + 5 + 9 + ... + 49 + 53 = (14/2)(1 + 53) = (7)(54) = 378

18

Chapter 1

sequences

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.3 1. Find the sum of the first six terms of An, where an = 2an–1 + an–2, a1 = 1, and a2 = 1.

2. Find the sum of 1 + 5 + 9 + ... + 49 + 53.

1.3

the sum of terms of an arithmetic sequence

3. The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the n th term and the value of the 50 th term

4. The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20 th term

5. An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term.

6. An arithmetic sequence has a its 5 th term equal to 22 and its 15 th term equal to 62. Find its 100 th term.

19

20

Chapter 1

sequences

7. Find the sum of all the integers from 1 to 1000.

8. Find the sum of the first 50 even positive integers.

1.3

the sum of terms of an arithmetic sequence

9. Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5.

10.

Find the sum S defined by 10 S = ∑ (2n + 1 / 2) n=1

21

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.3 1. Find the sum of the first six terms of An, where an = 2an–1 + an–2, a1 = 1, and a2 = 1. a3 = 2a3–1 + a3–2 = 2a2 + a1 = 2(1) + (1) = 2 + 1 = 3 a4 = 2a4–1 + a4–2 = 2a3 + a2 = 2(3) + (1) = 6 + 1 = 7 a5 = 2a5–1 + a5–2 = 2a4 + a3 = 2(7) + (3) = 14 + 3 = 17 a6 = 2a6–1 + a6–2 = 2a5 + a4 = 2(17) + (7) = 34 + 7 = 41 Now that I've found the values of the third through the sixth terms, I can find the value of the series; the sum is: 1 + 1 + 3 + 7 + 17 + 41 = 70 2. Find the sum of 1 + 5 + 9 + ... + 49 + 53. an = a1 + (n – 1)d 53 = 1 + (n – 1)(4) 53 = 1 + 4n – 4 53 = 4n – 3 56 = 4n 14 = n So there are 14 terms in this series. Now I have all the information I need: 1 + 5 + 9 + ... + 49 + 53 = (14/2)(1 + 53) = (7)(54) = 378

1.3

the sum of terms of an arithmetic sequence

19

3. The first term of an arithmetic sequence is equal to 6 and the common difference is equal to 3. Find a formula for the n th term and the value of the 50 th term Use the value of the common difference d = 3 and the first term a1 = 6 in the formula for the n th term given above an = a1 + (n - 1 )d = 6 + 3 (n - 1) =3n+3 The 50 th term is found by setting n = 50 in the above formula. a50 = 3 (50) + 3 = 153 4. The first term of an arithmetic sequence is equal to 200 and the common difference is equal to -10. Find the value of the 20 th term Use the value of the common difference d = -10 and the first term a1 = 200 in the formula for the n th term given above and then apply it to the 20 th term a20 = 200 + (-10) (20 - 1 ) = 10 5. An arithmetic sequence has a common difference equal to 10 and its 6 th term is equal to 52. Find its 15 th term. We use the n th term formula for the 6 th term, which is known, to write a6 = 52 = a1 + 10 (6 - 1 ) The above equation allows us to calculate a1. a1 = 2 Now that we know the first term and the common difference, we use the n th term formula to find the 15 th term as follows. a15 = 2 + 10 (15 - 1) = 142 6. An arithmetic sequence has a its 5 th term equal to 22 and its 15 th term equal to 62. Find its 100 th term. a5 = a1 + (5 - 1 ) d = 22 a15 = a1 + (15 - 1 ) d = 62

We obtain a system of 2 linear equations where the unknown are a1 and d. Subtract the right and left term of the two equations to obtain 62 - 22 = 14 d - 4 d Solve for d. d=4 Now use the value of d in one of the equations to find a1. a1 + (5 - 1 ) 4 = 22 Solve for a1 to obtain. a1 = 6 Now that we have calculated a1 and d we use them in the n th term formula to find the 100 th formula. a100 = 6 + 4 (100 - 1 )= 402 7. Find the sum of all the integers from 1 to 1000. The sequence of integers starting from 1 to 1000 is given by 1 , 2 , 3 , 4 , ... , 1000 The above sequence has 1000 terms. The first term is 1 and the last term is 1000 and the common difference is equal to 1. We have the formula that gives the sum of the first n terms of an arithmetic sequence knowing the first and last term of the sequence and the number of terms (see formula above). s1000 = 1000 (1 + 1000) / 2 = 500500 8. Find the sum of the first 50 even positive integers. The sequence of the first 50 even positive integers is given by 2 , 4 , 6 , ... The above sequence has a first term equal to 2 and a common difference d = 2. We use the n th term formula to find the 50 th term a50 = 2 + 2 (50 - 1) = 100 We now the first term and last term and the number of terms in the sequence, we now find the sum of the first 50 terms s50 = 50 (2 + 100) / 2 = 2550

9. Find the sum of all positive integers, from 5 to 1555 inclusive, that are divisible by 5. The first few terms of a sequence of positive integers divisible by 5 is given by 5 , 10 , 15 , ... The above sequence has a first term equal to 5 and a common difference d = 5. We need to know the rank of the term 1555. We use the formula for the n th term as follows 1555 = a1 + (n - 1 )d Substitute a1 and d by their values 1555 = 5 + 5(n - 1 ) Solve for n to obtain n = 311 We now know that 1555 is the 311 th term, we can use the formula for the sum as follows s311 = 311 (5 + 1555) / 2 = 242580 10.

Find the sum S defined by

10 S = ∑ (2n + 1 / 2) n=1 Let us first decompose this sum as follows 10 S = ∑ (2n + 1 / 2) n=1 10 10 = 2 ∑ n + ∑ (1 / 2) i=1 n=1 10(1+10)/2 = 55 The term ∑ (1 / 2) is the addition of a constant term 10 times and is given by 10(1/2) = 5 The sum S is given by S = 2(55) + 5 = 115

22

Chapter 1

sequences

1.4 GEOMETRIC SEQUENCE Sequence A Sequence is a set of things (usually numbers) that are in order.

Geometric Sequences In a Geometric Sequence each term is found by multiplying the previous term by a constant. 2, 4, 8, 16, 32, 64, 128, 256, ... Example This sequence has a factor of 2 between each number. Each term (except the first term) is found by multiplying the previous term by 2. In General we write a Geometric Sequence like this: {a, ar, ar2, ar3, ... } where: a is the first term, and r is the factor between the terms (called the "common ratio") Example 2. {1,2,4,8,...} The sequence starts at 1 and doubles each time, so a=1 (the first term) r=2 (the "common ratio" between terms is a doubling) And we get: {a, ar, ar2, ar3, ... } = {1, 1×2, 1×22, 1×23, ... } = {1, 2, 4, 8, ... } But be careful, r should not be 0: When r=0, we get the sequence {a,0,0,...} which is not geometric.

The Rule We can also calculate any term using the Rule: xn = ar(n-1) (We use "n-1" because ar0 is for the 1st term)

1.4

geometric sequence

10, 30, 90, 270, 810, 2430, ... 1. Example This sequence has a factor of 3 between each number. The values of a and r are: a = 10 (the first term) r = 3 (the "common ratio") The Rule for any term is: xn = 10 × 3(n-1) So, the 4th term is: x4 = 10×3(4-1) = 10×33 = 10×27 = 270 And the 10th term is: x10 = 10×3(10-1) = 10×39 = 10×19683 = 196830 A Geometric Sequence can also have smaller and smaller values: Example 1. 4, 2, 1, 0.5, 0.25, . . .This sequence has a factor of 0.5 (a half) between each number. Its Rule is xn = 4 × (0.5)n-1 Increasing the dimensions in geometry: a line is 1-dimensional and has a length of r in 2 dimensions a square has an area of r2 in 3 dimensions a cube has volume r3

23

24

Chapter 1

sequences

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.4 1. Find the tenth term and the n-th term of the following sequence: 1/2, 1, 2, 4, 8,...

2. Find the n-th term and the first three terms of the arithmetic sequence having a6 = 5 and d= 3/2.

1.4

geometric sequence

3. Find the n-th term and the first three terms of the arithmetic sequence having a4 = 93 anda8 = 65.

4. Find the n-th and the 26th terms of the geometric sequence with a5 = 5/4 and a12 = 160.

25

26

Chapter 1

sequences

5. The sum of the interior angles of a triangle is 180º, of a quadrilateral is 360º and of a pentagon is 540º. Assuming this pattern continues, find the sum of the interior angles of a dodecagon (12 sides). The sequence 180 360 540 ... ? Sides: 3 4 5 ... 12 Term: 1 2 3 ... ?

6. After knee surgery, your trainer tells you to return to your jogging program slowly. He suggests jogging for 12 minutes each day for the first week. Each week thereafter, he suggests that you increase that time by 6 minutes per day. How many weeks will it be before you are up to jogging 60 minutes per day?

1.4

geometric sequence

27

7. You complain that the hot tub in your hotel suite is not hot enough. The hotel tells you that they will increase the temperature by 10% each hour. If the current temperature of the hot tub is 75º F, what will be the temperature of the hot tub after 3 hours, to the nearest tenth of a degree? Starting temperature is 75º.

8. A culture of bacteria doubles every 2 hours. If there are 500 bacteria at the beginning, how many bacteria will there be after 24 hours? There are 500 bacteria to start, doubling every 2 hrs. 500 Start Term #

1000 after 2 hrs. 1

2000 4 hrs

... ...

? 24 hrs

2

...

?

28

Chapter 1

sequences

9. A mine worker discovers an ore sample containing 500 mg of radioactive material. It is discovered that the radioactive material has a half-life of 1 day. Find the amount of radioactive material in the sample at the beginning of the 7th day.500 mg of ore. Half-life of one day means that half of the amount remains after 1 day. Begin of day 1 500 mg End of day 1 250 mg

10.

Begin of day 2 250 mg End of day 2 125 mg

Begin of day 3 125 mg End of day 3 62.5 mg

... ...

Find the terms a2, a3, a4 and a5of a geometric sequence if a1 = 10 and the common ratio r = -1.

1.4

geometric sequence

11.

Find the 10th term of a geometric sequence if a1 = 45 and the common ratio r = 0.2.

12.

Find a20 of a geometric sequence if the first few terms of the sequence are given by -

, ,- ,

,...

29

30

Chapter 1

sequences

13.

Given the terms a10 =

and

of a geometric sequence, find the

exact value of the term a30 of the sequence.

14.

Find the sum 6 S = ∑ 3k-1 k=1

15.

Find the sum

1.4

geometric sequence

S =∑

16. Write the rational number 5.31313131. . . as the ratio of two integers.

31

NAME:______________________________ GRADE/SECTION:_____________________

SCORE:_____________ DATE:_______________

PRACTICE SET 1.4 1. Find the tenth term and the n-th term of the following sequence: 1/2, 1, 2, 4, 8,... The differences don't match: 2 – 1 = 1, but 4 – 2 = 2. So this isn't an arithmetic sequence. On the other hand, the ratios are the same: 2 ÷ 1 = 2, 4 ÷ 2 = 2, 8 ÷ 4 = 2. So this is a geometric sequence with common ratio r = 2 and a = 1/2. To find the tenth and n-th terms, I can just plug into the formula an = ar(n – 1):

an = (1/2) 2n–1 a10 = (1/2) 210–1 = (1/2) 29 = (1/2)(512) = 256

2. Find the n-th term and the first three terms of the arithmetic sequence having a6 = 5 and d= 3/2. The n-th term of an arithmetic sequence is of the form an = a + (n – 1)d. In this case, that formula gives me a6 = a + (6 – 1)(3/2) = 5. Solving this formula for the value of the first term of the sequence, I get a = –5/2. Then:

a1 = –5/2, a2 = –5/2 + 3/2 = –1, a3 = –1 + 3/2 = 1/2, and an = –5/2 + (n – 1)(3/2)

1.4

geometric sequence

25

3. Find the n-th term and the first three terms of the arithmetic sequence having a4 = 93 anda8 = 65. Since a4 and a8 are four places apart, then I know from the definition of an arithmetic sequence that a8 = a4 + 4d. Using this, I can then solve for the common difference d: 65 = 93 + 4d –28 = 4d –7 = d Also, I know that a4 = a + (4 – 1)d, so, using the value I just found for d, I can find the value of the first term a: 93 = a + 3(–7) 93 + 21 = a 114 = a Once I have the value of the first term and the value of the common difference, I can plug-n-chug to find the values of the first three terms and the general form of the n-th term: a1 = 114, a2 = 114 – 7 = 107, a3 = 107 – 7 = 100 an = 114 + (n – 1)(–7) 4. Find the n-th and the 26th terms of the geometric sequence with a5 = 5/4 and a12 = 160. These two terms are 12 – 5 = 7 places apart, so, from the definition of a geometric sequence, I know that a12 = ( a5 )( r7 ). I can use this to solve for the value of the common ratio r: 160 = (5/4)(r7) 128 = r7 2=r Since a5 = ar4, then I can solve for the value of the first term a: 5/4 = a(24) = 16a 5/64 = a Once I have the value of the first term and the value of the common ratio, I can plug each into the formulas, and find my answers: an = (5/64)2(n – 1) a26 = (5/64)(225) = 2 621 440

5. The sum of the interior angles of a triangle is 180º, of a quadrilateral is 360º and of a pentagon is 540º. Assuming this pattern continues, find the sum of the interior angles of a dodecagon (12 sides). The sequence 180 360 540 ... Sides: 3 4 5 ... Term: 1 2 3

? 12 ...

?

This sequence is arithmetic and the common difference is 180. The 12sided figure will be the 10th term in this sequence. Find the 10th term.

6. After knee surgery, your trainer tells you to return to your jogging program slowly. He suggests jogging for 12 minutes each day for the first week. Each week thereafter, he suggests that you increase that time by 6 minutes per day. How many weeks will it be before you are up to jogging 60 minutes per day? Adding 6 minutes to the weekly jogging time for each week creates the sequence: 12, 18, 24, ... This sequence is arithmetic.

This sequence is arithmetic.

7. You complain that the hot tub in your hotel suite is not hot enough. The hotel tells you that they will increase the temperature by 10% each hour. If the current temperature of the hot tub is 75º F, what will be the temperature of the hot tub after 3 hours, to the nearest tenth of a degree? Starting temperature is 75º. If the temperature is increased by 10%, the new temperature will be 110% of the original temperature. The common ratio will be 1.10. There are 4 terms. 75, after 1 hour, after 2 hours, after 3 hours (4 terms)

8. A culture of bacteria doubles every 2 hours. If there are 500 bacteria at the beginning, how many bacteria will there be after 24 hours? There are 500 bacteria to start, doubling every 2 hrs. 500 Start Term #

1000 after 2 hrs. 1

2000 4 hrs

... ...

? 24 hrs

2

...

?

Such bacterial growth is a geometric sequence with a common ratio of 2. The number of hours, however, is arithmetic with common difference of 2. Which term number is 24? Find out by observation or:

Now find the number of bacteria. The starting number and number of terms used may vary:

9. A mine worker discovers an ore sample containing 500 mg of radioactive material. It is discovered that the radioactive material has a half-life of 1 day. Find the amount of radioactive material in the sample at the beginning of the 7th day.500 mg of ore. Half-life of one day means that half of the amount remains after 1 day. Begin of day 1 500 mg End of day 1 250 mg

Begin of day 2 250 mg End of day 2 125 mg

Begin of day 3 125 mg End of day 3 62.5 mg

... ...

Decide to either work with the "beginning" of each day, or the "end" of each day, as each can yield the answer. Only the starting value and number of terms will differ. We will use "beginning":

10.

Find the terms a2, a3, a4 and a5 of a geometric sequence if a1 = 10 and the common ratio r = - 1. Use the definition of a geometric sequence a2 = a1 * r = 10 * (-1) = - 10 a3 = a2 * r = - 10 * (-1) = 10 a4 = a3 * r = 10 * (-1) = - 10 a5 = a4 * r = - 10 * (-1) = 10

11.

Find the 10 th term of a geometric sequence if a1 = 45 and the common ration r = 0.2. Use the formula an = a1 * rn-1 that gives the n th term to find a10 as follows a10 = a1 * rn-1 = 45 * 0.29 = 2.304 * 10-5

12.

Find a20 of a geometric sequence if the first few terms of the sequence are given by -1/2 , 1/4 , -1/8 , 1 / 16 , ... We first use the first few terms to find the common ratio r = a2 / a1 = (1/4) / (-1/2) = -1/2 r = a3 / a2 = (-1/8) / (1/4) = -1/2 r = a4 / a3 = (1/16) / (-1/8) = -1/2 The common ration r = -1/2. We now use the formula an = a1 * rn-1 for the n th term to find a20 as follows. a20 = a1 * r20-1 = (-1/2) * (-1/2)20-1 = 1 / (2020)

13.

Given the terms a10 = 3 / 512 and a15 = 3 / 16384 of a geometric sequence, find the exact value of the term a30 of the sequence. We first use the formula for the n th term to write a10 and a15 as follows a10 = a1 * r10-1 = 3 / 512 a15 = a1 * r15-1 = 3 / 16384 We now divide the terms a10 and a15 to write a15 / a10 = (a1 * r14 / a1 * r9) = (3 / 16384) / (3 / 512) Solve for r to obtain. r5 = 1 / 32 which gives r = 1/2 We now use a10 to find a1 as follows. a10 = 3 / 512 = a1(1/2)9 Solve for a1 to obtain. a1 = 3 We now use the formula for the n th term to find a30 as follows. a30 = 3(1/2)29 = 3 / 536870912

14.

Find the sum 6 S = ∑ 3k-1 k=1 We first rewrite the sum S as follows S = 1 + 3 + 9 + 27 + 81 + 243 = 364 Another method is to first note that the terms making the sum are those of an arithmetic sequence with a1 = 1 and r = 3 using the formula sn = a1 (1 - rn) / (1 - r) with n = 6. s6 = 1 (1 - 36) / (1 - 3) = 364

1.4

geometric sequence

31

15. Find the sum 10 S = ∑ 8*(1/4)i-1 i=1 An examination of the terms included in the sum are 8 , 8*((1/4)1 , 8*((1/4)2 , ... , 8*((1/4)9 These are the terms of a geometric sequence with a1 = 8 and r = 1/4 and therefore we can use the formula for the sum of the terms of a geometric sequence s10 = a1 (1 - rn) / (1 - r) = 8 * (1 - (1/4)10) / (1 - 1/4) = 10.67 (rounded to 2 decimal places) a. Write the rational number 5.31313131... as the ratio of two integers. We first write the given rational number as an infinite sum as follows 5.313131... = 5 + 0.31 + 0.0031 + 0.000031 + .... The terms making 0.31 + 0.0031 + 0.000031 ... are those of a geometric sequence with a1 = 0.31 and r = 0.01. Hence the use of the formula for an infinite sum of a geometric sequence S = a1 / (1 - r) = 0.31 / (1 - 0.01) = 0.31 / 0.99 = 31 / 99 We now write 5.313131... as follows 5.313131... = 5 + 31/99 = 526 / 99

32

Chapter 1

1.5

sequences

FINITE AND INFINITE GEOMETRIC SEQUENCE

If something is finite, then it has a limit or is bounded. a sequence is usually meant to be a progression of numbers with a clear starting point. Some sequences also stop at a certain number. In other words, they have a first term and a last term, and all the terms follow a specific order. This type of sequence is called a finite sequence.

Sum of Finite Geometric Progression The sum in geometric progression (also called geometric series) is given by → Equation (1)

Multiply both sides of Equation (1) by r will have → Equation (2)

Subtract Equation (2) from Equation (1)

The above formula is appropriate for GP with r < 1.0 Subtracting Equation (1) from Equation (2) will give

This formula is appropriate for GP with r > 1.0.

1.5

finite and infinite geometric sequence

33

Sum of Infinite Geometric Progression, IGP The number of terms in infinite geometric progression will approach to infinity (n = ∞). Sum of infinite geometric progression can only be defined at the range of -1.0 < (r ≠ 0) < +1.0 exclusive. From

For n → ∞, the quantity (a1 rn) / (1 - r) → 0 for -1.0 < (r ≠ 0) < +1.0, thus,

When the sequence goes on forever it is called an infinite sequence, otherwise it is a finite sequence Examples:  {1, 2, 3, 4, ...} is a very simple sequence (and it is an infinite sequence)  {20, 25, 30, 35, ...} is also an infinite sequence  {1, 3, 5, 7} is the sequence of the first 4 odd numbers (and is a finite sequence)  {4, 3, 2, 1} is 4 to 1 backwards  {1, 2, 4, 8, 16, 32, ...} is an infinite sequence where every term doubles  {a, b, c, d, e} is the sequence of the first 5 letters alphabetically  {f, r, e, d} is the sequence of letters in the name "fred"  {0, 1, 0, 1, 0, 1, ...} is the sequence of alternating 0s and 1s (yes they are in order, it is an alternating order in this case)

34

Chapter 1

sequences

.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.5 1. Find the sum of the first 8 terms of the geometric series if a1 = 1 and r = 2.

2. Find S10 of the geometric sequence 24, 12, 6,···.

3. Evaluate.

1.5

finite and infinite geometric sequence

4. Find the sum of the infinite geometric sequence 27, 18, 12, 8,···.

5. Find the sum of the infinite geometric sequence 8, 12, 18, 27,··· if it exists.

. 6. Find the sum of the geometric series 125 + 25 + 5 + 1 +...... ?

35

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 1.5 1. Find the sum of the first 8 terms of the geometric series if a1 = 1 and r = 2. (

=

)

=

=

=

= 255

2. Find S10 of the geometric sequence 24, 12, 6,···. First, find r.

=

r=

=

Now, find the sum: [

= =

( )

]

.

=

*

=

+

.

*

=

+

=

=

*

+

=

*

+

=

3. Evaluate.

(You are finding S10 for the series 3 – 6 + 12 – 24 + ···, whose common ratio is –2.) =

[

]

=

[

]

=

[

]

=

=

- 1023

4. Find the sum of the infinite geometric sequence 27, 18, 12, 8,···. First find r:

=

r=

=

Then find the sum:

=

=

=

=

=

27 .

=

81

5. Find the sum of the infinite geometric sequence 8, 12, 18, 27,··· if it exists. First find r:

=

r=

=

Since r = 3/2 is not less than one the series has no sum.

6. Find the sum of the geometric series 125 + 25 + 5 + 1 +...... ? Solution: The series is, 125 + 25 + 5 + 1 +..... a1 = 125 r= = The formula for the resultant sum of the Infinite Geometric Series is, =

=

=

=

=

125 .

=

36

Chapter 2

polynomial function

Chapter 2 POLYNOMIAL FUNCTIONS 2.1 2.2 2.3 2.4 2.5 2.6

LINEAR FUNCTIONS QUADRATIC FUNCTIONS POLYNOMIAL FUNCTIONS ON HIGHER DEGREE SYNTHETIC DIVISION REMAINDER AND FACTOR THEOREMS ZEROES AND GRAPHS OF POLYNOMIAL FUNCTIONS

Specific Objectives: at the end of the chapter the students should be able to;

          

Define polynomial functions , Recognize polynomial functions, Differentiate linear and quadratic as polynomial functions, Relate polynomial expressions from polynomial functions, Recall how to perform operations on polynomial expressions, Illustrate the synthetic division process for dividing polynomial expressions by a binomial, States the remainder theorem and provides a proof of the theorem, States the factor theorem and provides proof of the theorem, Factors polynomial functions using synthetic division, the remainder theorem, and the factor theorem, Determine the zeroes of polynomial functions. Solve problems involving factors and zeroes of polynomial functions,

Desargues, Girard [dayzah(r)g](1591--1661) Mathematician, born in Lyon, SC France. By 1626 he was in Paris, and took part as an engineer in the siege of La Rochelle in 1628. He founded the use of projective methods in geometry, inspired by the theory of perspective in art, and introduced the notion that parallel lines "meet at a point at infinity'. His work on the sections of a cone greatly influenced Pascal. From 1645 he began a new career as an architect in Paris and Lyons.

2.1

2.1.

linear function

37

LINEAR FUNCTIONS

Linear Function is a function f defined by an equation of the form y = mx + b, where m and b are real numbers. y = f(x) = mx + b Example: The amount of antifreeze needed to protect a radiator against to a temperature of – 180C about half the capacity of the radiator. How much antifreeze is needed to protect radiator with a capacity of 15L? Let x – the capacity of the radiator. y – the amount of antifreeze needed. The amount needed is half of the capacity

.

y =

x

y = x f(15) = (15) = 7.5 Example: Jasmine Baby-Sitting Service charges ₱ 140 plus ₱ 146 per hour. What is the cost of a nine-hour baby-sitting job? c(h) = ₱ 140 + ₱ 146(h) c(9) = ₱ 140 + ₱ 146(9) = ₱ 140 + ₱ 1314 c(9) = ₱ 1454 (1454, 9)

38

Chapter 2

polynomial function

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 2.1 1. The CEDBOR Truck Rental Company charges ₱ 1400 per day plus ₱ 18 per kilometer. Find the cost of renting a truck for one-day trip of 915 kilometers.

2. A cable television provider charges ₱ 1699 per month, plus ₱ 300 per month for additional for group of channels ordered. Find the total charge for a year in which additional group of channels are ordered.

3. The cost of renting a chain saw is ₱ 250 per day plus ₱ 50 for a liter of crude fuel. Find the cost of using the chain saw for 8.5 hours.

2.1

linear function

39

4. The airport parking garage charges ₱ 100 for the first hour and ₱ 28 for each additional hour. Find the cost of parking for 19 hours.

5. The cost of renting floor waxer is ₱ 170 per hour plus ₱ 220 for the wax. Find the cost of waxing a floor if the time involved was 8 hours.

6. Sally rents a compact car for ₱ 1160 per day plus ₱ 5.11 per kilometer. Compute the cost of a day trip of 753 kilometers.

7. A 50cm. spring will stretch (in cm) one third the weight (in kg.) attached to it. How long will the spring be if a 20kg. weight is attached?

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 2.1 1. The CEDBOR Truck Rental Company charges ₱ 1400 per day plus ₱ 18 per kilometer. Find the cost of renting a truck for one-day trip of 915 kilometers. c(k) = 1400 + 84k = 1400 + 18(915) = 1400 + 16470 = ₱ 17870 2. A cable television provider charges ₱ 1699 per month, plus ₱ 300 per month for additional for group of channels ordered. Find the total charge for a year in which additional group of channels are ordered. c(y) = = c(12) = =

1699y + 300y 1999y 1999(12) ₱ 23,988

3. The cost of renting a chain saw is ₱ 250 per day plus ₱ 50 for a liter of crude fuel. Find the cost of using the chain saw for 8.5 hours. c(h) = 50 + 250h c(8.5) = 50 + 250(8.5) = 50 + 2125 = ₱2175

4. The airport parking garage charges ₱ 100 for the first hour and ₱ 28 for each additional hour. Find the cost of parking for 19 hours. c(h) = 28(h - 1) + 100 c(19) = 28(19 - 1) + 100 = 28(18) + 100 = 504 + 100 = ₱ 604 5. The cost of renting floor waxer is ₱ 170 per hour plus ₱ 220 for the wax. Find the cost of waxing a floor if the time involved was 8 hours. c(h) = 220 + 170h c(8) = 220 + 170(8) = 220 + 1360 = ₱1580 6. Sally rents a compact car for ₱ 1160 per day plus ₱ 5.11 per kilometer. Compute the cost of a day trip of 753 kilometers. c(k) = 1160 + 5.11(k) c(753) = 1160 + 5.11(753) = 1160 + 3847.83 = ₱ 5007.83 7. A 50cm. spring will stretch (in cm) one third the weight (in kg.) attached to it. How long will the spring be if a 20kg. weight is attached? L(w) =

w + 50

L(20) = (20) + 50 = 6.67 + 50 = 56.67 cm.

40

Chapter 2

polynomial function

2.2 QUADRATIC FUNCTIONS Graph of Quadratic Functions

DEFINITION OF POLYNOMIAL FUNCTIONS

Let n be a nonnegative integer and let an, an – 1, . . .,a2, a1, a0 be real numbers with an 0. The function f(x) = anxn + an – 1xn – 1 + . . . + a2x2 + a1x + a0 is called a polynomial function of x with degree n. EFINITION OF QUADRATIC FUNCTIONS Let a, b, and c be real numbers with a called a quadratic function.

0. The function f(x) = ax2 + bx + c is

Parabola is the graph of quadratic function is a special type of “U”-shaped curve. Axis of symmetry or axis of parabola is characteristic where all parabolas are symmetric with respect to a line. Vertex of the parabola is the point where the axis intersects the parabola.

f(x) = ax2 + bx + c, a > 0 leading coefficient is positive opens upward vertex is low point

f(x) = ax2 + bx + c, a < 0 leading coefficient is negative opens downward vertex is high point

f(x) = ax2 , a > 0 leading coefficient is positive opens upward vertex is at the minimum point

f(x) = ax2 , a < 0 leading coefficient is negative opens downward vertex is at the maximum point

axis of symmetry parallel to y-axis

axis of symmetry parallel to y-axis

axis of symmetry is along the y-axis

axis of symmetry is along the y-axis

2.2

quadratic functions

41

STANDARD FORM OF A QUADRATIC FUNCTION STANDARD FORM OF A QUADRATIC FUNCTION The quadratic function f(x) = a(x - h)2 + k, a 0 is in standard form. The graph of f is a parabola whose axis is the vertical line x = h and whose vertex is the point (h, k). is a > 0, the parabola opens upward, and if a < 0 , the parabola opens downward. Example 1. Sketch the graph of f(x) = 2x2 - 4x + 3 and identify the vertex and axis of symmetry. Since : the first step in completing the square is to factor out any coefficient of x2 that is not 1. f(x) = 2x2 - 4x + 3 = 2(x2 - 2x) + 3 = 2(x2 - 2x + 1 - 1) + 3 multiply [2( )]2 = 1 = 2(x2 - 2x + 1) - 2(1) + 3 = 2(x - 1)2 - 2 + 3 f(x)= 2(x - 1)2 + 1 f(x) = a(x - h)2 + k, a 0 x - h =0 x = h x = 1 axis of parabola and y =k y = 1 vertex (1, 1) a > 0 concave-up

VERTEX OF A PARABOLA The vertex of the graph of f(x) = ax2 + bx + c is *−

b , 𝑎

f (−

𝑏 )+ 𝑎

42

Chapter 2

polynomial function

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.2 1. Compare the graphs of y = x2 and f(x) = 3x2.

2. Sketch the graph of f(x) = - x2 + 5x - 6 and identify the vertex and xintercepts.

3. Sketch the graph of f(x) = 2x2 + 4x + 6 and identify the vertex and the axis of parabola.

2.2

4. Compare the graphs of y = x2 and g(x) =

quadratic functions

x2.

1. 2. 3. 4.

5. Write the standard form of the equation of the parabola whose vertex is (2, 3) and that passes through the point (1, 0) as shown in the figure.

43

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.2 1. Compare the graphs of y = x2 and f(x) = 3x2.

2. Sketch the graph of f(x) = - x2 + 5x - 6 and identify the vertex and xintercepts. f(x) = - x2 + 5x - 6 vertex (− , ) = -( x2 - 5x) - 6 x-intercepts = -(x2 - 5x + - ) - 6 -(x2 - 5x + 6) = 0 = -(x2 - 5x + = -(x -

)2 +

= -(x -

)2 +

) - (- ) - 6 - 6

-(x - 3)(x - 2) = 0 x - 3 = 0 x - 2 = 0 x=3

x = 2

(3, 0)

(2, 0)

3. Sketch the graph of f(x) = 2x2 + 4x + 6 and identify the vertex and the axis of parabola. f(x) = 2x2 + 4x + 6 = 2(x2 + 2x) + 6 = 2(x2 + 2x + 1 - 1) + 6 = 2(x2 + 2x + 1) - 2(1) + 6 = 2(x + 1)2 - 2 + 6 = 2(x + 1)2 + 4 Vertex (-1, 4) axis of parabola x = -1

4. Compare the graphs of y = x2 and g(x) =

x2.

5.

6. 7. 8.

5. Write the standard form of the equation of the parabola whose vertex is (2, 3) and that passes through the point (1, 0) as shown in the figure.

f(x) = a(x - h)2 + k 0 = a(1 - 2)2 + 3 0 = a(-1)2 + 3 0=a+ 3 a = -3 f(x) = - 3 (x - 2) + 3

44

Chapter 2

polynomial function

2.3

POLYNOMIAL FUNCTIONS ON HIGHER DEGREE

GRAPHS OF POLYNOMIAL FUNCTIONS

Continuous graph is a polynomial functions

Discontinuous graph is not a polynomial function

Graphs with rounded turns is a polynomial functions

Graphs with Sharp Turns is not a polynomial functions

If n is odd, the graph of y = xn that crosses the axis at the x-intercepts

if n is even, the graph of y = xn touches the axis at the x-intercepts

THE LEADING COEFFICIENT TEST As x moves without bound to the left or to the right, the graph of the polynomial function f(x) = anxn + an – 1xn – 1 + . . . + a2x2 + a1x + a0 either rises or falls in the following behavior.

2.3

When n is odd:

polynomial functions on higher degree

45

If the leading coefficient is positive (an > 0), the graph fall to left If the leading coefficient is negative (an < 0), the graph rises to left where f(x) towards - as x towards - and rises to the right where f(x) towards as x towards - and falls to the right where f(x) towards as x towards . where f(x) towards - as x towards .

When n is even:

If the leading coefficient is positive(an > 0), the graph rises to left If the leading coefficient is negative (a n < 0), the graph falls to left where f(x) towards as x towards - and rises to the right where f(x) towards - as x towards - and fals to the right where f(x) towards as x towards . where f(x) towards - as x towards .

Example 1. Sketch the graph of the function f(x) = - x7

x -1 f(x) 1

0 0

1 -1

Example 2. Describe the right-hand and left-hand behavior of the graph of f(x) = - x5 + 6x. If the leading coefficient is negative (an < 0), the graph rises to left where f(x) towards as x towards and falls to the right where f(x) towards - as x towards .

46

Chapter 2

polynomial function

The Intermediate Value Theorem INTERMEDIATE VALUE THEOREM Let a and b be real numbers such that a < b. If f is a polynomial function such that f(a) f(b), then, in the interval [a, b], f takes on every value between f(a) and f(b).

f(b)

f(c) = d f(a) 0

𝑎

𝑐

b

Let f(x) = anxn + an – 1xn – 1 + . . . + a2x2 + a1x + a0 be a polynomial function The following statements are equivalent. ZERO:

k is a zero of the polynomial function f.

FACTOR:

x - k is a factor of the polynomial f(x).

SOLUTION: k is a solution of the polynomial equation f(x) = 0 If k is real number, then the following is also equivalent. X-INTERCEPT: k is an x-intercept of the graph of the polynomial function f.

2.3

polynomial functions on higher degree

Zeroes of Higher Degree Polynomial Functions

The following statements are true for a polynomial function f of degree n,  The graph of f has, at most, n – 1 turning points. It might be increasing or decreasing.  The function f has, at most, n real zeros.

REAL ZEROS OF POLYNOMIAL FUNCTIONS If f is a polynomial function and a is a real number, the following statements are equivalent. 1. 2. 3. 4.

x = a is a zero of the function f. x = a is a solution of the polynomial equation f(x) = 0. (x - a) is a factor of the polynomial f(x). (a, 0) is an x-intercept of the graph off.

Example 1. Find all the real zeros of f(x) = - 2x4 + 2x2. Number of turning points n - 1 = 4 - 1 = 3 f(x) = - 2x4 + 2x2 = - 2x2(x2 - 1) = -2x2(x + 1)(x - 1) 0 = -2x2(x + 1)(x - 1) (0, 0) (-1, 0) (1, 0)

47

48

Chapter 2

polynomial function

REPEATED ZEROS A factor (x - a)2, k > 1, yields a repeated zero x = a of multiplicity of k 1. If k is odd, the graph crosses the x-axis at x = a, 2. If k is even, the graph touches the axis at x = a.

Example 2. Sketch the graph of f(x) = 6x6 - 4x5 By leading coefficient test highest degree is even. Eventually left and right rises. Find the zeros of the polynomial 2x5( 3x - 2) = 0 x=0 x = (0, 0)

( , 0)

0

1

-1

Additional points x f(x)

0 0

1/6 1/3 1/2 2/3 -0.00039 -0.00275 -0.01563 0

2.3

polynomial functions on higher degree

49

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.3

1. Sketch the graph of a function h(x) = (x - 1)4.

x -1 0 f(x) 16 1

1 16

2. Describe the right-hand and left- hand behavior of the graph of the function f(x) = x4 - 5x2 + 4

50

Chapter 2

polynomial function

3. Determine the right-hand and left-hand behavior of the graph of the polynomial function f(x) = x3 + 5x.

Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type, and leading coefficient. 4. f(x) =

x2 - 3x4 - 7

Function - polynomial function Standard Form – Degree – Type – Leading Coefficient 5. f(x) = x3 + 3x Function Standard Form Degree – Type – Leading Coefficient -

2.3

polynomial functions on higher degree

6. f(x) = 6x2 + 2x-1 + x Function Standard Form Degree – Type – Leading Coefficient 7. f(x) = − 0.5x +

x2 − √

Function Standard Form Degree – Type – Leading Coefficient 8. Evaluate the polynomial function f(x) = 2x4 − 8x2 + 5x − 7. If x = 2.

51

52

Chapter 2

polynomial function

9. Find all the real zeros of the polynomial function f(x) = x2 - 25. Determine the multiplicity of each zero.

10.

Find all the real zeros of the polynomial function f(x) = x4 - x3 - 20x2. Determine the multiplicity of each zero.

11.

Find a polynomial function that has the given zeros (0, 10).

12.

Determine a polynomial of degree n that has the given zeros x = -3, 0, 1 of degree n = 3.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.3

1. Sketch the graph of a function h(x) = (x - 1)4.

x -1 0 f(x) 16 1

1 16

2. Describe the right-hand and left- hand behavior of the graph of the function f(x) = x4 - 5x2 + 4

Since the degree is even Leading coefficient is positive Right and left-hand behavior rises up

3. Determine the right-hand and left-hand behavior of the graph of the polynomial function f(x) = x3 + 5x. Since the degree is odd Leading coefficient is positive Left-hand falls and right-hand rises. Decide whether the function is a polynomial function. If it is, write the function in standard form and state its degree, type, and leading coefficient. 4. f(x) =

x2 - 3x4 - 7

Function - polynomial function Standard Form - f(x) = - 3x4 +

x2 - 7

Degree – 4 Type – quadratic function Leading Coefficient - − 3) 5. f(x) = x3 + 3x Function - not a polynomial function because the term 3x does not have variable base and an exponent that is a whole number Standard Form Degree – Type – Leading Coefficient -

6. f(x) = 6x2 + 2x-1 + x

Function - not a polynomial function because the term 2x-1 has an exponent that is not a whole number Standard Form Degree – Type – Leading Coefficient 7. f(x) = − 0.5x +

x2 − √

Function - a polynomial function Standard Form - f(x) =

x2 − 0.5x − √

Degree – 2 Type – quadratic function Leading Coefficient 8. Evaluate the polynomial function f(x) = 2x4 − 8x2 + 5x − 7. If x = 2.

f(x) = 2x4 − 8x2 + 5x − 7 f(2) = 2(2)4 − 8(2)2 + 5(2) − 7 = 2(256) − 8(4) + 10 − 7 = 512 − 32 + 10 − 7 = 522 − 39 = 483

9. Find all the real zeros of the polynomial function f(x) = x2 - 25. Determine the multiplicity of each zero. f(x) = x2 - 25 = (x + 5)(x - 5) (x + 5)(x - 5) = 0 x+ 5 = 0 x - 5 = 0 x = -5 x = 5 odd multiplicity 10.

Find all the real zeros of the polynomial function f(x) = x4 - x3 - 20x2. Determine the multiplicity of each zero. f(x) = x4 - x3 - 20x2 = x2(x2 - x - 20) = x2(x - 5)(x + 4) x2(x - 5)(x + 4) = 0 x = 0 , x - 5 = 0, x + 4 = 0 x = 0, x = 5, x = -4 even multiplicity

11.

Find a polynomial function that has the given zeros (0, 10). x (x - 10) = 0 x2 - 10x = 0 f(x) = x2 - 10x

12.

Determine a polynomial of degree n that has the given zeros x = -3, 0, 1 of degree n = 3. x(x + 3)(x - 1) = 0 (x2 + 3x)(x - 1) = 0 x3 + 2x2 - 3x = 0 f(x) = x3 + 2x2 - 3x

2.4

SYNTHETI C DIVISION

2.4

synthetic division

LONG DIVISION OF POLYNOMIALS THE DIVISION ALGORITHM if f(x) and d(x) are polynomials such that d(x) 0,and the degree of d(x) is less than or equal to the degree of f(x), there exist unique polynomials q(x) and r(x) such that, f(x) = d(x)q(x) + r(x) Remainder Quotient Divisor where r(x) = 0 or the degree of r(x) is less than Dividend the degree of d(x). If the remainder r(x) is zero, d(x) divides evenly into f(x).

Example 1. Divide 4x3 - 17x2 + 4x - 2 by x - 1, and use the result to factor the polynomial completely.

x - 1√ − 3 4x − − −

4x2 − 13x + 9 − 2 4x 13x2 + 4x 13x2 13x 9x − 2 9x − 9 7

4x2 9 (x - 1)(4x - 9) - 4x - 9x -13x

4x3 - 17x2 + 4x - 2 = (x - 1)2(4x - 9)

53

54

Chapter 2

polynomial function

Synthetic Division

SYVTHETIC DIVISION (FOR A CUBIC FUNNCTION) To divide ax3 + bx2 + cx + d by x - k consider the following steps: k

a b c

d

numerical coefficients of the dividend

ka a

vertical pattern: add terms. r

remainder

diagonal pattern: multiply by k

Example 1. Use synthetic division to divide x5 - 8x3 - x + 2 by x + 2. -2 1 0 0 -8 -1 2 -2 4 -8 32 -62 1-2 4 -16 31 -60

4

3

2

= x - 2x + 4x - 16x + 31 -

Example 2. Use synthetic division to divide 6x3 + 4x2 + 4x - 5 by x − 2. -2 6 4 4 - 5 -12 16 - 40 6 -8 20 - 45

2.4

synthetic division

55

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.4 1. Use long division to verify that y1 =

equivalent to

2. Divide 2x6 - 34x4 - 26x2 - 90 by x + 4 using synthetic division.

3. Express the function in the form f(x) = (x - k)q(x) + r for k = 4 of the function x3 - x2 - 14x + 11 and demonstrate that f(k) = r.

56

Chapter 2

polynomial function

4. Find each the function value using synthetic division then verify using another method. f(x) = 3x3 + 5x2 - 10x + 1 a. f(2) b. f( )

5. Simplify the rational expression

.

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 2.4 1. Use long division to verify that y1 =

equivalent to

y2 = x 2 - 8 + x2 - 5 x2 + 2 √ − − 4 2 x + 2x - 5x2 - 1 - 5x2 - 10 9 y1 y2 2. Divide 2x6 - 34x4 - 26x2 - 90 by x + 4 using synthetic division. -4 2 0 -34 -8 32

0 -26 0 -90 -8 32 -24 96

2 -8 -2 -8 6 -24 6 = 2x5 - 8x4 - 2x3 - 8x2 + 6x - 24 + 3. Express the function in the form f(x) = (x - k)q(x) + r for k = 4 of the function x3 - x2 - 14x + 11 and demonstrate that f(k) = r. 4

1 -1 -14 11 4 12 -8 1 3 -2 3 f(k) = (x - k)q(x) + r = (x - 4)(x2 + 3x - 2) + 3 f(k) = x3 - x2 - 14x + 11 f(4) = (4)3 - (4)2 - 14(4) + 11 = 64 - 16 - 56 + 11 = 75 - 72 f(4) = 3

4. Find each the function value using synthetic division then verify using another method. f(x) = 3x3 + 5x2 - 10x + 1 a. f(2) b. f( ) 2 3 5 -10

1

3 5 -10 1

6 22 24

-

3 11 12 25

3

-

3x2 + 11x + 12 x - 2√

−

3x2 x -

3x3 - 6x2

-

−

√ 3x2 -

x2

11x2 - 10x

x2 - 10x

11x2 - 22x

x2

-

x

+ 1

12x - 24

-

x

+

5. Simplify the rational expression 1 3 -1 -1 -2 1 2 -3

x

+

12x + 1 25

-1

x -

-3 3 0

= x2 + 2x - 3 =(x + 3)(x - 1)

= (x + 3)(x - 1)

-

.

2.5

The remainder and factor theorems

57

2.5 THE REMAINDER AND FACTOR THEOREMS

THE REMAINDER THEOREM If a polynomial f(x) is divided by x - k, the remainder is r = f(k) THE FACTOR THEOREM A polynomial f(x) has a factor (x – k) if and only if f(k) = 0.

USES OF THE REMAINDER IN SYNTHETIC DIVISION The remainder r, obtained in the synthetic division of f(x) by x k, provides the following information. 1. The remainder r gives the value of f at x = k. That is, r = f(k). 2. If r = 0, (x - k) is a factor of f(x). 3. If r = 0, (k, 0) is an x-intercept of the graph of f.

Example 1. Use the remainder theorem top evaluate the following Function. f(x) = 6x3 + 4x2 + 4x - 5 at x = -2. -2 6 4 -12 6 -8

4 -5 16 -40 20 - 45

f(-2) = f(-2) = = = = =

- 45 since f(x) = r 6(-2)3 + 4(-2)2 + 4(-2) - 5 6(-8) + 4(4) - 8 - 5 -48 + 16 - 13 -61 + 16 -45

58

Chapter 2

polynomial function

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.5 1. Use the remainder theorem to evaluate the function at x = -2. f(x) = 3x3 + 8x2 + 5x − 7

2. Use the remainder theorem to evaluate the function at x = 2. f(x) = 2x3 + 2x2 − 5x + 2

3. Use the remainder theorem to evaluate the function at x = 4. f(x) = x3 - x2 − 14x + 11

2.5

the remainder and factor theorems

4. Use the remainder theorem to evaluate the function at x = -2. f(x) = x3 - 5x2 − 11x + 8

5. Use the remainder theorem to evaluate the function at x = 2. f(x) = x3 - 7x + 6

6. Use the remainder theorem to evaluate the function at x = -5. f(x) = 3x3 + 5 x2 - 10x + 1

59

60

Chapter 2

polynomial function

7. Show that (x - 2) and (x + 3) are factors of f(x) = 2x4 + 7x3 - 4x2 - 27x - 18

8. Show that (x + 2) and (x - 1) are factors of f(x) = 2x3 + x2 - 5x + 2

9. Show that (x - 5) and (x + 4) are factors of f(x) = x4 - 4x3 - 15x2 + 58x - 40

2.5

the remainder and factor theorems

10.

Show that (x + 2) and (x - 4) are factors of f(x) = 8x4 - 14x3 - 71x2 - 10x + 24

11.

Show that (x + 3) and (x - 2) are factors of f(x) = 3x3 + 2x2 - 19x + 6

12.

Show that (2x + 1) and (3x - 2) are factors of f(x) = 6x3 + 41x2 - 9x - 14

61

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.5 1. Use the remainder theorem to evaluate the function at x = -2. f(x) = 3x3 + 8x2 + 5x − 7 -2 3 8 5 -7 -6 -4 - 2 3 2 1 -9 Since the remainder is r = -9, then f(-2) = -9 Check: f(-2) = 3(-2)3 + 8(-2)2 + 5(-2) − 7 = 3(-8) + 8(4) − 10 − 7 = - 24 + 32 - 17 = -9 2. Use the remainder theorem to evaluate the function at x = 2. f(x) = 2x3 + 2x2 − 5x + 2 2 2 2 -5 2 4 12 14 2 6 7 16 Since the remainder is r = 16, then f(2) = 16 Check: f(2) = 2(2)3 + 2(2)2 - 5(2) + 2 = 2(8) + 2(4) − 10 + 2 = 16 + 8 - 8 = 16 3. Use the remainder theorem to evaluate the function at x = 4. f(x) = x3 - x2 − 14x + 11 4 1 -1 -14 11 4 12 -8 1 3 -2 3 Since the remainder is r = 3, then f(4) = 3 Check: f(4) = (4)3 - (4)2 - 14(4) + 11 = 64 - 16 - 56 + 11 = 75 - 72 = 3

4. Use the remainder theorem to evaluate the function at x = -2. f(x) = x3 - 5x2 − 11x + 8 -2 1 -5 -11 8 -2 14 -6 1 -7 3 2 Since the remainder is r = 2, then f(-2) = 2 Check: f(-2) = (-2)3 - 5(-2)2 - 11(-2) + 8 = -8 - 5(4) + 22 + 8 = -8 - 20 + 22 + 8 = -28 + 30 = 2 5. Use the remainder theorem to evaluate the function at x = 2. f(x) = x3 - 7x + 6 2 1 -7 6 4 -6 1 -3 0 Since the remainder is r = 0, then f(2) = 0 Check: f(2) = (2)3 - 7(2) + 6 = 8 - 14 + 6 = 14 - 14 = 0 6. Use the remainder theorem to evaluate the function at x = -5. f(x) = 3x3 + 5 x2 - 10x + 1 -5 3

5 -10 1 -15 50 -200 3 -10 40 -199 Since the remainder is r = -199, then f(-5) = -199 Check: f(-5) = 3(-5)3 + 5(-5)2 -10(-5) + 1 = 3(-125) + 5(25) + 50 + 1 = -375 + 125 + 50 + 1 = -375 + 176 = - 199

7. Show that (x - 2) and (x + 3) are factors of f(x) = 2x4 + 7x3 - 4x2 - 27x - 18 2 2

7 -4 -27 -18 4 22 36 18 2 11 18 9 0

-3 2 11 18 -6 -15 2 5 3

f(2) = 0, (x - 2) is a factor

9 -9 0

f(-3) = 0, (x + 3) is a factor

8. Show that (x + 2) and (x - 1) are factors of f(x) = 2x3 + x2 - 5x + 2 -2 2 1 -4 2 -3

-5 6 1

1 2

1 -1 0

-3 2 2 -1

2 -2 0

f(-2) = 0, (x + 2) is a factor

f(1) = 0, (x - 1) is a factor

9. Show that (x - 5) and (x + 4) are factors of f(x) = x4 - 4x3 - 15x2 + 58x - 40 5 1 1 -4 1

-4 5 1

1 -4 1 -3

-15 58 -40 5 -50 40 -10 8 0 -10 12 2

8 -8 0

f(5) = 0, (x - 5) is a factor

f(-4) = 0, (x + 4) is a factor

10.

Show that (x + 2) and (x - 4) are factors of f(x) = 8x4 - 14x3 - 71x2 - 10x + 24 -2 8 8

-14 -71 -10 24 -16 60 22 -24 -30 -11 12 0

4 8 -30 32 8 2 11.

f(4) = 0, (x - 4) is a factor

Show that (x + 3) and (x - 2) are factors of f(x) = 3x3 + 2x2 - 19x + 6 -3 3

2 -9 -7

3

2 3 -7 6 3 -1

12.

-11 12 8 -12 -3 0

f(-2) = 0, (x + 2) is a factor

-19 6 21 -6 2 0 2 -2 2

f(-3) = 0, (x + 3) is a factor

0

f(2) = 0, (x - 2) is a factor

Show that (2x + 1) and (3x - 2) are factors of f(x) = 6x3 + 41x2 - 9x - 14 -

6

41

-9

-14

6

-3 -19 38 -28

14 0

6

38 4

-28 28

6

42

0

f(− ) = 0, (2x + 1) is a factor

f( ) = 0, (3x - 2) is a factor

62

Chapter 2

polynomial function

2.6 ZEROES AND GRAPHS OF POLYNOMIAL FUNCTIONS The Fundamental Theorem of Algebra THE FUNDAMENTAL THEOREM OF ALGEBRA If f(x) is a polynomial of degree n, where n > 0, then f has ate least one zero in the complex number system.

LINEAR FACTORIZATION THEOREM If f(x) is a polynomial of degree n, where n > 0, then f has precisely n linear factors f(x) = an(x - c1)(x - c2) . . . (x - cn), where c1,c2, . . . cn are complex numbers Zeros of Polynomial Functions 1. The first-degree polynomial f(x) = x - 5 has exactly one zero: x = 5 2. counting multiplicity, the second-degree polynomial function f(x) = x2 - 5x + 6 = (x - 3)(x - 2) where zeros are x = 2 and x = 3 3. The third-degree polynomial function f(x) = x3 + 9x = x(x2 + 9) where zeros are x = 0 , x = 3i and x = - 3i 4. The fourth-degree polynomial function f(x) = x4 - 16 = (x2 - 4)(x2 + 4)= (x + 2)(x - 2)(x + 2i)(x - 2i) where zeros are x = -2, x = 2, x = -2i, and x = 2i

2.5

the remainder and factor theorems

The Rational Zero Test THE RATIONAL ZERO TEST If the polynomial f(x) = anxn + an – 1xn – 1 + . . . + a2x2 + a1x + a0 has integer coefficients, every rational zero of f has the form 𝑝 Rational zero = where p and q have no common factors other 𝑞

-1

than1, and p = a factor of the constant term a0 q = a factor of the leading coefficient an.

Example 1. Find the rational zeros of f(x) = x4 - x3 + x2 - 3x - 6. possible zeros are the factors of constant = 6 1, , , through test among the possible zeros it is -1, and 2 satisfies.

1 -1 1 -3 -6 = x4 - x3 + x2 - 3x - 6 -1 2 -3 6 1 -2 3 -6 0 = x3 - 2x2 + 3x - 6 2

1 -2 3 -6 2 0 6 1 0 3 0 = x2 + 3

f(x) = x4 - x3 + x2 - 3x - 6 = (x + 1)(x - 2)(x2 + 3) = (x + 1)(x - 2)(x + 3i)(x - 3i)

CONJUGATE PAIRS COMPLEX ZEROS OCCUR IN CONJUGATE PAIRS Let f(x) be a polynomial function that has real coefficients. If a + bi , where b 0, is a zero of the function, the conjugate a - bi is also a zero of the function.

-1

0 -1 -2 -3 -4 -5 -6 -7 -8

1

2 3

Example 1. Find the fourthdegree polynomial function with real coefficients that has -2, -2, and 2i as zeros. f(x) = a(x + 2)(x + 2)(x - 2i)(x + 2i) let a = 1 = (x2 + 4x + 4)(x2 + 4) = x4 + 4x3 + 8x2 + 16x + 16

63

64

Chapter 2

polynomial function

Factoring Polynomials FACTORS OF A POLYNOMIAL Every polynomial of degree n > 0 with real coefficients can be written as the product of linear and quadratic factors with real coefficients, where the quadratic factors have no real zeros. Other Test for Zeroes of Polynomials DESCARTES’S RULE OF SIGNS Let f(x) = anxn + an – 1xn – 1 + . . . + a2x2 + a1x + a0 be the polynomial with real coefficients and a0 ≠ 0. 1. The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integer. 2. The number of negative real zeros of f is either equal to the number of variations in sign of f(-x) or less than that number by an even integer. A variation in sign means that two consecutive coefficients that have opposite signs. Example 1. Describe the possible real zeros of f(x) = 6x3 - 7x2 + 5x - 8. + to + to - to f(x) = 6x3 - 7x2 + 5x - 8 -

3 variations

to +

expected zeros - 3 or 1 positive zeros

-

to -

f(-x) = 6(-x)3 - 7(-x)2 + 5(-x) = -6x3 - 7x2 - 5x - 8

8 no variations

- to -

UPPER AND LOWER BOUND RULES Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x - c, by using synthetic division. 1. If c > 0 and each number in the last row is either positive or zero, c is an upper bound for the real zeros of f. 2. If c < 0 and the number in the last row are alternately positive and negative(zero entries count as positive or negative), c is a lower bound for the real zeros of f.

2.5

the remainder and factor theorems

65

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.6 1. Find all the zeros of the function g(x) = (x - 3)(x + 6)3.

2. Find all the zeros of the function f(x) = x3 - 6x2 + 11x - 6.

3. List all the possible zeros of the polynomial equation 2a4 + 7a3 - 26a2 + 23a - 6 = 0.

66

Chapter 2

polynomial function

4. Determine one of the exact zeros of the function f(x) = x4 - 3x2 + 2, use synthetic division to verify your result , and then factor the polynomial completely.

5. Find the polynomial function with integer coefficient 1, 5i, -5i.

6. Find all the zeros of the function g(x) = x2 - 4x3 + 8x2 - 16x + 16 and write the polynomial as a product of linear factors.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 2.6 1. Find all the zeros of the function g(x) = (x - 3)(x + 6)3.

g(x) = (x - 3)(x + 6)3 = (x - 3)(x + 6)(x + 6)2 = (x - 3)(x + 6)(x - 6i)(x + 6i) x = 3 and x = -6 2. Find all the zeros of the function f(x) = x3 - 6x2 + 11x - 6.

possible zeros 1, , 3, f(x) = x3 - 6x2 + 11x - 6. 3 variation f(-x) = - x3 - 6x2 - 11x -6 no variation 1

1 -6 11 -6 1 -5 6 1 -5 6 0 = x2 - 5x + 6 by factoring = (x - 3)(x - 2)

zeros are x = 1, x = 3 and x = 2 3. List all the possible zeros of the polynomial equation 2a4 + 7a3 - 26a2 + 23a - 6 = 0.

Possible

1,

,

3,

4. Determine one of the exact zeros of the function f(x) = x4 - 3x2 + 2, use synthetic division to verify your result , and then factor the polynomial completely. Possible zeros = 1, 1

1 1

-1

1 1

0 1 1

-3 1 -2

0 -2 -2

2 -2 0 = x3 + x2 - 2x - 2

1 -1 0

-2 0 -2

-2 2 0 = x2 - 2

f(x) = x4 - 3x2 + 2 = (x + 1)(x - 1)(x + 2i)(x - 2i) x = -1 and x = 1

5. Find the polynomial function with integer coefficient 1, 5i, -5i. f(x) = (x + 1)(x + 5i)(x - 5i) = (x + 1)(x2 + 5) = x3 + x2 + 5x + 5 6. Find all the zeros of the function g(x) = x2 - 4x3 + 8x2 - 16x + 16 and write the polynomial as a product of linear factors. 2

1 1

2

1 1

x= 2

-4 2 -2

8 -4 4

-16 8 -8

-2 2 0

4 0 4

-8 8 0

16 -16 0

g(x) = x2 - 4x3 + 8x2 - 16x + 16 = (x + 2)(x + 2)(x2 + 2) = (x + 2)2(x + 2i)(x - 2i)

3.0

circles

67

Chapter 3 CIRCLES 3.1 3.2 3.3 3.4

PARTS OF A CIRCLE RELATIONS AMONG PARTS OF A CIRCLE THEOREMS RELATING THE PARTS OF A CIRCLE THEOREMS ON SECANT AND TANGENT LINES AND SEGMENTS

Specific Objectives: at the end of the chapter the students should be able to;  Define the parts of a circle ,  Illustrate the parts of a circle,  Derives the relation among chords, arcs, central angle, and inscribed angles,  State and prove the theorems relating chords, arcs, central angle and inscribed angle,  Define secant and tangent lines and segments, and sector of a circle,  State and prove the theorems on secant and tangent lines and segments,  Solve problems involving parts of a circles. Euler, Leonhard [oyler](1707--83) Mathematician, born in Basel, Switzerland. He studied mathematics there under Jean Bernoulli, and became professor of physics (1731) and then of mathematics (1733) at the St Petersburg Academy of Sciences. In 1738 he lost the sight of one eye. In 1741 he moved to Berlin as director of mathematics and physics in the Berlin Academy, but returned to St Petersburg in 1766, soon afterwards losing the sight of his other eye. He was a giant figure in 18th-c mathematics, publishing over 800 different books and papers, on every aspect of pure and applied mathematics, physics and astronomy. His Introduction in analyzing infinitorum (1748) and later treatises on differential and integral calculus and algebra remained standard textbooks for a century and his notations, such as e and π have been used ever since. For the princess of Anhalt-Dessau he wrote Lettres à une princessed' Allemagne (1768 -72), giving a clear non-technical outline of the main physical theories of the time. He had aprodigious memory, which enabled him to continue mathematical work and to compute complex calculations in his head when he was totally blind. He is without equal in the use of algorithms to solve problems.

68

Chapter 3

circles

3.1

PARTS OF A CIRCLE

A circle is the set of coplanar points at a given distance from a given point in the plane. Center is the point in a plane.

radius diameter c

A radius of a circle is a segment determined by the center and a point on the circle. It is also used to mean the length of this segment. A diameter of a circle is the segment that contains the center and has endpoint on the circle. It is also used to mean the length of this segment, The interior of a circle is the set of points whose distance from the center is less than the radius The exterior of a circle is the set of points in the plane whose distance from the center is greater than the radius secant line

c tangent line

A secant to a circle is a line lies in the plane of a circle and contains an interior point of a circle intersects the circle in exactly two points. A tangent to a circle is a line in the plane of the circle that intersects the circle in exactly one point. Point of tangency is the point of intersection.

3.1

parts of a circle

69

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 3.1 B G

D A

C H E

F

8. Name the three radii of a circle C.

9. Name a diameter of circle C.

10.

Name the secant of circle C.

11.

If CD = 8 units, what is the length of BF.

12. Point G is in the exterior of circle C. Point A is in the interior of circle C. What is the relation between CG and CA.

70

Chapter 3

circles

13.

If point H lies on circle C and CD = 5 units, what is the length of CH.

14.

If line GB is tangent line to the circle C, what is the point of tangency?

Draw and label the following in circle C. 15.

A radius CA

16.

A secant BD

17.

A diameter EF

18.

A tangent FG

19.

A segment CH, such that CH > CA.

20.

A segment CI, such that CI < CA.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 3.1 B G

D A

C H E

F

1. Name the three radii of a circle C. CD, BC, CF 2. Name a diameter of circle C. BF 3. Name the secant of circle C. BE 4. If CD = 8 units, what is the length of BF. 16 units 5. Point G is in the exterior of circle C. Point A is in the interior of circle C. What is the relation between CG and CA. Since CD = radius, then CG > CD and CA < CD.

6. If point H lies on circle C and CD = 5 units, what is the length of CH. 5 units 7. If line GB is tangent line to the circle C, what is the point of tangency? Point B

G

F

I

E

A

C

B

Draw and label the following in circle C. 8. A radius CA 9. A secant BD 10.

A diameter EF

11.

A tangent FG

12.

A segment CH, such that CH > CA.

13.

A segment CI, such that CI < CA.

H

F D

3.2

3.2

relations among parts of a circle

71

RELATIONS AMONG PARTS OF A CIRCLE

A chord is a segment whose endpoints lie on a circle. Congruent circles are circles with congruent radii

A polygon inscribed in a circle is a polygon whose vertices lie on the circle.

Acute Triangle

Right Triangle Obtuse Triangle

A polygon circumscribed about a circle is a polygon whose sides are tangent to the circle.

A

C B

A central angle of a circle is an angle whose vertex is the center of the circle. 1

72

Chapter 3

circles

A minor arc of a circle is the set of points on the circle which lie on a central angle or in the interior of the central angle.

A V

C B

A semicircle is the union of the endpoints of a diameter and the points of the circle in a given half-plane formed by the line containing the diameter.

U

A

C

B

The degree measure of semicircle is 1800.

V

The degree measure of a minor arc is the measure of its central angle.

A U

C

1

V B

Arc Addition Postulate If V is on AB, then mAVB = mAV + mVB An angle inscribed in an arc is an angle whose sides contain the endpoints of the arc and whose vertex is a point on the arc other than the endpoints. An intercepted arc is an arc whose endpoints lie on different rays of an angle and whose other points lie in the interior of the angle

1

A

B

3.2

relations among parts of a circle

73

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 3.2 1. What is the longest chord of a circle?

2. When are two circles congruent?

The following numbers should refer to the illustration below

A

D C

U B

3. If CV

DE, then VE

4. If CU AB, CV

V E

________.

DE, and CU

CV, then AB

________.

74

Chapter 3

circles

5. If DE

AB, CU

AB, and CV

DE, then CU

_________.

6. If CU

AB, CV

DE, and CU

CV, why BU

VD

The following numbers should refer to the illustration below

D C A

7. If AC = 10 and CB = 6, Solve for AE.

B

E

3.2

8. If AD = 12 and CB = 3, Solve for AE

9. If AE = 24 and CB = 5, find AC.

relations among parts of a circle

75

76

Chapter 3

circles

10.

If AD = 30 and AE = 24, Find CB

Tell whether the statement is always, sometimes or never true. 11.

Two circles with congruent radii are congruent circles.

12.

A circle is congruent to itself.

13.

A line perpendicular to a chord of a circle bisects the chord.

14.

If a radius of a circle is perpendicular to a chord of the circle, then the radius bisects the chord.

15.

In a circle, two chords the same distance from the center are congruent.

16.

A chord of a circle is a segment.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 3.2 1. What is the longest chord of a circle? diameter 2. When are two circles congruent? When they have same diameter or radius. The following numbers should refer to the illustration below

A

D

U B

3. If CV VE

DE, then VE

V E

________.

VD

4. If CU AB, CV AB

C

DE

DE, and CU

CV, then AB

________.

5. If DE CU

AB, CU

AB, and CV

DE, then CU

_________.

DE, and CU

CV, why BU

VD

CV

6. If CU

AB, CV

According to the theorem that states, in the same circle or in congruent circles, if two chords are the same distance from the center, then the chords are congruent.

The following numbers should refer to the illustration below

D C A

7. If AC = 10 and CB = 6, Solve for AE. AB = √ =√ = √ = 8

− −

Since AE = 2AB Then AE = 2 (8) = 16

B

E

8. If AD = 12 and CB = 3, Solve for AE Since AC = AD = (12) =6 AB = √ − =√ − = √ = 5.2 Since AE = 2AB Then AE = 2 (5.2) = 10.4

9. If AE = 24 and CB = 5, find AC. Since AB = AE = (24) = 12 AC = √ =√ =√ = √ = 13

10.

If AD = 30 and AE = 24, Find CB Since AC = AD = (30) = 15 Since AB = AE = (24) = 12 CB = √ − =√ − =√ − = √ = 9

Tell whether the statement is always, sometimes or never true. 11.

Two circles with congruent radii are congruent circles.

12.

always A circle is congruent to itself.

13.

always A line perpendicular to a chord of a circle bisects the chord.

14.

sometimes If a radius of a circle is perpendicular to a chord of the circle, then the radius bisects the chord.

15.

always In a circle, two chords the same distance from the center are congruent.

16.

always A chord of a circle is a segment. always

3.3

3.3

theorems relating the parts of a circle

77

THEOREMS RELATING THE PARTS OF A CIRCLE

Theorem TL1 : In the plane of a circle, if a line is perpendicular to a radius at a point on the circle, then the line is tangent to the circle. A B

C

Theorem BC: If a line through the center of a circle is perpendicular to a chord, it bisect the chord. B l

C

A D

Theorem CC: In the same circle or in congruent circles, if two chords are the same distance from the center, then the chords are congruent. A C

B D

Theorem SD: In the same circle or in congruent circle, if two chords are congruent, then they are the same distance from the center. A C

B

D

78

Chapter 3

circles

Theorem CC: A circle can be circumscribed about any triangle. A l1

l2

C

B

D

Theorem CI: A circle can be inscribed in any triangle. U

M N

A C B

V

D

W

Theorem IC: if two chords intersect in a circle, then the product of the lengths of the segments on one chord is equal to the product of the lengths of the segments on the other.

A

B D

E

C

F

3.3

theorems relating the parts of a circle

79

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 3.3 The following numbers should refer to the illustration below A

C B

D

Equilateral triangle ABD is inscribed in circle C. 1. If BD = 12, then CB = __________.

2. If CD = 10, then AD = ___________.

3. Draw these figures using a compass and a ruler only in your workbook in a provided space.

a.

80

Chapter 3

b.

c.

d.

circles

3.3

theorems relating the parts of a circle

81

4. Construction of chord in a circle. a. Construct a chord of length 3cm in a circle of radius 2cm

Step

Construction

1

Draw a circle of radius 2cm with O as center.

2

Mark any point P on the circle

3

With P as center and 3cm as radius, draw an arc to cut the circle at Q

4

Join P and Q ( PQ is the chord of length 3cm)

Note that this chord divides the circle into 2 regions called segments (PSQ and PTQ). PSQ is smaller in size as compared to PTQ. Note that the diameter POR (4cm) is the chord of maximum length. Definition: The region bounded by the chord and an arc is called ‘segment’ of the circle. Observations: 1. Diameter is the chord having maximum length. 2. A chord cuts the circle into 2 segments. 3. Diameter cuts the circle into 2 equal segments (semi circles).

82

Chapter 3

circles

5. Finding distance between chord and center of the circle. a. Construct a chord of length 4cm in a circle of radius 2.5cm and measure the distance between the center of the circle and the chord.

Step Construction 1

Draw a circle of radius 4cm with O as center

2

Mark any point P on the circle

3

With P as center and 4cm as radius, draw an arc to cut the circle at Q

4

Join P and Q ( PQ is the chord of length 4cm)

5

Bisect the line PQ to find its mid-point M. (With P and Q as centers, draw arcs of radius more than half the length of PQ on both sides of PQ and let these arcs meet at R and S). Let RS Meet PQ at M.

6

OM is the distance between chord PQ and center O. We notice that OM=1.5cm

3.3

theorems relating the parts of a circle

83

6. Construct few more chords of length 4cm and measure their distance from the center. Considering the properties draw the corresponding figure. Observations: 1. As the length of the chord increases, the distance between the chord and the center decreases and becomes 0 for the diameter. 2. As the length of the chord decreases, the distance between the chord and the center increases and becomes R (radius) finally. 3. Chords of equal lengths are equidistant from the center.

No 1

Figure

Properties of angle at circumference ASBA is a minor segment.

ACB is angle at the circumference and is formed by the minor segment ASBA.

2

Minor arc(ASB) subtends acute angle(ACB) at the circumference. ASBOA is a semi circle.

ACB is angle at the circumference and is formed by the semi circle ASBOA. Note that ACB = 900.

3

Semi circle subtends a right angle at the circumference. ASBA is a major segment. ACB is angle at the circumference and is formed by the major segment ASBA.

Major arc subtends obtuse angle at the circumference.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 3.3 The following numbers should refer to the illustration below A

C B

D

Equilateral triangle ABD is inscribed in circle C. 1. If BD = 12, then CB = __________. CB = 6 2. If CD = 10, then AD = ___________. AD = 20 3. Draw these figures using a compass and a ruler only in your workbook in a provided space.

a.

b.

c.

d.

4. Construction of chord in a circle. b. Construct a chord of length 3cm in a circle of radius 2cm

Step

Construction

1

Draw a circle of radius 2cm with O as center.

2

Mark any point P on the circle

3

With P as center and 3cm as radius, draw an arc to cut the circle at Q

4

Join P and Q ( PQ is the chord of length 3cm)

Note that this chord divides the circle into 2 regions called segments (PSQ and PTQ). PSQ is smaller in size as compared to PTQ. Note that the diameter POR (4cm) is the chord of maximum length. Definition: The region bounded by the chord and an arc is called ‘segment’ of the circle. Observations: 1. Diameter is the chord having maximum length. 2. A chord cuts the circle into 2 segments. 3. Diameter cuts the circle into 2 equal segments (semi circles).

5. Finding distance between chord and center of the circle. b. Construct a chord of length 4cm in a circle of radius 2.5cm and measure the distance between the center of the circle and the chord.

Step Construction 1

Draw a circle of radius 4cm with O as center

2

Mark any point P on the circle

3

With P as center and 4cm as radius, draw an arc to cut the circle at Q

4

Join P and Q ( PQ is the chord of length 4cm)

5

Bisect the line PQ to find its mid point M. (With P and Q as centers, draw arcs of radius more than half the length of PQ on both sides of PQ and let these arcs meet at R and S). Let RS Meet PQ at M.

6

OM is the distance between chord PQ and center O. We notice that OM=1.5cm

6. Construct few more chords of length 4cm and measure their distance from the center. Considering the properties draw the corresponding figure. Observations: 1. As the length of the chord increases, the distance between the chord and the center decreases and becomes 0 for the diameter. 2. As the length of the chord decreases, the distance between the chord and the center increases and becomes R (radius) finally. 3. Chords of equal lengths are equidistant from the center.

No 1

Figure

Properties of angle at circumference ASBA is a minor segment.

ACB is angle at the circumference and is formed by the minor segment ASBA.

2

Minor arc(ASB) subtends acute angle(ACB) at the circumference. ASBOA is a semi circle.

ACB is angle at the circumference and is formed by the semi circle ASBOA. Note that ACB = 900.

3

Semi circle subtends a right angle at the circumference. ASBA is a major segment. ACB is angle at the circumference and is formed by the major segment ASBA.

Major arc subtends obtuse angle at the circumference.

84

Chapter 3

circles

3.4

THEOREMS ON SECANT AND TANGENT LINES AND SEGMENTS

Theorem TL2 : If a line tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency.

C

l

A

B

A common tangent is a line that is tangent to each of two coplanar circles.

common internal tangents

C

Common external tangents do not intersect the segment joining the centers of the circles.

Common external tangents D

Common internal tangents do intersect the segment joining the centers of the circles.

Tangent circles are two coplanar circles that are tangent to the same line at the same point.

F

E

3.4

theorems on secant and tangent lines and segments

Theorem SC: The measure of an angle formed by two secants which intersect in the interior of a circle is one-half the sum of the measures of the arcs intercepted by the angle and its vertical angle.

Theorem AS: The measure of an angle formed by two secants that intersect in the exterior of a circle is one-half the the difference of the measures of of the intercepted arcs.

85

U V

1 W

2

3

C D

3

2

X

1 E

F

Theorem TS: The measure of an angle formed by a tangent and a secant that intersect at the point of tangency is one-half the measure of the intercepted arc.

A

C D1

D2

B

E D3

86

Chapter 3

circles

Theorem IE: The measure of an angle formed by a secant and a tangent that intersect in the exterior of a circle is one-half the difference of the measures of the intercepted arc.

Theorem IT: The measure of an angle formed by two intersecting tangents is onehalf the difference of the measures of the intercepted arc.

A

B

1 C

D

E

A E

C

B

1 D

Theorem SC: If two secants drawn to a circle from an exterior point, the product of the lengths of one secant and its external secant segment is equal to the product of the lengths of the other secant and its external secant segment.

Theorem CE: If the tangent and secant are drawn to a circle from an exterior point of the circle, the square of the length of the tangent segment is equal to the product of the lengths of the secant segment and its external secant segment.

A

B

F D

C

E

A

C D

E B

3.4

theorems on secant and tangent lines and segments

87

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 3.4 1. Illustrate by drawing two circles meeting in one common external tangent.

2. Illustrate by drawing two circles meeting in two common external tangent.

3. How many common tangents do two internally tangent circles have?

4. How many common tangents do two externally tangent circles have?

5. Illustrate by drawing two circles meeting in two common external tangent and one common internal tangent.

88

Chapter 3

circles

6. Illustrate by drawing two circles meeting in two common external tangent and two common internal tangent.

7. Illustrate by drawing two circles meeting with no common tangent.

The following numbers should refer to the illustration below. A

D

C E B

In circle C, AD is a tangent, and AB is a diameter, BE is a secant, and BE intersect AD at D. 8. If BD = 13 and AD = 5, solve for AB.

3.4

theorems on secant and tangent lines and segments

9. If BD = 20 and AD = 16, solve for AB.

10.

If BD = 12 and AD = 6, solve for CB.

11.

If BD = 6 √ and AD = 6, solve for AC.

12.

If CB = 4 and AD = 6, solve for BD.

89

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 3.4 1. Illustrate by drawing two circles meeting in one common external tangent.

2. Illustrate by drawing two circles meeting in two common external tangent.

3. How many common tangents do two internally tangent circles have? 2 4. How many common tangents do two externally tangent circles have? 2 5. Illustrate by drawing two circles meeting in two common external tangent and one common internal tangent.

6. Illustrate by drawing two circles meeting in two common external tangent and two common internal tangent.

7. Illustrate by drawing two circles meeting with no common tangent.

The following numbers should refer to the illustration below. A

D

C E B

In circle C, AD is a tangent, and AB is a diameter, BE is a secant, and BE intersect AD at D. 8. If BD = 13 and AD = 5, solve for AB. AB = √ =√ = √ = 12

− −

9. If BD = 20 and AD = 16, solve for AB. AB = √ − =√ − = √ = 12 10. If BD = 12 and AD = 6, solve for CB. AB = √ − =√ − = √ = 10.39 Since CB = AB = (10.39) CB = 5.20 11.

If BD = 6 √ and AD = 6, solve for AC. AB = √

√

−

=√ − = √ − = √ = 6 Since CB = AB

12.

= (6) CB = 3 If CB = 4 and AD = 6, solve for BD. Since CB = 4 is the radius of a circle Then AB = 8 is the diameter of a circle BD = √ =√ =√ = √ = 10

90

Chapter 4

geometric figures on the rectangular coordinate plane

Chapter 4 GEOMETRIC FIGURES ON THE RECTANGULAR COORDINATE PLANE 4.1 4.2 4.3 4.4

DISTANCE FORMULA RULE OF DISTANCE FORMULA ON GEOMETRIC PROPERTIES EQUATION OF A CICLE PROBLEMS INVOLVING GEOMETRIC FIGURES IN THE COORDINATE PLANE

Specific Objectives: at the end of the chapter the students should be able to;      

Derive the distance formula between two points on the plane, Apply the distance formula to prove some geometric properties, State the center-radius form of the equation of the circle, Find the center and radius of a circle given its equation and vice versa, Sketch the graph of a circle on the coordinate plane, Solve problems involving geometric figures in the coordinate plane,

Gödel or Goedel, Kurt [goedl](1906--78) Logician and mathematician, born in Brünn, Moravia (now Brno, Czech Republic). He studied and taught in Vienna, then emigrated to the USA in 1940 and joined the Institute of Advanced Study at Princeton. He became a US citizen in 1948. He stimulated significant work in mathematical logic and propounded one of the most important proofs in modern mathematics: Gödel's proof, published in 1931 with reference to Russells Principia mathemetica, showed that any formal logical system adequate for number theory must contain propositions not provable in that system .

4.1

4.1

DISTANCE FORMULA

distance formula

91

The distance d between A(x1, y1) and B(x2, y2), apply the Pythagorean theorem to right triangle ABC

y B(x2, y2)

d

A(x1, y1)

C(x2, y1)

x

(AB)2 = (AC)2 + (BC)2 d2 = (x2 - x1)2 + (y2 - y1)2 d= √

−

−

equation is called distance formula.

92

Chapter 4

geometric figures on the rectangular coordinate plane

Example: Find the distance between (-3, 6) and (4, -2) Let (x1, y1) = (-3, 6) and (x2, y2) = (4, -2) d= √ − = √ = √ = √ = √ = 10.63

− − − −

Example: Find the distance between (-4, -5) and (-5, -6) Let (x1, y1) = (-4, -5) and (x2, y2) = (-5, -6) d= √ − = √ − = √ − = √ = √ = 1.41

− − −

Example: Consider the figure to find the distance Let (x1, y1) = (1, 3) and (x2, y2) = (6, 10)

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

d= √ = √ = √ = √ = √ = 8.6

− −

− −

4.1

distance formula

93

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 4.1 1. Find the distance between (4, 7) and (5, 3)

2. Find the distance between (-6, 6) and (-6, 7)

3. Find the distance between (-5, -8) and (-6, -4)

94

Chapter 4

geometric figures on the rectangular coordinate plane

4. Find the distance between (5, -8) and (6, -5)

5. Find the distance between (5, 7) and (7, 6)

6. Find the distance between (-4, 9) and (-7, 5)

4.1

7. Consider the figure to find the distance y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

8. Consider the figure to find the distance y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

distance formula

95

96

Chapter 4

geometric figures on the rectangular coordinate plane

9. Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

10.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

11.

2

3

4

5

6

7

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

7

4.1

distance formula

97

98

Chapter 4

geometric figures on the rectangular coordinate plane

12.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

7

4.1

13.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

14.

2

3

4

5

6

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

distance formula

99

100

Chapter 4

geometric figures on the rectangular coordinate plane

15.

Find the distance between (-24, 39) and (-27, 35)

16.

Find the distance between (-40, 90) and (-70, 50)

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 4.1 1. Find the distance between (4, 7) and (5, 3) Let (x1, y1) = (4, 7) and (x2, y2) = (5, 3) d= √ − − = √ − − = √ − = √ = √ = 4.12 2. Find the distance between (-6, 6) and (-6, 7) Let (x1, y1) = (-6, 6) and (x2, y2) = (-6, 7) d= √ − − = √ − − = √ = √ = 3. Find the distance between (-5, -8) and (-6, -4) Let (x1, y1) = (-5, -8) and (x2, y2) = (-6, -4) d= √ − − = √ − − = √ − = √ = √ = 4.12

4. Find the distance between (5, -8) and (6, -5) Let (x1, y1) = (5, -8) and (x2, y2) = (6, -5) d= √ − − = √ − − = √ = √ = √ = 3.16

5. Find the distance between (5, 7) and (7, 6) Let (x1, y1) = (5, 7) and (x2, y2) = (7, 6) d= √ = √ = √ = √ =

− −

− − −

6. Find the distance between (-4, 9) and (-7, 5) Let (x1, y1) = (-4, 9) and (x2, y2) = (-7, 5) d= √ − = √ − = √ − = √ = √ =5

− − −

7. Consider the figure to find the distance y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

Let (x1, y1) = (5, 1) and (x2, y2) = (6, 10) d= √ − − = √ − − = √ = √ = √ = 9.06 8. Consider the figure to find the distance y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

Let (x1, y1) = (5, 1) and (x2, y2) = (1, 8) d= √ − − = √ − − = √ − = √ = √ = 8.06

9. Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

Let (x1, y1) = (1, 7) and (x2, y2) = (6, 10) d= √ − − = √ − − = √ = √ = √ = 10.3

10.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

7

Let (x1, y1) = (1, 7) and (x2, y2) = (7, 3) d= √ − − = √ − − = √ − = √ = √ = 7.21 11.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

7

Let (x1, y1) = (1, 2) and (x2, y2) = (7, 3) d= √ − − = √ − − = √ = √ = √ = 6.08 12.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

7

Let (x1, y1) = (1, 3) and (x2, y2) = (7, 3) d= √ − − = √ − − = √ = √ =

13.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

Let (x1, y1) = (2, 1) and (x2, y2) = (2, 10) d= √ − − = √ − − = √ = √ = 14.

Consider the figure to find the distance

y 10 9 8 7 6 5 4 3 2 1

x 1

2

3

4

5

6

Let (x1, y1) = (1, 0) and (x2, y2) = (6, 10) d= √ − − = √ − − = √ = √ = √ = 11.18 15.

Find the distance between (-24, 39) and (-27, 35) Let (x1, y1) = (-24, 39) and (x2, y2) = (-27, 35) d= √ − = √ − = √ − = √ = √ =5

16.

− − −

Find the distance between (-40, 90) and (-70, 50) Let (x1, y1) = (-40, 90) and (x2, y2) = (-70, 50) d= √ − = √ − = √ − = √ = √ = 50

− − −

4.2

4.2

rule of distance formula on geometric properties

RULE OF DISTANCE FORMULA ON GEOMETRIC PROPERTIES

Let P(x, y) be any point on the circle, C(h, k) be the center of the circle, r be the radius of the circle. Considering the distance formula we can write the equation of the circle in standard form. d = √ (r = √ or

− − h

y − y y − k )2

r2 = 𝐱 − 𝐡 𝟐 𝐲 − 𝐤 𝟐 𝐱 − 𝐡 𝟐 𝐲 − 𝐤 𝟐 = r2

Notice: If the radius is zero, then we only have a single point (h, k) which is a point circle or no circle.

Let P(x, y) be any point on the circle, C(h, k) = (0. 0) be the center of the circle, r be the radius of the circle.

Considering the distance formula we can write the equation of the circle with center at the origin. d = √ (r = √ or

− −

r2 = 𝐱 𝟐 𝐲𝟐 x2 + y2 = r2

y − y y − )2

101

102

Chapter 4

geometric figures on the rectangular coordinate plane

r2 = − h y − k 𝟐 or 𝐱 − 𝐡 𝐲 − 𝐤 𝟐 = r2 the standard form is convenient in that it shows at a glance the center and radius of the circle. General form is often the form in which equation is given. Changing an equation from general to standard form in order to identify the center and radius of the circle. Thus, the general form of the equation of a circle is x2 + y2 + Ax + By + C = 0, x2 + y2 + Ax + By = - C x2 + Ax + y2 + By = -C x2 + Ax + (x2 +

𝐴

𝐴

)2

+ y2 + By +

)2 + (y2 +

𝐵

)2 =

A

𝐵

B

)2

=-C +

𝐴

)2

+

𝐵

)2

C

Three outcomes is expected: 1. If the fraction on the right is positive, then it has a square root, the graph is then the circle with center at (h, k) = (radius at r =

𝐴

, -

𝐵

) −

√

c

2. If the fraction on the right is 0, then the graph is the Point at (-

𝐴

, -

𝐵

)

3. If the fraction on the right is negative, then the graph is the empty set, because the left-hand side can never be negative.

4.2

rule of distance formula on geometric properties

103

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 4.2 1. Find the standard form of the equation of the circle whose center is at (2, 1) and whose radius is 3 and draw the circle.

y 5 4 3 2 1 -3

-2

-1

0

1

2

3

4

5

-1 -2 -3 -4 -5

2. Find the standard form of the equation of the circle with center at (-1, 1) , diameter 3 and draw the circle.

104

Chapter 4

geometric figures on the rectangular coordinate plane

y 5 4 3

2 1 -4

-3

-2

-1

0

1

2

3

4

x

-1 -2 -3 -4 -5

3. Find the radius of a circle, given that the center is at (2, –3) and the point (–1, –2) lies on the circle.

4.2

rule of distance formula on geometric properties

4. The equation of the circle with radius 3 centered at (1,2) is,

5. Find the center and radius of the circle x2 + y2 - 4x + 6y - 12 = 0.

6. Find the standard form of the equation of the circle whose center is at (2, 1) which passes through the point (-2, -3).

105

106

Chapter 4

geometric figures on the rectangular coordinate plane

y 6 5 4 3 2 1 -4

-3

-2

-1

0

1

2

3

4

5

6

7

x

-1 -2 -3 -4 -5 7. Find the equation of the circle whose center is at the origin and radius 2. And draw the circle. 3 2 1 -4

-3

-2

-1

0 -1 -2 -3

1

2

3

4

x

NAME:______________________________ GRADE /SECTION:_____________________

SCORE:_____________ DATE:_______________

PRACTICE SET 4.2 1. Find the standard form of the equation of the circle whose center is at (2, 1) and whose radius is 3 and draw the circle. Let C(h, k) = (2, 1) and r = 3 r2 = − − (x - 2)2 + (y - 1)2 = 32 (x - 2)2 + (y - 1)2 = 9 y 5 4 3 2 1 -3

-2

-1

0

1

2

3

4

5

-1 -2 -3 -4 -5

2. Find the standard form of the equation of the circle with center at (-1, 1) , diameter 3 and draw the circle. Let C(h, k) = (-1, 1) and r = 3 r2 = − − 2 2 (x + 1) + (y - 1) = 32 (x + 1)2 + (y - 1)2 = 9

y 5 4 3

2 1 -4

-3

-2

-1

0

1

2

3

4

x

-1 -2 -3 -4

-5

3. Find the radius of a circle, given that the center is at (2, –3) and the point (– 1, –2) lies on the circle. The radius is the distance between the center and any point on the circle, so I need to find the distance

Then the radius is sqrt(10), or about 3.16, rounded to two decimal places.

4. The equation of the circle with radius 3 centered at (1,2) is (x - 1)2 + (y - 2)2 = 9 5. Find the center and radius of the circle x2 + y2 - 4x + 6y - 12 = 0 We must get the equation into standard form by completing the squares. We have x2 - 4x + y2 + 6y = 12 -4/2 = -2 and 6/2 = 3 (-2)2 = 4 and 32 = 9 Adding and subtracting we have x2 - 4x + 4 - 4 + y2 + 6y + 9 - 9 = 12 (x - 2)2 + (y + 3)2 = 12 + 4 + 9 = 25 = 52 So that the center of the circle is (2,-3) and the radius is 5. 6. Find the standard form of the equation of the circle whose center is at (2, 1) which passes through the point (-2, -3).

r= √ − = √ − − = √ − = √ = √ = 5.66

− − − −

r2 = − − 2 2 (x - 2) + (y - 1) = (5.66)2 (x - 2)2 + (y - 1)2 = 31.70

y 5 4 3 2 1 -4

-3

-2

-1

0

1

2

3

4

5

6

7

x

-1 -2 -3 -4 -5 7. Find the equation of the circle whose center is at the origin and radius 2. And draw the circle. x2 + y2 = r2

3

x2 + y2 = 22

2

x 2 + y2 = 4

1 -4

-3

-2

-1

0 -1 -2

1

2

3

4

x

4.3

4.3

equation of a circle

EQUATION OF A CICLE

Circle in standard form.

d = √ (r = √

or

r2 = −

− − h

y − y y − k )2

−

− −

= r2

Notice: If the radius is zero, then we only have a single point (h, k) which is a point circle or no circle. Circle with center at the origin. d = √ (r = √

− −

y − y y − )2

or

r2 = x2 + y2 = r2

or

x2 + y2 + Ax + By + C = 0, x2 + y2 + Ax + By = - C

Circle in general form

107

108

Chapter 4

geometric figures on the rectangular coordinate plane

Example: What is the equation of the circle pictured on the graph below?

Since the radius of this this circle is 1, and its center is The origin, this picture's equation is (y−0)2+(x−0)2=12 y2+x2=1 Example: Look at the graph below, can you express the equation of the circle in standard form?

Since the radius of this this circle is 1, and its center is (1,0) , this circle's equation is (y−0)2+(x−1)2=12 y2+(x−1)2=1 Example: Look at the graph below, can you express the equation of the circle in standard form?

Since the radius of this this circle is 2, and its center is (3,1) , this circle's equation is (x−3)2+(y−1)2=22 (x−3)2+(y−1)2=4

4.3

equation of a circle

109

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 4.3 1. Find the center and radius of the circle having the following equation: 4x2 + 4y2 – 16x – 24y + 51 = 0. The given equation. Move the loose number over to the other side. Group the x-stuff together. Group the y-stuff together. Whatever is multiplied on the squared terms (it'll always be the same number), divide it off from every term. This is the complicated step. You'll need space inside your groupings, because this is where you'll add the squaring term. Take the x-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation, as shown. Do the same with the y-term coefficient. Convert the left side to squared form, and simplify the right side. Read off the answer from the rearranged equation.

110

Chapter 4

geometric figures on the rectangular coordinate plane

2. Find the center and radius of the circle with the following equation: 100x2 + 100y2 – 100x + 240y – 56 = 0. The given equation. Move the loose number over to the other side. Group the x-stuff and y-stuff together. Divide off by whatever is multiplied on the squared terms. Take the coefficient on the x-term, multiply by one-half, square, and add inside the x-stuff and also to the other side. Do the same with the y-term. Convert the left-hand side to squared form, and simplify the righthand side. If necessary, fiddle with signs and exponents to make your equation match the circle equation's format. Read off the answer.

4.3

equation of a circle

111

3. Plot (x - 4)2 + (y - 2)2 = 25 7 6 5 4 3 2 1 -2

-1

0

1

2

3

4

5

6

7

8

-1 -2 -3 4. (x-2)2 + (y-3)2 = 4. (a) Find the center and radius of the circle. (b) Graph the circle. 5 4 3 2 1 -4

-3

-2

-1

0 -1

1

2

3

4

x

9

112

Chapter 4

geometric figures on the rectangular coordinate plane

5. (x+1)2 + (y-2)2 = 9. the circle.

(a) Find the center and radius of the circle. (b) Graph

5 4 4 2 1 -4

-3

-2

-1

0

1

2

3

4

x

-1

6. 2x2 + 2y2 = 8. circle.

(a) Find the center and radius of the circle. (b) Graph the

3 2 1 -4

-3

-2

-1

0 -1 -2 -3

1

2

3

4

x

4.3

equation of a circle

7. x2 + y2 - 6x + 4y + 9 = 0. (a) Find the center and radius of the circle. (b) Graph the circle.

1 -4

-3

-2

-1

0

1

2

3

4

x

-1 -2 -3 -4 -5 8. x2 + y2 - 6x + 2y + 4 = 0. (a) Find the center and radius of the circle. (b) Graph the circle.

3 2 1 -2

-1

0 -1 -2 -3

1

2

3

4

5

6

x

113

114

Chapter 4

geometric figures on the rectangular coordinate plane

9. Convert x2 + y2 - 4x -6y + 8 = 0 into center-radius form.

10.

Write the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).

11.

Write the equation for the circle shown below if it is shifted 3 units to the right and 4 units up.

4.3

equation of a circle

115

y 9 8 7 6 5 4 3 2 1 -5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

-1 -2 -3 -4 -5 12.

Convert

into center-radius form.

x

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 4.3 1. Find the center and radius of the circle having the following equation: 4x2 + 4y2 – 16x – 24y + 51 = 0. The given equation. Move the loose number over to the other side. Group the x-stuff together. Group the y-stuff together. Whatever is multiplied on the squared terms (it'll always be the same number), divide it off from every term. This is the complicated step. You'll need space inside your groupings, because this is where you'll add the squaring term. Take the x-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation, as shown. Do the same with the y-term coefficient. Convert the left side to squared form, and simplify the right side. Read off the answer from the The center is at (h, k) = (x, y) = (2, 3). rearranged equation. The radius is r = √ =

2. Find the center and radius of the circle with the following equation: 100x2 + 100y2 – 100x + 240y – 56 = 0. The given equation. Move the loose number over to the other side. Group the x-stuff and y-stuff together. Divide off by whatever is multiplied on the squared terms. Take the coefficient on the x-term, multiply by one-half, square, and add inside the x-stuff and also to the other side. Do the same with the y-term. Convert the left-hand side to squared form, and simplify the righthand side. If necessary, fiddle with signs and exponents to make your equation match the circle equation's format. Read off the answer.

The center is at ( 1/2, – 6/5 ) and the radius is 3/2.

3. Plot (x - 4)2 + (y - 2)2 = 25 The formula for a circle is (x - a)2 + (y - b)2 = r2 So the center is at (4,2) And r2 is 25, so the radius is √25 = 5

So we can plot: The Center: (4,2) Up: (4,2+5) = (4,7) Down: (4,2-5) = (4,-3) Left: (4-5,2) = (-1,2) Right: (4+5,2) = (9,2) 4. (x-2)2 + (y-3)2 = 4. (a) Find the center and radius of the circle. (b) Graph the circle. (a) Center: (h= 2, k= 3) = ( 2, 3 ) and since r2 = 4 => r = 4 = 2 (b) The graph is

radius r=2

5. (x+1)2 + (y-2)2 = 9. (a) Find the center and radius of the circle. (b) Graph the circle. The standard form is: (x - h)2 + (y-k)2 = r2

(a) (b)

(x - (-1))2 + (y-2)2 = (3)2 Center: (h= -1, k= 2) = ( -1, 2 ) and radius r=3 since r2= 9 > r=9=3 The graph

6. 2x2 + 2y2 = 8. (a) Find the center and radius of the circle. (b) Graph the circle. x 2 + y2 = 4 (x - h)2 + (y - k)2 = r2 (x - 0)2 + (y - 0)2 = (2)2 (a) Center: (h= 0, k= 0) = ( 0, 0 ) and radius r = 2 since r2 = 4 => r = 4 = 2 (b) The graph is

7. x2 + y2 - 6x + 4y + 9 = 0. (a) Find the center and radius of the circle. (b) Graph the circle. (x2 - 6x + ?1 ) + (y2 + 4y + ?2 ) = -9 + ?1 + ?2 (x2 - 6x + 9 ) + (y2 + 4y + 4 ) = -9 + 9 + 4 ( x - 3 )2 + ( y + 2 )2 = 4 ( x - 3 )2 + ( y - (-2) )2 = 4 a) Center: (h= 3, k= -2) = ( 3, -2 ) and radius r = 2 since r2 = 4 => r = 4 = 2 (b) The graph is

8. x2 + y2 - 6x + 2y + 4 = 0. (a) Find the center and radius of the circle. (b) Graph the circle. (x2 - 6x + ?1 ) + (y2 + 2y + ?2 ) = -4 + ?1 + ?2 (x2 - 6x + 9 ) + (y2 + 2y + 1 ) = -4 + 9 + 1 ( x - 3 )2 + ( y + 1 )2 = 4 2 2 ( x - 3) + ( y - (-1) ) = 4 (a) Center: (h= 3, k= -1) = ( 3, -1 ) and radius r = 2 since r2 = 4 => r = 4 = 2 (b) The graph is

9. Convert x2 + y2 - 4x -6y + 8 = 0 into center-radius form.

10.

Write the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).

Find the center by using the midpoint formula.

11.

Write the equation for the circle shown below if it is shifted 3 units to the right and 4 units up.

y 9 8 7 6 5 4 3 2 1 -5

-4

-3

-2

-1

0

1

2

3

4

5

6

7

8

-1 -2 -3 -4 -5 12.

Convert

into center-radius form.

Center: (-3/2, 2)

Radius: 1/2

x

116

Chapter 4

4.4

PROBLEMS INVOLVING GEOMETRIC FIGURES IN THE COORDINATE PLANE

Example: Given: Prove:

geometric figures on the rectangular coordinate plane

is isosceles

Read the question carefully. The word isosceles, by definition, tells you that you are looking for two congruent sides. Since congruent implies "of equal length" and the word length implies "distance", you will use the Distance Formula. Draw a neat, labeled graph for the problem.

State the formula that you will be using. Since we are looking for two sides of equal length, you can STOP when you find the two sides. Look at the figure before you begin and choose the two sides that you "think" are of equal length. The triangle is isosceles because it has two End with a concluding sentence congruent sides. stating why you know the triangle is isosceles.

4.4

problems involving geometric figures in the coordinate plane

117

Example: Read the question carefully. The word trapezoid, by definition, tells you that you are looking for a figure with ONLY ONE set of parallel sides. Lines are parallel when they have the same slope. You will use the Slope Formula. Draw a neat, labeled graph for the problem.

State the formula that you will be using. We are looking for one set of parallel sides and one set on nonparallel sides. Be sure to state the connection between the slopes and the sides being parallel or non-parallel.

End with a concluding sentence stating why you know the figure is

118

Chapter 4

geometric figures on the rectangular coordinate plane

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 4.4 1. Find the perimeter and area of the Triangle.

4.4

problems involving geometric figures in the coordinate plane

2. Find the perimeter and area of the Trapezoid.

3. Find the perimeter and area of the following parallelogram.

119

120

Chapter 4

geometric figures on the rectangular coordinate plane

4. Find the perimeter and area of the following pentagon.

5. Determine the shape of the given points {(1,-6),(4,-6),(5,-9),(0,-9)} y 5

4 3 2 1 -4

-3

-2

-1

0 -1 -2 -3 -4 -5

1

2

3

4

x

4.4

problems involving geometric figures in the coordinate plane

6. Determine the shape of the given points {(-8,-4),(-6,-2),(-4,-4),(-5,-6), (-7,-6)}

-4

-3

-2

-1

y 5 4 3 2 1 0 -1 -2 -3 -4 -5

1

2

3

4

x

7. Determine the shape of the given points {(-4,7),(-2,4),(-6,1)} y 5 4 3 2 1 -4

-3

-2

-1

0 -1 -2 -3 -4 -5

1

2

3

4

x

121

122

Chapter 4

geometric figures on the rectangular coordinate plane

8. Determine the shape of the given points {(3,-1),(7,-2),(7,-5),(3,-4)} y 5 4 3

2 1 -4

-3

-2

-1

0 -1

1

2

3

4

x

-2 -3 -4 -5 9. Determine the shape of the given points {(-1,2),(2,2),(3,0),(2,-2),(-1,-2), (-2,0)} y 5 4 3 2 1 -4

-3

-2

-1

0 -1 -2 -3 -4 -5

1

2

3

4

x

4.4

10.

problems involving geometric figures in the coordinate plane

Determine the shape of the given points {(7,2),(7,-1),(10,-1)} y 5 4 3 2 1 -4

11.

-3

-2

-1

0 -1 -2 -3 -4 -5

1

2

3

4

x

Determine the shape of the given points {(2,7),(6,7),(4,5),(8,5)} y 5 4 3 2 1 -4

-3

-2

-1

0 -1 -2 -3 -4 -5

1

2

3

4

x

123

124

Chapter 4

geometric figures on the rectangular coordinate plane

12.

Determine the shape of the given points {(4,0),(0,4),(-4,0), (0, -4)} y 5 4 3 2 1 -4

13.

-3

-2

-1

0 -1 -2 -3 -4 -5

1

2

3

4

x

Determine the shape of the given points {(3,0),(0,2),(-3,0),(0,-2)} y 5 4 3 2 1 -4

-3

-2

-1

0 -1 -2 -3 -4 -5

1

2

3

4

x

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 4.3 1. Find the perimeter and area of the following Triangle.

Left side: bottom = 4, vertical side = 6, hypotenuse = Right side: bottom =2, vertical side = 6, hypotenuse = Bottom side: = 6 P = 7.2 + 6.3 + 6 = 19.5 units A = Bh = x6x6 = 18 square units

2. Find the perimeter and area of the following Trapezoid.

find the hypotenuse of the triangle top = 1, vertical side = 4, hypotenuse = Right side: We will need to find the hypotenuse of the triangle Top side = 2, vertical side = 4, hypotenuse = P = 6 + 3 + 4.1 + 4.5 = 17.6 units Left triangle = Bh = x1x4 = 2 square units Right triangle = Bh = x2x4 = 4 square units Rectangle = l x w = 3 x 4 = 12 square units A = 2 + 4 + 12 = 18 square units

3. Find the perimeter and area of the following parallelogram.

Vertical side is

x 6 = 3, horizontal side is

P = 4.24 x 4 = 16.96 units A=

= 18 square units

x 6 = 3; hypotenuse =

4. Find the perimeter and area of the following pentagon.

Bottom side = 2, vertical side = 2, hypotenuse = The top right triangle’s hypotenuse also Now for the bottom left triangle: Top = 1, vertical side = 3, hypotenuse = The bottom right triangle’s hypotenuse is also P = 2.83 + 3.16 + 2 + 3.16 + 2.83 = 13.98 or 5. Determine the shape of the given points {(1,-6),(4,-6),(5,-9),(0,-9)}

6. Determine the shape of the given points {(-8,-4),(-6,-2),(-4,-4),(-5,-6), (-7,-6)}

7. Determine the shape of the given points {(-4,7),(-2,4),(-6,1)}

8. Determine the shape of the given points {(3,-1),(7,-2),(7,-5),(3,-4)}

9. Determine the shape of the given points {(-1,2),(2,2),(3,0),(2,-2),(-1,-2),(2,0)}

10.

Determine the shape of the given points {(7,2),(7,-1),(10,-1)}

11.

Determine the shape of the given points {(2,7),(6,7),(4,5),(8,5)}

12.

Find the equation of the circle with center (4, -1) and which passes through (-1, 2).

Now, plug in (-1, 2) for and and solve for .

Substituting in 34 for 13.

, the equation is

Determine if the point (-2, -2) are on (x + 1)2 + (y - 5)2 = 50.

(-2, -2) is on the circle 14.

The radius of your circle is the distance between the points (−1,4) and (3,−2). Using the Distance Formula: d= √ − − = √ − − = √ − = √ = √ = 7.21

5.0

measures of position

125

Chapter 5 MEASURES OF POSITION 5.1

5.2 5.3

MEASURES OF POSITION 5.1.1 QUARTILES 5.1.2 DECILES 5.1.3 PERCENTILES PROBLEMS INVOLVING MEASURES OF POSITION CONSTRUCT BOX PLO

Specific Objectives: at the end of the chapter the students should be able to;  Define the following measures of position: quartiles, deciles, and percentiles,  Describe the following measures of position: quartiles, deciles, and percentiles,  Relate quartile, deciles, and percentiles,  interpret quartile, deciles, and percentiles,  calculate specified percentiles of a set of data,  use measure of position to describe a set of data and infer some information about the data, Hamilton, Sir William Rowan (1805--65) Mathematician, the inventor of quaternions, born in Dublin, Ireland. At the age of nine he knew 13 languages, and at 15 he had read Newton's Principia, and begun original investigations. In 1827, while still an undergraduate, he was appointed professor of astronomy at Dublin and Irish Astronomer Royal; in 1835 he was knighted. His first published work was on optics, and he then developed a new approach to dynamics which became of importance in the 20th-c development of quantum mechanics. He introduced quaternions as a new algebraic approach to three-dimensional geometry, and they proved to be the seed of much modern algebra.

126

Chapter 5

measures of position

5.1

MEASURES OF POSITION: 5.1.1 Quartiles Just like the median divides the set of observation into two equal parts when arranged in the numerical order, in the same way quartile divides the set of observation into 4 equal parts. The value of the middle term, between the first term and median is known as first or Lower Quartile and is denoted as Q1. The value of middle term between the last term and the median is known as third or Upper Quartile and is denoted as Q3.The median itself is known as the Second Quartile and is denoted as Q2 When the set of observation is arranged in an ascending order, then the Lower quartile is given as:

Q3 =

(

𝟏 𝒕𝒉

𝒏 𝟒

)

term

If the solution is a decimal number then, Lower quartile Q1 is given by rounding to the nearest whole integer. The Second quartile, which is the median of the set of observation is given as:

Q3 =

(

𝟏 𝒕𝒉

𝒏

)

𝟐

term

The Upper quartile is given as, Q3 =

(

𝟏] 𝒕𝒉

𝟑[𝒏 𝟒

)

term

5.1

measures of position

127

If the solution is a decimal number then, Upper quartile Q3 is given by rounding to the nearest whole integer. The lower and the upper quartile value helps us to find the measure of dispersion in the set of observation, which is called as 'inter-quartile range', it is denoted as IQR and it is the difference between upper and lower quartile. IQR = Q3 − Q1

Example: Find the median, lower quartile, upper quartile and inter-quartile range of the following data set of scores: 19, 22, 24, 20, 24, 27, 25, 24, 30 ? Solution: First, lets arrange of the values in an ascending order: 19, 20, 22, 24, 24, 24, 25, 27, 30 Now lets calculate the Median, Median = (

)th term

=(

)th term

= 5th term = 24 Lower quartile = ( =( =(

)th term

)th term )th term

= 2.5th term

128

Chapter 5

measures of position

Find the average of 2nd and 3rd term = = = 21 Upper quartile = (

=( =( =(

[

]

[

]

[

=(

[

]

)th term

)th term

)th term

]

)th term

)th term

= 7.5th term (lets find the average of 7th and 8th term) = = = 26 Inter - quartile = Upper quartile - lower quartile = 26 - 21 =5

5.1

measures of position

129

Example: Find the first quartile, second quartile and third quartile of the given information of the following sequence 4, 77, 16, 59, 93, 88 ? Solution: First, lets arrange of the values in an ascending order : 4, 16, 59, 77, 88, 93 Given n = 6 Lower quartile = ( =(

)th term

)th term

= ( )th term = 1.7th term Here we can consider the 2nd term (rounding 1.7 to nearest whole integer) from the set of observation. ⇒ 2nd term = 16 Lower quartile = 16 Upper quartile = ( =( =( =( =(

[

]

[

]

[

]

)th

)th term

)th term

[ ]

)th term

)th term

= 5.25th term Here we can consider the 5th term (rounding 5.25 to nearest whole integer) from the set of observation. ⇒ 5.25th = 88 Upper quartile = 88 Inter - quartile = Upper quartile - lower quartile = 88 - 16 = 72

130

Chapter 5

measures of position

The quartiles are the three values of the variable that divide an ordered data set into four equal parts. Q1, Q2 and Q3 determine the values for 25%, 50% and 75% of the data. Q2 coincides with the median. Calculating Quartiles 1. Order the data from smallest to largest. 2. Look for the place that occupies every quartile by means of the expression

, k = 1, 2, 3

, in the table of cumulative

frequencies. Odd Number of Data

2, 5, 3, 6, 7, 4, 9 2, 3, 4, 5, 6, 7, 9 Q1 Q2 Q3

Even Number of Data

2, 5, 3, 4, 6, 7, 1, 9 1, 2, 3, 4, 5, 6, 7, 9 2.5 4.5 6.5 Q1 Q2 Q3

Calculation of Quartiles for Grouped Data

𝒌

Qk = Li +

𝑵 𝟒

𝑭𝒊 𝒇𝒊

𝟏

𝒂𝒊 k = 1, 2, 3

5.1

measures of position

131

Li is the lower limit of the quartile class. N is the sum of the absolute frequency. Fi-1 is the absolute frequency immediately below the quartile class. ai is the width of the class containing the quartile class. The quartiles are independent of the widths of the classes.

Example:

Calculate the quartiles of the distribution for the following table: fi 8 10 16 14 10 5 2 65

[50, 60) [60, 70) [70, 80) [80, 90) [90, 100) [100, 110) [110, 120)

Fi 8 18 34 48 58 63 65

Calculating the First Quartile = 16.25 Q1 = 60 +

.

10

.

10

= 68.25 Calculating the Second Quartile = 32.5 Q2 = 70 + = 79.06 Calculation of the Third Quartile = 48.75 Q3 = 90 + = 90.75

. 10

132

Chapter 5

measures of position

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 5.1.1 1. A year ago, Angela began working at a computer store. Her supervisor asked her to keep a record of the number of sales she made each month. The following data set is a list of her sales for the last 12 months: 34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37. Use Angela's sales records to find: a. the median b. the range c. the upper and lower quartiles d. the interquartile range a. the median

b. Range = difference between the highest and lowest values c. Lower quartile = value of middle of first half of data Q1

d. Upper quartile = value of middle of second half of data Q3

5.1

measures of position

e. Interquartile range = Q3 – Q1 2. Find the IQR for the following data set: 3, 5, 7, 8, 9, 11, 15, 16, 20, 21.

3. Find the intervals in which the first, second and third quartiles lie. Test Scores Frequency Cumulative Frequency 76-80 3 3 81-85 86-90 91-95

7 6 4

10 16 20

133

134

Chapter 5

measures of position

4. Find the First, Second and Third Quartiles of the data set below using the cumulative frequency curve. Age (years) 10 11 12 13 14 15 16 17

Frequency 5 10 27 18 6 16 38 9

Solution: Age (years) Frequency Cumulative Frequency

5.1

measures of position

135

5. What is the upper quartile for the following set of data's, 30, 31, 32, 33, 34.

6. What is the lower quartile for the following set of data's, 30, 31, 32, 33, 34.

136

Chapter 5

measures of position

7. Find the quartile for the following data: 4, 7, 3, 8, 20, 18, 5

8. For the following data of scores: 12, 22, 24, 27, 16, 19, 15, find first quartile, second quartile, third quartile and Interquartile Range.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 5.1.1 1. A year ago, Angela began working at a computer store. Her supervisor asked her to keep a record of the number of sales she made each month. The following data set is a list of her sales for the last 12 months: 34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37. Use Angela's sales records to find: a. the median b. the range c. the upper and lower quartiles d. the interquartile range Answers a. The values in ascending order are: 1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57. Median = (12th + first) ÷ 2 = 6.5th value = (sixth + seventh observations) ÷ 2 = (24 + 28) ÷ 2 = 26 b. Range = difference between the highest and lowest values = 57 – 1 = 56 c. Lower quartile = value of middle of first half of data Q1 = the median of 1, 11, 15, 19, 20, 24 = (third + fourth observations) ÷ 2 = (15 + 19) ÷ 2 = 17 d. Upper quartile = value of middle of second half of data Q3 = the median of 28, 34, 37, 47, 50, 57 = (third + fourth observations) ÷ 2 = (37 + 47) ÷ 2 = 42

e. Interquartile range = Q3–Q1 = 42 – 17 = 25 2. Find the IQR for the following data set: 3, 5, 7, 8, 9, 11, 15, 16, 20, 21. Put the numbers in order. 3, 5, 7, 8, 9, 11, 15, 16, 20, 21. Make a mark in the center of the data: 3, 5, 7, 8, 9, | 11, 15, 16, 20, 21. Place parentheses around the numbers above and below the mark you made in Step 2 –it makes Q1 and Q3 easier to spot. (3, 5, 7, 8, 9), | (11, 15, 16, 20, 21). Find Q1 and Q3 Q1 is the median (the middle) of the lower half of the data, and Q3 is the median (the middle) of the upper half of the data. (3, 5, 7, 8, 9), | (11, 15, 16, 20, 21). Q1 = 7 and Q3 = 16. Subtract Q1 from Q3. 16 – 7 = 9. 3. Find the intervals in which the first, second and third quartiles lie. Test Scores

Frequency

76-80

3

Cumulative Frequency 3

81-85 86-90 91-95

7 6 4

10 16 20

If there are a total of 20 scores, the first quartile will be located (25% · 20 = 5) five values up from the bottom. This puts the first quartile in the interval 81-85. In a similar fashion, the second quartile will be located (50% · 20 = 10) ten values up from the bottom in the interval 81-85. The third quartile will be located (75% · 20 = 15) fifteen values up from the bottom in the interval 86-90.

4. Find the First, Second and Third Quartiles of the data set below using the cumulative frequency curve. Age (years) Frequency 10 5 11 10 12 27 13 18 14 6 15 16 16 38 17 9 Solution: Age (years) Frequency Cumulative Frequency 10 5 5 11 10 15 12 27 42 13 18 60 14 6 66 15 16 82 16 38 120 17 9 129

5. What is the upper quartile for the following set of data's, 30, 31, 32, 33, 34. Given: Data = 30, 31, 32, 33, 34. To find: Upper quartile Formula: Median, Odd = `(n + 1)/2` Even = `n/2` + 1 Find Median: Median = `(n + 1)/2` = `(5 + 1)/2` = `6/2` =3 Therefore, 32 is known as the median value. Find Upper quartile: Upper quartile - Values which is greater than the median. Therefore, 33, 34 is known as the upper quartile. Result: Upper Quartile = 33, 34 Thus, this is the obtained result for solving the above problem. 6. What is the lower quartile for the following set of data's, 30, 31, 32, 33, 34. Given: Data = 30, 31, 32, 33, 34. To find: Lower quartile Formula: Median, Odd = `(n + 1)/2` Even = `n/2` + 1 Find Median: Median = `(n + 1)/2` = `(5 + 1)/2` = `6/2` =3 Therefore, 32 is known as the median value. Step 5:Find Lower quartile: Lower quartile - Values which is lower than the median. Therefore, 30, 31 is known as the lower quartile. Result: Lower Quartile = 30, 3

7. Let us find the quartile for the following data: 4, 7, 3, 8, 20, 18, 5. First let us arrange the given data in ascending order 3, 4, 5, 7, 8, 18, 20 Number of Observation (n) = 7 Lower quartile, Q1 = (n+14)th term = (7+14)th term = 2nd term i.e. 4 Second Quartile, Q2 = (n+12)th term = (7+12)th term = 4th term i.e. 7 Upper Quartile, Q3 = (3(n+1)4)th term = (3(7+1)4)th term = (3(8)4)th term = 6th term i.e. 18 8. For the following data of scores: 12, 22, 24, 27, 16, 19, 15, find first quartile, second quartile, third quartile and Interquartile Range. Solution: Let us arrange the values in ascending order:12, 15, 16, 19, 22, 24, 27 Given n = 7 Lower quartile = (n+14)th term = (7+14)th term = (84)th term = 2nd term i.e. 15 Second Quartile = (n+12)th term = (7+12)th term = (82)th term = 4th term i.e. 19 Lower Quartile = (3(n+1)4)th term = (3(7+1)4)th term = (3(8)4)th term = 6th term i.e. 24 Inter-quartile range = Upper quartile - Lower Quartile = 15 - 24 = -9

5.1

5.1.2

measures of position

137

DECILES

Deciles are the nine values of a variable, which divides its distribution into ten parts with equal frequencies. For this, it is very important to arrange the set of data in ascending or descending order, which means the values have to be sorted out before dividing. Deciles are all the nine values of the variable that can divide an ordered data into ten equal parts. As there are nine deciles, lets denote each of them as D1, D2, D3, D4, D5, D6,D7,D8,andD9. notation is defined as: D1 is denoted as the first decile under which 10% of the total population lies. D2 is denoted as the second decile under which 20% of the total population lies. D3 is denoted as the third decile under which 30% of the total population lies. D4 is denoted as the fourth decile under which 40% of the total population lies. D5 is denoted as the fifth decile under which 50% of the total population lies. D6 is denoted as the sixth decile under which 60% of the total population lies. D7 is denoted as the seventh decile under which 70% of the total population lies. D8 is denoted as the eighth decile under which 80% of the total population lies. D9 is denoted as the ninth decile under which 90% of the total population lies. Hence, the deciles can determine the values for 10%, 20%, 30%.......90% of the data. It has the capacity to show the spread of data as well as the range of a group of numbers. It shows whether a certain element varies or tends toward the same number. It is applied to show if the numbers are average or extreme in several cases.

138

Chapter 5

measures of position

Given below are the uses of deciles: Splitting this way, makes it easy to discover the smallest as well as the largest values of the given distribution. This type of grouping can be mostly seen in financial fields for academical as well as statistical studies. It is very useful when dealing with large amounts of data, which includes the timely results for standardized tests in schools, etc. It helps the government to find how the income in a country is distributed, how much of the total income is earned by low wage earning groups and by high wage earning groups. If both the groups earn the same proportion of the income, then there is income equality. So, deciles help to find inequality too. When the number of datas is very small and if its frequency can be divided completely by 10, then it is very simple to calculate deciles. First of all, arrange the values of the data set into ascending order and split the data into 10 group. Then the first decile will be the first set, the second decile will be the second set and so on. But when the frequencies are very large and can’t be equally divided by 10, then we use the following formula: 𝒌

Dk = Li +

𝑵

𝑭𝒊

𝟒

𝟏

𝒇𝒊

𝒂𝒊 k = 1, 2, 3, . . . 9

where, li is the lower limit of the class selected as decile. N is the total frequency, Fi -1 is the cumulative frequency just below the decile class ai is the class width fi is the frequency of the decile class.

5.1

measures of position

Steps for Finding Deciles: Step 1: Arrange the numbers in ranked order (ascending). Step 2: Use the following formula to find the position of each of the 9 deciles in the cumulative frequency distribution table of the given data. Dkth position = , where k is the decile to be found, or k = 1, 2… 9 and N is the total number of frequency. Step 3: The following formula can be used to find each of the decile

Dk = Li + Example:

k = 1, 2, 3, . . . 9 Consider the following data Class [0,10) [10,20) [20,30) [30,40) [40,50) [50,60)

Frequency 5 21 15 4 10 8

The first step is to find the cumulative frequency table: Class [0,10) [10,20) [20,30) [30,40) [40,50) [50,60)

Frequency 5 21 15 4 10 8

Here N = 63

Cumulative frequency 5 26 41 45 55 63

139

140-

Chapter 5

measures of position

Step 1:For First Decile: Here k = 1 So

= = 6.3 So the class is [10, 20)

D1 = Li + = 10 +

= 10 = 10 = 10 = 10

+ +

+ + 0.62 = 10.62 Step 2: For second Decile: Here k = 2 So

= = 12.6

So the class is [10, 20)

D2 = Li + = 10 +

= 10 = 10 = 10 = 10

+ +

+ + 3.62 = 13.62

5.1

Step 3:For third Decile: Here k = 3 So = = 18.9 So the class is [10, 20)

D3 = Li + = 10 +

= 10 + = 10 + = 10 + = 10 + 6.62

= 16.62

Step 4: For fourth Decile: Here k = 4 So = = 25.2 So the class is [10, 20)

D4 = Li + = 10 +

= 10 = 10 = 10 = 10

+ +

+ + 9.62 = 19.62

measures of position

141

142

Chapter 5

measures of position

Step 5:For fifth Decile: Here k = 5 So = = 31.5 So the class is [20, 30)

D5 = Li + = 20 +

= 20 = 20 = 20 = 20

+ +

+ + 3.67 = 23.67 Step 6: For sixth Decile: Here k = 6 So = = 37.8 So the class is [20, 30)

D6 = Li + = 20 +

= = = =

20 + 20 +

20 + 20 + 7.87 = 27.87

5.1

Step 7: For seventh Decile: Here k = 7 So = = 44.1 So the class is [30, 40)

D7 = Li + = 30 +

= = = =

30 + 30 +

30 + 30 + 7.75 = 37.75 Step 8: For eighth Decile: Here k = 8 So = = 50.4 So the class is [40, 50)

D8 = Li + = 40 +

= 40 + = 40 + = 40 + = 40 + 5.4 = 45.4

measures of position

143

144

Chapter 5

measures of position

Step 9:For ninth Decile: Here k = 9 So = = 56.7 So the class is [50, 60)

D9 = Li + = 50 +

= = = =

50 + 50 +

50 + 50 + 2.12 = 52.12 Deciles are similar to quartiles. But while quartiles sort data into four quarters, deciles sort data into ten equal parts: The 10th, 20th, 30th, 40th, 50th, 60th, 70th, 80th, 90th and 100th percentiles. A decile rank assigns a number to a decile: Decile Rank 1 2 3 4 5 6 7 8 9

Percentile 10th 20th 30th 40th 50th 60th 70th 80th 90th

5.1

measures of position

145

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 5.1.2 1. Find the deciles for the following data. 3, 15, 24, 28, 33, 35, 38, 42, 43, 38, 36, 34, 29, 25, 17, 7, 34, 36, 39, 44, 31, 26, 20, 11, 13, 22, 27, 47, 39, 37, 34, 32, 35, 28, 38, 41, 48, 15, 32, 60, 56, 13. Cumulative Frequency is calculated using the formula, Fi = Fi-1 + fi Class

Frequency fi

3 - 10 10 - 17 17 - 24 24 - 31 31 - 38 38 - 45 45 - 52 52 - 59 59 - 66

2 5 3 7 12 9 2 1 1

Calculation of the First Decile

Cumulative Frequency Fi 2 7 10 17 29 38 40 41 42

Calculation of the Second Decile

Calculation of the Third Decile

146

Chapter 5

measures of position

Calculation of the Fourth Decile

Calculation of the Fifth Decile

Calculation of the Sixth Decile

Calculation of the Seventh Decile

Calculation of the Eigth Decile

Calculation of the Ninth Decile

5.1

measures of position

2. Calculate the deciles of the distribution for the following table: fi Fi 50 - 60 8 8 Calculation of the First Decile 60 - 70 10 18 70 - 80 16 34 80 - 90 14 48 90 -100 10 58 100 - 110 5 63 110 - 120 2 65 65

Calculation of the Second Decile

Calculation of the Third Decile

147

148

Chapter 5

measures of position

Calculation of the Fourth Decile

Calculation of the Fifth Decile

Calculation of the Sixth Decile

Calculation of the Seventh Decile

Calculation of the Eigth Decile

Calculation of the Ninth Decile

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 5.1.2 1. Find the deciles for the following data. 3, 15, 24, 28, 33, 35, 38, 42, 43, 38, 36, 34, 29, 25, 17, 7, 34, 36, 39, 44, 31, 26, 20, 11, 13, 22, 27, 47, 39, 37, 34, 32, 35, 28, 38, 41, 48, 15, 32, 60, 56, 13. Cumulative Frequency is calculated using the formula, Fi = Fi-1 + fi Class

Frequency fi Cumulative Frequency Fi 2 2 5 7 3 10 7 17 12 29 9 38 2 40 1 41 1 42

3 - 10 10 - 17 17 - 24 24 - 31 31 - 38 38 - 45 45 - 52 52 - 59 59 - 66

Step 2:For second Decile: Here k = 2

Step 1:For First Decile: Here k = 1 So

𝐾

𝑁

𝑘

𝑁

= = 4.2 So the class is [3, 10) D1 = L i +

𝑓𝑖

𝑎𝑖

𝐾

𝑁

𝑘

𝑁

= = 8.4 So the class is [10, 17) D1 = L i +

–

= 3 + = 3 + = 3 + 14.7 = 17.7

= 10 + = = = =

𝐹𝑖 𝑓𝑖

= 10 +

= 3 + = 3 +

𝐹𝑖

So

Step 3:For third Decile: Here k = 3 So

𝐾

𝑁

𝑘

𝑁

= = 12.6 So the class is [17, 24) 𝑎𝑖

D1 = L i +

𝐹𝑖 𝑓𝑖

= 17 + –

10 + 10 + 10 + 8.96 18.96

= 17 + = = = =

–

17 + 17 + 17 + 13.09 30.09

𝑎𝑖

Step 4:For fourth Decile: Here k = 4 So

𝐾

𝑁

𝑘

𝑁

So

𝐹𝑖 𝑓𝑖

𝑎𝑖

–

= 24 +

= = = =

Step 7:For seventh Decile: Here k = 7 𝐾

𝑁

𝑘

𝑁

= = 29.4 So the class is [38, 45) D7 = L i +

𝐹𝑖 𝑓𝑖

= 38 + = 38 + = = = =

𝑘

𝑁

D5 = L i +

𝐹𝑖 𝑓𝑖

So

𝑎𝑖

= = = =

Step 8:For eighth Decile: Here k = 8 𝐾

𝑁

𝑘

𝑁

D8 = L i +

𝐹𝑖 𝑓𝑖

= 38 + –

38 + 38 + 38 + 0.311 38.311

= 38 + = = = =

𝑘

𝑁

D6 = L i +

𝐹𝑖 𝑓𝑖

𝑎𝑖

–

= 31 +

= = 33.6 So the class is [38, 45) 𝑎𝑖

𝑁

= 31 +

31 + 31 + 31 + 2.33 33.333

So

𝐾

= = 25.2 So the class is [31, 38)

–

= 31 +

24 + 24 + 24 + 6.8 30.8

So

𝑁

= 31 +

= 24 + = = = =

𝐾

Step 6:For sixth Decile: Here k = 6

= = 21 So the class is [31, 38)

= = 16.8 So the class is [24, 31) D4 = L i +

Step 5:For fifth Decile: Here k = 5

31 + 31 + 31 + 4.783 35.783

Step 9:For eighth Decile: Here k = 9 So

𝐾

𝑁

𝑘

𝑁

= = 37.8 So the class is [38, 45) 𝑎𝑖

D9 = L i +

𝐹𝑖 𝑓𝑖

= 38 + –

38 + 38 + 38 + 3.578 41.578

= 38 + = = = =

–

38 + 38 + 38 + 6.844 44.844

𝑎𝑖

3. Calculate the deciles of the distribution for the following table: [50, 60) [60, 70) [70, 80) [80, 90) [90, 100) [100, 110) [110, 120)

fi

Fi

8 10 16 14 10 5 2 65

8 18 34 48 58 63 65

Calculation of the Second Decile

Calculation of the First Decile

Calculation of the Third Decile

Calculation of the Fourth Decile

Calculation of the Fifth Decile

Calculation of the Sixth Decile

Calculation of the Seventh Decile

Calculation of the Eighth Decile

Calculation of the Ninth Decile

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5.1.3 PERCENTILES Percentiles are defined as the values that divide the whole series into 100 equal parts. So, there are 99 quartiles namely first percentile denoted by P1, second percentile denoted by P2 …... and 99th percentile denoted by P99. 50th percentile is nothing but Median. Since it denotes the position of the item in the series, it is a positional average. A percentile is a comparison score between a particular score and the scores of the rest of a group. It shows the percentage of scores that a particular score surpassed. The percentile rank is calculated using the formula

where P is the desired percentile N is the number of data points. Example 1:If the scores of a set of students in a math test are 20,30, 15 and 75, what is the percentile rank of the score 30? Arrange the numbers in ascending order and give the rank ranging from 1 to the lowest to 4 to the highest.

Use the formula:

Therefore, the score 30 has the 60th percentile.

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Example 2: Determine the 35th percentile of the scores 7, 3, 12, 15, 14, 4, and 20. Arrange the numbers in ascending order and give the rank ranging from 1 to the lowest to 7 to the highest.

Use the formula:

The integer part of R is 2, calculate the score corresponding to the ranks 2 and 3. They are 4 and 7. The product of the difference and the decimal part is 0.8(7 – 4) = 2.4. Therefore, the 35th percentile is 2 + 2.4 = 4.4. Example 3: The scores for student are 40, 41, 42, 43, 45, 47, 49, 50, 51, 51, 54, 57, 59, 60, 61, 63, 67, 69, 70, 71. Find out the percentile for score 51. Formula for percentile with 'x' values is x 100 = =

x 100 x 100

= 45 The score 51 is in 45th percentile

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NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 5.1.3 1. The scores for student are 21, 24, 26, 27, 28, 28, 31, 32, 34, 36, 39, 40, 42, 44, 45, 46, 49, 53, 57, 60. Find out the percentile for score 28.:

2. Find the 14th percentile of the following data X 2 8 16 25 38

F 13 5 7 9 10

X 2 8 16 25 38

F 13 5 7 9 10

Cumulative Frequency 13 13 + 5 = 18 18 + 7 = 25 25 + 9 = 34 34 + 10 = 44

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4. Find the 60th percentile of the following data: 26, 17, 31, 20, 30, 39, 30 and 39

5. Find the 15th percentile of the following data: 5, 8, 17, 28, 35, 44, and 54

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measures of position

6. Find the 70th percentile of the following data X F Solution: 12 7 14 11 15 14 17 2 21 1 X

F

12 14 15 17 21

7 11 14 2 1

Cumulative Frequency 7 7 + 11 = 18 18 + 14 = 32 32 + 2 = 34 34 + 1 = 35

7. Find the 48th percentile of the following data:Solution:

Class 10-20 20-30 30-35 35-50 50-65

F 7 3 2 2 1

Class 10-20 20-30 30-35 35-50 50-65

F 7 3 2 2 1 N = 15

Cumulative Frequency 7 7 + 3 = 10 10 + 2 = 12 12 + 2 = 14 14 + 1 = 15

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8. The scores for student are 40, 45, 49, 53, 61, 65, 71, 79, 85, 91. What is the percentile for score 71?

9. The scores for student are 42, 46, 52, 61, 68, 79, 85, 92. What is the percentile for score 61?

10. If Mark graduated 25th out of a class of 150 students, then 125 students were ranked below Mark. Mark's percentile rank would be:

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11.

The math test scores were: 50, 65, 70, 72, 72, 78, 80, 82, 84, 84, 85, 86, 88, 88, 90, 94, 96, 98, 98, 99. Find the percentile rank for a score of 84 on this test.

12.

The math test scores were: 50, 65, 70, 72, 72, 78, 80, 82, 84, 84, 85, 86, 88, 88, 90, 94, 96, 98, 98, 99. Find the percentile rank for a score of 86 on this test.

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13.

Solve for the 25th, 75th, and 85th percentile rank

Interval 69 - 76 77 - 84 85 - 92 93 - 100

Frequency 1 4 4 1

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 5.1.3 1. The scores for student are 21, 24, 26, 27, 28, 28, 31, 32, 34, 36, 39, 40, 42, 44, 45, 46, 49, 53, 57, 60. Find out the percentile for score 28.: The number of scores for student is 20 and the number of below 28 is 4. The 28 is repeated in two times. Formula for percentile with 'x' values is

x 100 =

x 100

= x 100 = 25. The score 28 is in 25th percentile. 2. Find the 14th percentile of the following data X F 2 13 8 5 16 7 25 9 38 10 X 2 8 16 25 38

F 13 5 7 9 10

Cumulative Frequency 13 13 + 5 = 18 18 + 7 = 25 25 + 9 = 34 34 + 10 = 44

Here, N = 44 k th percentile, P14 = Size of kN+1th100 = Size of 14(45)100 item = Size of 6.3th item = 2 Answer: 14th percentile = 2 3. Find the 60th percentile of the following data: 26, 17, 31, 20, 30, 39, 30 and 39 First we have to arrange the numbers in the ascending order. 17, 20, 26, 30, 30, 31, 39, 39 Here, n = 8 Pk = k n+1th100 item = 60(8+1)100 item = 5.4th item = 5th item + 0.4(6th item-5th item) = 30 + 0.4(31 - 30) = 30 + 0.4 = 30.4 Answer: 60th percentile = 30.4

4. Find the 15th percentile of the following data: 5, 8, 17, 28, 35, 44, and 54 The numbers are arranged in the increasing order. Here, n = 7 Pk = k n+1th100 item = 15(7+1)100 item = 1.2th item = 1st item + 0.2(2nd item – 1st item) = 5 + 0.2(8 - 5) = 5 + 0.6 = 5.6 Answer: 15th percentile = 5.6

5. Find the 70th percentile of the following data X F Here, N = 35 kth percentile, 12 7 Pk = Size of kN+1th100 item 14 11 = Size of 70(36)100th item 15 14 = Size of 25.2th item = 15 17 2 Answer: 70th percentile = 15 21 1 X 12 14 15 17 21

F 7 11 14 2 1

Cumulative Frequency 7 7 + 11 = 18 18 + 14 = 32 32 + 2 = 34 34 + 1 = 35

6. Find the 48th percentile of the following data:Class F 48N100=48(15)100 = 7.2(which lies in 20-30) 10-20 7 P48 class = 20-30 20-30 3 l = 20, cf = 7, c = 10, f = 3 30-35 2 Therefore P48 = l + (kN100−cf)cf = 20 + (7.2−7)103 35-50 2 = 20 + 0.67 = 20.67 50-65 1 Answer: 48th percentile = 20.67

Class 10-20 20-30 30-35 35-50 50-65

F 7 3 2 2 1 N = 15

Cumulative Frequency 7 7 + 3 = 10 10 + 2 = 12 12 + 2 = 14 14 + 1 = 15

7. The scores for student are 40, 45, 49, 53, 61, 65, 71, 79, 85, 91. What is the percentile for score 71? No. of. scores below 71 = 6 Total no. of. scores = 10 The formula for percentile is given as, Percentile = No. of values below xTotal no. of values × 100 Percentile of 71 = 610 × 100 = 0.6 × 100 = 60 8. The scores for student are 42, 46, 52, 61, 68, 79, 85, 92. What is the percentile for score 61? No. of. scores below 61 = 3 Total no. of. scores = 8 The formula for percentile is given as, Percentile = No. of values below xTotal no. of values × 100. Percentile of 61 = 38 × 100 = 0.375 × 100 = 37.5 = 37 9. If Mark graduated 25th out of a class of 150 students, then 125 students were ranked below Mark. Mark's percentile rank would be:

10. The math test scores were: 50, 65, 70, 72, 72, 78, 80, 82, 84, 84, 85, 86, 88, 88, 90, 94, 96, 98, 98, 99. Find the percentile rank for a score of 84 on this test. Be sure the scores are ordered from smallest to largest. Locate the 84. Solution Using Formula:

Solution Using Visualization: Since there are 2 values equal to 84, assign one to the group "above 84" and the other to the group "below 84". 50, 65, 70, 72, 72, 78, 80, 82, 84, | 84, 85, 86, 88, 88, 90, 94, 96, 98, 98, 99

The score of 84 is at the 45th percentile for this test. 11. The math test scores were: 50, 65, 70, 72, 72, 78, 80, 82, 84, 84, 85, 86, 88, 88, 90, 94, 96, 98, 98, 99. Find the percentile rank for a score of 86 on this test. Be sure the scores are ordered from smallest to largest. Locate the 86. Solution Using Formula:

Solution Using Visualization:

Since there is only one value equal to 86, it will be counted as "half" of a data value for the group "above 86" as well as the group "below 86". 50, 65, 70, 72, 72, 78, 80, 82, 84, 84, 85, 8|6, 88, 88, 90, 94, 96, 98, 98, 99

The score of 86 is at the 58th percentile for this test.

12. Solve for the 25th, 75th, and 85th percentile rank

Interval 69 - 76 77 - 84 85 - 92 93 - 100

Frequency 1 4 4 1

a) 10 · 25% = 2.5 2.5 values up from the bottom puts it in the interval 77 - 84 b) 10 · 75% = 7.5 7.5 values up from the bottom puts it in the interval 85 - 92 c) 5 students received a grade of 85% or higher. What percent of 10 is 5? n% (10) = 5 10 10 n% = .5 or n = 50% 50% of the students received a pencil

5.2

5.2

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157

PROBLEMS INVOLVING MEASURES OF POSITION

Standard Scores (z-scores) The standard score is obtained by subtracting the mean and dividing the difference by the standard deviation. The symbol is z, which is why it's also called a z-score.

where z is the z-score, X is the value of the element, μ is the population mean, and σ is the standard deviation. Here is how to interpret z-scores.    





A z-score less than 0 represents an element less than the mean. A z-score greater than 0 represents an element greater than the mean. A z-score equal to 0 represents an element equal to the mean. A z-score equal to 1 represents an element that is 1 standard deviation greater than the mean; a z-score equal to 2, 2 standard deviations greater than the mean; etc. A z-score equal to -1 represents an element that is 1 standard deviation less than the mean; a z-score equal to -2, 2 standard deviations less than the mean; etc. If the number of elements in the set is large, about 68% of the elements have a z-score between -1 and 1; about 95% have a z-score between -2 and 2; and about 99% have a z-score between -3 and 3.

The mean of the standard scores is zero and the standard deviation is 1. This is the nice feature of the standard score - no matter what the original scale was, when the data is converted to its standard score, the mean is zero and the standard deviation is 1.

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Example: How well did Althea perform in her mathematics coursework compared to the other 50 students? To answer this question, we can re-phrase it as: What percentage (or number) of students scored higher than Althea and what percentage (or number) of students scored lower than Althea? First, let's reiterate that Althea scored 70 out of 100, the mean score was 60, and the standard deviation was 15 (see below). Score Mean Mathematics

(X) 70

µ 60

Standard Deviation s 15

In terms of z-scores, this gives us:

Percentiles, Deciles, Quartiles Percentiles (100 regions) The kth percentile is the number which has k% of the values below it. The data must be ranked. 1. 2. 3. 4.

Rank the data Find k% (k /100) of the sample size, n. If this is an integer, add 0.5. If it isn't an integer round up. Find the number in this position. If your depth ends in 0.5, then take the midpoint between the two numbers.

It is sometimes easier to count from the high end rather than counting from the low end. For example, the 80th percentile is the number which has 80% below it and 20% above it. Rather than counting 80% from the bottom, count 20% from the top.

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159

Note: The 50th percentile is the median. If you wish to find the percentile for a number (rather than locating the kth percentile), then 1. 2. 3. 4.

Take the number of values below the number Add 0.5 Divide by the total number of values Convert it to a percent

Deciles (10 regions) The percentiles divide the data into 100 equal regions. The deciles divide the data into 10 equal regions. The instructions are the same for finding a percentile, except instead of dividing by 100 in step 2, divide by 10. Quartiles (4 regions) The quartiles divide the data into 4 equal regions. Instead of dividing by 100 in step 2, divide by 4. Note: The 2nd quartile is the same as the median. The 1st quartile is the 25th percentile, the 3rd quartile is the 75th percentile. The quartiles are commonly used (much more so than the percentiles or deciles). The TI-82 calculator will find the quartiles for you. Some textbooks include the quartiles in the five number summary. Hinges The lower hinge is the median of the lower half of the data up to and including the median. The upper hinge is the median of the upper half of the data up to and including the median. The hinges are the same as the quartiles unless the remainder when dividing the sample size by four is three (like 39 / 4 = 9 R 3).

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The statement about the lower half or upper half including the median tends to be confusing to some students. If the median is split between two values (which happens whenever the sample size is even), the median isn't included in either since the median isn't actually part of the data. Example 1: sample size of 20 The median will be in position 10.5. The lower half is positions 1 10 and the upper half is positions 11 - 20. The lower hinge is the median of the lower half and would be in position 5.5. The upper hinge is the median of the upper half and would be in position 5.5 starting with original position 11 as position 1 - this is the original position 15.5. Illustration

The upper and lower hinges are descriptive statistics of a set of data values, where N is of the form N = 4n + 5 with n = 0, 1, 2, .... The hinges are obtained by ordering the data in increasing order a1 , ..., an , and writing them out in the shape of a "w" as illustrated above. The values at the bottom legs are called the hinges H1 and H2(and the central peak is the statistical median). In this ordering, H1

an + 2 =

(1)

M

a2n + 3 =

(2)

H2

a3n + 4 =

(3)

For N of the form 4n + 5 , the hinges H1and H2are identical to the quartiles Q1andQ2 . The difference H2 - H1 is called the H-spread.

5.2

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161

Five Number Summary The five number summary consists of the minimum value, lower hinge, median, upper hinge, and maximum value. Some use the quartiles instead of the hinges. Interquartile Range (IQR) The interquartile range is the difference between the third and first quartiles. That's it: Q3 - Q1 Outliers Outliers are extreme values. There are mild outliers and extreme outliers. The Bluman text does not distinguish between mild outliers and extreme outliers and just treats either as an outlier. Extreme Outliers Extreme outliers are any data values which lie more than 3.0 times the interquartile range below the first quartile or above the third quartile. x is an extreme outlier if , x < Q1 - 3 (IQR) or x > Q3 + 3 (IQR) Mild Outliers Mild outliers are any data values which lie between 1.5 times and 3.0 times the interquartile range below the first quartile or above the third quartile. x is a mild outlier if, or

Q1 – 3(IQR) 1.82)

In this case, we want the area to the right of 1.82. This is not what is given in the table. We can use the identity P(z > 1.82) = 1 - P(z < 1.82) reading the table gives P(z < 1.82) = .9656 Our answer is P(z > 1.82) = 1 - .9656 = .0344

6. Find P(-1.18 < z < 2.1)

Solution Once again, the table does not exactly handle this type of area. However, the area between 1.18 and 2.1 is equal to the area to the left of 2.1 minus the area to the left of 1.18. That is P(-1.18 < z < 2.1) = P(z < 2.1) - P(z < -1.18) To find P(z < 2.1) we rewrite it as P(z < 2.10) and use the table to get P(z < 2.10) = .9821. The table also tells us that P(z < -1.18) = .1190 Now subtract to get P(-1.18 < z < 2.1) = .9821 - .1190 = .8631

7. From the sample size of 21, Determine the median is in position, the lower half is positions, the upper half is positions, the lower hinge , the upper hinge. The median is in position 11. The lower half is positions 1 - 11 The upper half is positions 11 - 21. The lower hinge is the median of the lower half and would be in position 6. The upper hinge is the median of the upper half and would be in position 16.

8. Find the median, the first and the third quartile and solve for the interquartile range 1,2,5,6,7,9,12,15,18,19,27 Find the median (How to find a median). 1,2,5,6,7,9,12,15,18,19,27 Place parentheses around the numbers above and below the median. Not necessary statistically–but it makes Q1 and Q3 easier to spot. (1,2,5,6,7),9,(12,15,18,19,27) Find Q1 and Q3 Q1 can be thought of as a median in the lower half of the data. Q3 can be thought of as a median for the upper half of data. (1,2,5,6,7), 9, ( 12,15,18,19,27). Q1=5 and Q3=18. Subtract Q1 from Q3 to find the interquartile range. 18-5=13. 9. Consider the following numbers: 1, 3, 4, 5, 5, 6, 7, 11. Find the median, the first and the third quartile and solve for the interquartile range.

Q1 is the middle value in the first half of the data set. Since there are an even number of data points in the first half of the data set, the middle value is the average of the two middle values; that is, Q1 = (3 + 4)/2 or Q1 = 3.5.

Q3 is the middle value in the second half of the data set. Again, since the second half of the data set has an even number of observations, the middle value is the average of the two middle values; that is, Q3 = (6 + 7)/2 or Q3 = 6.5.

The interquartile range is Q3 minus Q1, so IQR = 6.5 - 3.5 = 3.

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Find the median, lower quartile and upper quartile for the following data: 11, 4, 6, 8, 3, 10, 8, 10, 4, 12 and 31. Answer Ordering the data, we get 3, 4, 4, 6, 8, 8,10, 10, 11, 12 and 31. The median is the (11 + 1) ÷ 2 = 6th value. The lower quartile is the (11 + 1) ÷ 4 = 3rd value. The upper quartile is the 3 (11 + 1) ÷ 4 = 9th value. Therefore, the median is 8, the lower quartile is 4, and the upper quartile is 11. 3, 4, 4, 6, 8, 8, 10, 10, 11, 12, 31

11.

Find the interquartile range. A survey was carried out to find the number of pets owned by each child in a class. The results are shown in the table: Number of petsFrequency 0 3 1 5 2 2 3 7 4 10 5 3 6 1 >6 0 Answer Remember that there is a total of 31 children in the class. The lower quartile is the 8th value, which is 1. The upper quartile is the 24th value, which is 4. Therefore, the interquartile range is 4 - 1 = 3. Note that the interquartile range ignores extreme values. The range includes extreme values. Look at this set of data: 1, 5, 7, 8, 9, 12, 13, 15, 17, 18, 35. The interquartile range is 17 - 7 = 10. The range is 35 - 1 = 34.

5.2

5.3

problems involving measures of position

CONSTRUCT A BOX PLOT

Box and Whiskers Plot A graphical representation of the five number summary. A box is drawn between the lower and upper hinges with a line at the median. Whiskers (a single line, not a box) extend from the hinges to lines at the minimum and maximum values.

167

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Reading a Box-and-Whisker Plot

A box and whisker plot (sometimes called a boxplot) is a graph that presents information from a five-number summary. It does not show a distribution in as much detail as a stem and leaf plot or histogram does, but is especially useful for indicating whether a distribution is skewed and whether there are potential unusual observations (outliers) in the data set. Box and whisker plots are also very useful when large numbers of observations are involved and when two or more data sets are being compared. (See the section on fivenumber summaries for more information.) Box and whisker plots are ideal for comparing distributions because the center, spread and overall range are immediately apparent.

5.2

problems involving measures of position

NAME:______________________________ GRADE/SECTION:_____________________

169

SCORE:_____________ DATE:_______________

PRACTICE SET 5.2 1-3 Refer to the situation below: Olympia and Miguel works at a computer store. He also recorded the number of sales he made each month. In the past 12 months, he sold the following numbers of computers: 51, 17, 25, 39, 7, 49, 62, 41, 20, 6, 43, 13 . 1. Give a five-number summary of Miguel’s and Olympia's sales.

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2. Make two box and whisker plots, one for Olympia's sales and one for Miguel's.

3. Briefly describe the comparisons between their sales.

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171

4. Consider the boxplot below.

2 4 6 8 10 12 14 Which of the following statements are true? I. The distribution is skewed right. II. The interquartile range is about 8. III. The median is about 10. (A) I only (B) II only (C) III only (D) I and III (E) II and III

16

18

5. The oldest person in math wizard is 90. The youngest person is 15.The median age of the residents is 44, the lower quartile is 25, and the upper quartile is 67.Represent this information with a box-and-whisker plot.

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6. Draw a box-and-whisker plot for the following data set: 4.3, 5.1, 3.9, 4.5, 4.4, 4.9, 5.0, 4.7, 4.1, 4.6, 4.4, 4.3, 4.8, 4.4, 4.2, 4.5, 4.4

7. Draw the box-and-whisker plot for the following data set: 77, 79, 80, 86, 87, 87, 94, 99

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173

8. Give the five-number summary of the following data set: 79, 53, 82, 91, 87, 98, 80, 93

9. Find the outliers, if any, for the following data set: 10.2, 14.1, 14.4. 14.4, 14.4, 14.5, 14.5, 14.6, 14.7,14.7, 14.7, 14.9, 15.1, 15.9, 16.4

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10.

Find the outliers and extreme values, if any, for the following data set, and draw the box-and-whisker plot. Mark any outliers with an asterisk and any extreme values with an open dot. 21, 23, 24, 25, 29, 33, 49

11.

Let's start by making a box-and-whisker plot (also known as a "box plot") of the geometry test scores we saw earlier: 90, 94, 53, 68, 79, 84, 87, 72, 70, 69, 65, 89, 85, 83, 72.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 5.3 1-3 Refer to the situation below: Olympia and Miguel works at a computer store. He also recorded the number of sales he made each month. In the past 12 months, he sold the following numbers of computers: 51, 17, 25, 39, 7, 49, 62, 41, 20, 6, 43, 13 . 1. Give a five-number summary of Miguel’s and Olympia's sales. First, put the data in ascending order. Then find the median. 6, 7, 13, 17, 20, 25, 39, 41, 43, 49, 51, 62. Median = (12th + 1st) ÷ 2 = 6.5th value = (sixth + seventh observations) ÷ 2 = (25 + 39) ÷ 2 = 32 There are six numbers below the median, namely: 6, 7, 13, 17, 20, 25. Q1 = the median of these six items = (6 + 1 ) ÷ 2= 3.5th value = (third + fourth observations) ÷ 2 = (13 + 17) ÷ 2 = 15 Here are six numbers above the median, namely: 39, 41, 43, 49, 51, 62. Q3 = the median of these six items = (6 + 1) ÷ 2= 3.5th value = (third + fourth observations) ÷ 2 = 46 The five-number summary for Carl's sales is 6, 15, 32, 46, 62. Using the same calculations, we can determine that the five-number summary for Olympia is 1, 17, 26, 42, 57.

2. Make two box and whisker plots, one for Olympia's sales and one for Miguel's.

Salesperson Olympia

Salesperson Miguel

0

10

20

30

40

50

60

70

3. Briefly describe the comparisons between their sales. Answers Please note that box and whisker plots can be drawn either vertically or horizontally. Miguel's highest and lowest sales are both higher than Olympia's corresponding sales, and Miguel's median sales figure is higher than Olympia's. Also, Miguel's interquartile range is larger than Olympia's. These results suggest that Miguel consistently sells more computers than Olympia does.

4. Consider the boxplot below.

2 4 6 8 10 12 14 16 18 Which of the following statements are true? I. The distribution is skewed right. II. The interquartile range is about 8. III. The median is about 10. (A) I only (B) II only (C) III only (D) I and III (E) II and III Solution The correct answer is (B). Most of the observations are on the high end of the scale, so the distribution is skewed left. The interquartile range is indicated by the length of the box, which is 18 minus 10 or 8. And the median is indicated by the vertical line running through the middle of the box, which is roughly centered over 15. So the median is about 15. 5. The oldest person in math wizard is 90. The youngest person is 15.The median age of the residents is 44, the lower quartile is 25, and the upper quartile is 67.Represent this information with a box-and-whisker plot. Solution

6. Draw a box-and-whisker plot for the following data set: 4.3, 5.1, 3.9, 4.5, 4.4, 4.9, 5.0, 4.7, 4.1, 4.6, 4.4, 4.3, 4.8, 4.4, 4.2, 4.5, 4.4 first step is to order the set. This gives me: 3.9, 4.1, 4.2, 4.3, 4.3, 4.4, 4.4, 4.4, 4.4, 4.5, 4.5, 4.6, 4.7, 4.8, 4. 9, 5.0, 5.1 The first number I need is the median of the entire set. Since there are seventeen values in this list, I need the ninth value: 3.9, 4.1, 4.2, 4.3, 4.3, 4.4, 4.4, 4.4, 4.4, 4.5, 4.5, 4.6, 4.7, 4.8, 4. 9, 5.0, 5.1 The median is Q2 = 4.4. And then the "whiskers" are drawn to the endpoints: 7. Draw the box-and-whisker plot for the following data set: 77, 79, 80, 86, 87, 87, 94, 99 My first step is to find the median. Since there are eight data points, the median will be the average of the two middle values: (86 + 87) ÷ 2 = 86.5 = Q2 This splits the list into two halves: 77, 79, 80, 86 and 87, 87, 94, 99. Since the halves of the data set each contain an even number of values, the sub-medians will be the average of the middle two values. Copyright © Elizabeth Stapel 2004-2011 All Rights Reserved Q1 = (79 + 80) ÷ 2 = 79.5 Q3 = (87 + 94) ÷ 2 = 90.5 The minimum value is 77 and the maximum value is 99, so I have: min: 77, Q1: 79.5, Q2: 86.5, Q3: 90.5, max: 99 Then my plot looks like this:

As you can see, you only need the five values listed above (min, Q1, Q2, Q3, and max) in order to draw your box-and-whisker plot. This set of five values has been given the name "the five-number summary".

8. Give the five-number summary of the following data set: 79, 53, 82, 91, 87, 98, 80, 93 The five-number summary consists of the numbers I need for the box-and-whisker plot: the minimum value, Q1 (the bottom of the box), Q2 (the median of the set), Q3 (the top of the box), and the maximum value (which is also Q4). So I need to order the set, find the median and the sub-medians, and then list the required values in order. ordering the list: 53, 79, 80, 82, 87, 91, 93, 98, so the minimum is 53 and the maximum is 98 finding the median: (82 + 87) ÷ 2 = 84.5 = Q2 lower half of the list: 53, 79, 80, 82, so Q1 = (79 + 80) ÷ 2 = 79.5 upper half of the list: 87, 91, 93, 98, so Q3 = (91 + 93) ÷ 2 = 92 five-number summary: 53, 79.5, 84.5, 92, 98 9. Find the outliers, if any, for the following data set: 10.2, 14.1, 14.4. 14.4, 14.4, 14.5, 14.5, 14.6, 14.7,14.7, 14.7, 14.9, 15.1, 15.9, 16.4

To find out if there are any outliers, I first have to find the IQR. There are fifteen data points, so the median will be at position (15 + 1) ÷ 2 = 8. Then Q2 = 14.6. There are seven data points on either side of the median, so Q1 is the fourth value in the list and Q3 is the twelfth: Q1 = 14.4 and Q3 = 14.9. Then IQR = 14.9 – 14.4 = 0.5. Outliers will be any points below Q1 – 1.5×IQR = 14.4 – 0.75 = 13.65 or above Q3 + 1.5×IQR = 14.9 + 0.75 = 15.65. Then the outliers are at 10.2, 15.9, and 16.4.

10.

Find the outliers and extreme values, if any, for the following data set, and draw the box-and-whisker plot. Mark any outliers with an asterisk and any extreme values with an open dot. 21, 23, 24, 25, 29, 33, 49 To find the outliers and extreme values, I first have to find the IQR. Since there are seven values in the list, the median is the fourth value, so Q2 = 25. The first half of the list is 21, 23, 24, so Q1 = 23; the second half is 29, 33, 49, so Q3 = 33. Then IQR = 33 – 23 = 10. The outliers will be any values below 23 – 1.5×10 = 23 – 15 = 8 or above 33 + 1.5×10 = 33 + 15 = 48. The extreme values will be those below 23 – 3×10 = 23 – 30 = –7 or above 33 + 3×10 = 33 + 30 = 63. So I have an outlier at 49 but no extreme values, I won't have a top whisker because Q3 is also the highest non-outlier, and my plot looks like this:

11.

Let's start by making a box-and-whisker plot (also known as a "box plot") of the geometry test scores we saw earlier: 90, 94, 53, 68, 79, 84, 87, 72, 70, 69, 65, 89, 85, 83, 72. This is also called quartile 2 (Q2). This is the lower quartile (Q1). This is the upper quartile (Q3).

Extreme values = 53 and 94.

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Chapter 6 BASIC COMBINATIONAL CONCEPTS 6.1 6.2 6.3

COUNT OCCURENCES OF AN EVENT FUNDAMENTAL COUNTING PRINCIPLE PERMUTATIONS AND COMBINATIONS

Specific Objectives: at the end of the chapter the students should be able to;  Counts the number of occurrences of an event like grid table, a tree diagram and systematic listing,  State the fundamental counting principle,  Use the fundamental counting principle to count number of arrangements or ways that a task can be carried out,  Recognize groupings that require order and grouping that do not require order,  Derive and use the formula for finding the permutation and combination of n object taken r at a time,  Explain the relationship of a permutation to a combination of n object taken r at a time.

Lebesgue, Henri (Léon) [luhbeg](1875--1941) Mathematician, born in Beauvais, N France. He studied at the Ecole Normale Supérieure, and taught at Rennes, Poitiers, the Sorbonne, and the Collège de France. Following the work of Emile Borel and René Baire (1874--1932), he developed the theory of measure and integration which bears his name, and applied it to many problems of analysis, in particular to the theory of Fourier series.

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COUNT OCCURENCES OF AN EVENT:

Tree diagrams

Tree diagrams allow us to see all the possible outcomes of an event and calculate their probability. Each branch in a tree diagram represents a possible outcome. If two events are independent, the outcome of one has no effect on the outcome of the other. Example: If we toss two coins, getting heads with the first coin will not affect the probability of getting heads with the second. A tree diagram which represent a coin being tossed three times looks like this

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From the tree diagram, we can see that there are eight possible outcomes. To find out the probability of a particular outcome, we need to look at all the available paths (set of branches). The sum of the probabilities for any set of branches is always 1. Also note that in a tree diagram to find a probability of an outcome we multiply along the branches and add vertically. The probability of three heads is: P (H H H) = 1/2 × 1/2 × 1/2 = 1/8 P (2 Heads and a Tail) = P (H H T) + P (H T H) + P (T H H)

= 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 + 1/2 × 1/2 × 1/2 = 1/8 + 1/8 + 1/8 = 3/8 Example:   

Bag A contains three red marbles and four blue marbles. Bag B contains five red marbles and three blue marbles. A marble is taken from each bag in turn. Find the missing probabilities for the tree diagram:

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Answer

Example:

What is the probability of getting a blue bead followed by a red? Answer Multiply the probabilities together:

  

Example:

P (blue and red) = 4/7 × 5/8 = 20/56 = 5/1 What is the probability of getting a bead of each colour? Answer P (blue and red or red and blue) = P (blue and red) + P (red and blue)

  

= 4/7 × 5/8 + 3/7 × 3/8 = 20/56 + 9/56 = 29/56

6.1

NAME:______________________________ GRADE/SECTION:_____________________

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SCORE:_____________ DATE:_______________

PRACTICE SET 6.1 1. Here is a tree diagram for the toss of a coin: There are two "branches" (Heads and Tails)  

The probability of each branch is written on the branch The outcome is written at the end of the branch

2. You are off to soccer, and love being the Goalkeeper, but that depends who is the Coach today: with Coach Sam the probability of being Goalkeeper is 0.5 with Coach Alex the probability of being Goalkeeper is 0.3 Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6). So, what is the probability you will be a Goalkeeper today?

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3. Show the sample space for tossing one penny and rolling one die. (H = heads, T = tails)

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4. A family has three children. How many outcomes are in the sample space that indicates the sex of the children? Assume that the probability of male (M) and the probability of female (F) are each 1/2.

5. A coin and a die are thrown at random. Find the probability of: a) getting a head and an even number b) getting a head or tail and an odd number

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6. Clare tossed a coin three times. a) Draw a tree diagram to show all the possible outcomes. b) Find the probability of getting: (i) Three tails. (ii) Exactly two heads. (iii) At least two heads.

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7. A spinner is labeled with three colors: Red, Green and Blue. Marcus spun the spinner once and tossed a coin once. a) Draw a tree diagram to list all the possible outcomes. b) Calculate the probability of getting blue on the spinner and head on the coin. c) Calculate the probability of red or green on the spinner and tail on the coin.

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8. Julia spins 2 spinners; one of which is labeled 1, 2 and 3, and the other is labeled 4, 5 and 6.

a) Draw a tree diagram for the experiment. b) What is the probability that the spinners stop at “3” and “4”? c) Find the probability that the spinners do not stop at “3” and “4”. d) What is the probability that the first spinner does not stop at “1”?

6.1

9.

count occurrences of an event

Box A contains 3 cards numbered 1, 2 and 3. Box B contains 2 cards numbered 1 and 2. One card is removed at random from each box. a) Draw a tree diagram to list all the possible outcomes. b) Find the probability that: (i) the sum of the numbers is 4 (ii) the sum of the two numbers is even. (iii) the product of the two numbers is at least 5. (iv) the sum is equal to the product.

185

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10.

A bag contains 4 cards numbered 2, 4, 6, 9. A second bag contains 3 cards numbered 2, 3, 6. One card is drawn at random from each bag. a) Draw a tree diagram for the experiment. b) With the help of the tree diagram, calculate the probability that the two numbers obtained: (i) have different values. (ii) are both even. (iii) are both prime. (iv) have a sum greater than 5. (v) have a product greater than 16.

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11.

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The letters ABCD, can be put into a different order: DCBA or BADC. How many different combinations of the letters ABCD can you make? To answer this question, obviously, you have to make a list.

the game of dominoes is played with black and white tiles. Each tile is divided into two halves and on each half a number from 0-6 is represented in the form of dots. Each tile contains a pair of numbers and each pair appears only once in a complete set. How many tiles are there in a complete set of tiles? The solution to this problem can take the form of a tree diagram

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Melike and Funda are good friends. When Melike got the flu, her doctor wrote a prescription for twenty 30mg pills, and told her to take two a day, one after breakfast and one after dinner. Three days later, Funda also got the flu. Her doctor prescribed thirty 20mg pills, and told her to take a pill every four hours between 9:00 and 21:00. Whose medicine was finished first?

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Find the three smallest two-digit numbers that have an odd number of factors. Number 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Prime Factors

Factors

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 6.1 1. Here is a tree diagram for the toss of a coin: There are two "branches" (Heads and Tails)  

The probability of each branch is written on the branch The outcome is written at the end of the branch

Now we can see such things as:    

The probability of "Head, Head" is 0.5×0.5 = 0.25 All probabilities add to 1.0 (which is always a good check) The probability of getting at least one Head from two tosses is 0.25+0.25+0.25 = 0.75 ... and more

2. You are off to soccer, and love being the Goalkeeper, but that depends who is the Coach today: with Coach Sam the probability of being Goalkeeper is 0.5 with Coach Alex the probability of being Goalkeeper is 0.3 Sam is Coach more often ... about 6 out of every 10 games (a probability of 0.6). So, what is the probability you will be a Goalkeeper today?

An 0.4 chance of Alex as Coach, followed by an 0.3 chance gives 0.12. Now we add the column: 0.3 + 0.12 = 0.42 probability of being a Goalkeeper today (That is a 42% chance) Check One final step: complete the calculations and make sure they add to 1:

0.3

+ 0.3 + 0.12 + 0.28 = 1

3. Show the sample space for tossing one penny and rolling one die. (H = heads, T = tails) By following the different paths in the tree diagram, we can arrive at the sample space. Sample space: { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 } The probability of each of these outcomes is 1/2 • 1/6 = 1/12

[The Counting Principle could also verify that this answer yields the correct number of outcomes: 2 • 6 = 12 outcomes.]

4. A family has three children. How many outcomes are in the sample space that indicates the sex of the children? Assume that the probability of male (M) and the probability of female (F) are each 1/2. Sample space: There are 8 outcomes in the sample space. { MMM MMF The probability of each MFM outcome is MFF 1/2 • 1/2 • 1/2 = 1/8. FMM FMF FFM FFF }

[Again, the Counting Principle could also verify that this result yields the correct number of outcomes: 2 • 2 • 2 = 8 outcomes.] 5. A coin and a die are thrown at random. Find the probability of: a) getting a head and an even number b) getting a head or tail and an odd number Solution: We can use a tree diagram to help list all the possible outcomes.

From the diagram, n(S) = 12 a) Let A denote the event of a head and an even number. A = ((H, 2), (H, 4), (H, 6)} and n(A) = 3 b) Let B denote the event a head or tail and an odd number. B = {(H, 1), (H, 3), (H, 5), (T, 1), (T, 3), (T, 5)} 6. Clare tossed a coin three times. a) Draw a tree diagram to show all the possible outcomes. b) Find the probability of getting: (i) Three tails. (ii) Exactly two heads. (iii) At least two heads. Solution: a) A tree diagram of all possible outcomes.

b) The probability of getting: (i) Three tails. Let S be the sample space and A be the event of getting 3 tails. n(S) = 8; n(A) = 1 P(A) = ii) Exactly two heads. Let B be the event of getting exactly 2 heads. n(B) = 3 P(B) = iii) At least two heads. Let C be the event of getting at least two heads. n(C) = 4 P(C) =

7. A spinner is labeled with three colors: Red, Green and Blue. Marcus spun the spinner once and tossed a coin once. a) Draw a tree diagram to list all the possible outcomes. b) Calculate the probability of getting blue on the spinner and head on the coin. c) Calculate the probability of red or green on the spinner and tail on the coin. Solution: a) A tree diagram of all possible outcomes.

b) The probability of getting blue on the spinner and head on the coin. Let S be the sample space and A be the event of getting blue and head n(S) = 6 ; n(A) = 1 P(A) = c) The probability of red or green on the spinner and tail on the coin. Let B be the event of getting red or green and tail n(B) = 2 P(A) =

8. Julia spins 2 spinners; one of which is labeled 1, 2 and 3, and the other is labeled 4, 5 and 6.

a) Draw a tree diagram for the experiment. b) What is the probability that the spinners stop at “3” and “4”? c) Find the probability that the spinners do not stop at “3” and “4”. d) What is the probability that the first spinner does not stop at “1”? Solution: a) Tree diagram for the experiment.

b) The probability that the spinners stop at “3” and “4” n(S ) = 9 Probability that the spinners stop at (3,4) = c) The probability that the spinners do not stop at “3” and “4” Probability that the spinners do not stop at (3,4) = d) The probability that the first spinner does not stop at “1” Probability that the first spinner stop at “1” = Probability that the first spinner does not stop at “1” =

9. Box A contains 3 cards numbered 1, 2 and 3. Box B contains 2 cards

numbered 1 and 2. One card is removed at random from each box. a) Draw a tree diagram to list all the possible outcomes. b) Find the probability that: (i) the sum of the numbers is 4 (ii) the sum of the two numbers is even. (iii) the product of the two numbers is at least 5. (iv) the sum is equal to the product. Solution: a) A tree diagram of all possible outcomes.

b) The probability that: (i) the sum of the numbers is 4. Let S be the sample space and A be the event that the sum is 4. n(S) = 6; n(A) = 2 P(A) = (ii) the sum of the two numbers is even. Let B be the event that the sum is even. n(B) = 3 P(B) = (iii) the product of the two numbers is at least 5. Let C be the event that the product of the two numbers is at least 5. n(C) = 1 P(C) = (iv) the sum is equal to the product. Let D be the event that the sum of the two numbers is equal to the product. n(D) = 1 P(D) =

10. A bag contains 4 cards numbered 2, 4, 6, 9. A second bag contains 3 cards

numbered 2, 3, 6. One card is drawn at random from each bag. a) Draw a tree diagram for the experiment. b) With the help of the tree diagram, calculate the probability that the two numbers obtained: (i) have different values. (ii) are both even. (iii) are both prime. (iv) have a sum greater than 5. (v) have a product greater than 16. Solution: a) A tree diagram of all possible outcomes.

b) The probability that the two numbers obtained: (i) have different values. Let S be the sample space and A be the event that the two values are different n(S) = 12 ; n(A) = 10 P(A) = (ii) are both even. Let B be the event that both values are even. n(B) = 6 P(B) = (iii) are both prime. Let C be the event that both values are prime. n(C) = 2 P(C) = (iv) have a sum greater than 5. Let D be the event that the sum of both values is greater than 5. n(D) = 10 P(D) = (v) have a product greater than 16. Let E be the event that the product of both values is greater than 16. n(E) = 6 P(E) =

11.

The letters ABCD, can be put into a different order: DCBA or BADC. How many different combinations of the letters ABCD can you make? To answer this question, obviously, you have to make a list. Teach your students to make a SYSTEMATIC list. For example: ABCD BACD CABD DABC ABDC BADC CADB DACB ACBD BCAD CBAD DBAC ACDB BCDA CBDA DBCA ADBC BDAC CDAB DCAB ADCB BDCA CDBA DCBA By making a SYSTEMATIC list, students will see every possible combination. (Later, perhaps, they will learn that the number of permutations of size 4 taken from a set of 4 can be represented by the formula 4 * 3 * 2 * 1 = 24).

12.

the game of dominoes is played with black and white tiles. Each tile is divided into two halves and on each half a number from 0-6 is represented in the form of dots. Each tile contains a pair of numbers and each pair appears only once in a complete set. How many tiles are there in a complete set of tiles? The solution to this problem can take the form of a tree diagram

13.

Melike and Funda are good friends. When Melike got the flu, her doctor wrote a prescription for twenty 30mg pills, and told her to take two a day, one after breakfast and one after dinner. Three days later, Funda also got the flu. Her doctor prescribed thirty 20mg pills, and told her to take a pill every four hours between 9:00 and 21:00. Whose medicine was finished first?

14.

Find the three smallest two-digit numbers that have an odd number of factors.

Number 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Prime Factors 11 2, 3 13 2, 7 3, 5 2 17 2, 3, 19 2, 5 3, 7 2, 11 23 2, 3 5 2, 13 3 2, 7 29 3, 5 31 2 3, 11 2, 17 5, 7 2, 3 37 2, 19

Factors 1, 11 1,2,3,4,6,12 1, 13 1, 2, 7, 14 1, 3, 5, 15 1, 2, 4, 4, 8, 16 1, 17 1, 2, 3, 6, 9, 18 1, 19 1, 2, 4, 5, 10, 20 1, 3, 7, 21 1, 2, 11, 22 1, 23 1, 2, 3, 4, 6, 8, 12, 24 1, 5, 5, 25 1, 2, 13, 26 1, 3, 9, 27 1, 2, 4, 7, 14, 28 1, 29 1, 3, 5, 6, 10, 30 1, 31 1, 2, 4, 8, 16, 32 1, 3, 11, 33 1, 2, 17, 34 1, 5, 7, 35 1, 2, 3, 4, 9, 12, 18, 36 1, 37 1, 2, 19, 38

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6.2

FUNDAMENTAL COUNTING PRINCIPLE

The Fundamental Counting Principle: If there are a ways for one activity to occur, and b ways for a second activity to occur, then there are a • b ways for both to occur.

Examples: roll a die and flip a coin There are 6 ways to roll a die and 2 ways to flip a coin. There are 6 • 2 = 12 ways to roll a die and flip a coin. Example: draw two cards from a standard deck of 52 cards without replacing the cards There are 52 ways to draw the first card. There are 51 ways to draw the second card. There are 52 • 51 = 2,652 ways to draw the two cards. Example: a coin is tossed five times There are 2 ways to flip each coin. There are 2 • 2 • 2 • 2 •2 = 32 arrangements of heads Example: a die is rolled four times There are 6 ways to roll each die. There are 6 • 6 • 6 • 6 = 1,296 possible outcomes. Example: The sample space for the experiments of flipping a coin and rolling a die are S = { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}. Sure enough, there are twelve possible ways. The fundamental counting principle allows us to figure out that there are twelve ways without having to list them all out.

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Example: you have 3 shirts and 4 pants. That means 3×4=12 different outfits. Example: There are 6 flavors of ice-cream, and 3 different cones. That means 6×3=18 different single-scoop ice-creams you could order. It also works when you have more than 2 choices: Example: You are buying a new car. There are 2 body styles: sedan or hatchback There are 5 colors available: Black red white blue sky blue There are 3 models: GL (standard model), SS (sports model with bigger engine) SL (luxury model with leather seats) You can count the choices, or just do the simple calculation: Total Choices = 2 × 5 × 3 = 30 factorial notation: the notation, n!, used to represent the product of the first n natural numbers. n! is read as “n factorial.” Example:

n! = n × (n - 1) × (n - 2) × (n - 3) × ... × 3 × 2 × 1 Note: By definition, 0! = 1. Another thing you must remember is that n! is only defined if n is a whole number. This means numbers like 1.5! and (-2)! are undefined.

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NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 6.2 1. A pair of dice is rolled once. How many possible outcomes are there? What is the probability of rolling doubles?

2. How many ways can six different books be positioned on a book shelf?

3. How many ways can the letters ABCDEFG be arranged?

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4. On an English test, a student must write two essays. For the first essay, the student must select from topics A, B, and C. For the second essay, the student must select from topics 1, 2, 3, and 4. How many different ways can the student select the two essay topics?

5. How many three-letter combinations can be made from A, D and L.

6. How many different license plates can be generated if the first 4 characters have to be letters and the last 3 characters have to be numbers?

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7. Determine the number of distinguishable arrangements for MISSISSIPPI .

8. How many different numbers can be formed by using all of the following digits? 2, 2, 4, 4, and 5

9. Marion has to walk five blocks east and three blocks north. How many routes can she take?

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fundamental counting principle

10.

How many different two-digit numbers can be made from the first 5 natural numbers if repetition is allowed.

11.

How many different three letter combinations can be made using the vowels a, e, i, and o if repetition is allowed.

12.

Evaluate the expression using your understanding of factorial.

13.

Evaluate the expression using your understanding of factorial.

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14.

Evaluate the expression using your understanding of factorial.

15.

Loreta has 3 different math books and 4 different geography books. Determine the number of ways all 7 books can be arranged on the shelf if  there are no restrictions

 the math books are on the left and the geography books are on the right .

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fundamental counting principle

 the books in the same subject must be kept together

16.

A father, mother, 2 boys, and 3 girls are asked to line up for a photograph. Determine the number of ways they can line up if a. there are no restrictions

b. the parents stand together

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c. the parents do not stand together

d. all the females stand together

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17.

Evaluate 10P3 using pencil and paper and a graphing calculator.

18.

Solve 2(nP2) = 60 for n.

19.

Solve nP2 = 72 for n.

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20.

Solve nP3 = 42n for n.

21.

How many possible letter arrangements can be formed using all of the letters of COUPLES if the consonants and vowels must alternate?

6.2

fundamental counting principle

22.

How many even, 4-digit numbers can be formed using 2, 3, 4, and 5?

23.

How many different ways can 4 math books and 5 other books be arranged on a shelf if the math books are kept together?

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24.

How many arrangements are possible of all of the letters of CANTER if the vowels cannot be adjacent?

25.

Aaron has 3 math books, 4 science books, and 5 history books. How many different ways can these books be arranged on a shelf if books of the same subject must be kept together?

26.

If a committee has eight members. a. How many ways can the committee members be seated in a row?

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b. How many ways can the committee select a president, vicepresident and secretary.

27.

Determine the number of distinguishable arrangements for each of the following words. SASKATOON

28.

How many different numbers can be formed by using all of the following digits? 2, 2, 4, 4, and 5

29.

A tray contains 4 different cookies. How many ways can you select and arrange 3 cookies from the tray?

NAME:______________________________ GRADE /SECTION:_____________________

SCORE:_____________ DATE:_______________

PRACTICE SET 6.2 1. A pair of dice is rolled once. How many possible outcomes are there? What is the probability of rolling doubles? By fundamental counting principle. 6 sides on die 1 x 6 sides on die 2 = 36 The probability of rolling doubles: There are 6 sets of doubles(1,1: 2,2: 4.4: 5,5: 6,6) =

6 chances of rolling doubles 36 total outcomes

The probability of rolling doubles is

or .167 or 16.7% chance.

2. How many ways can six different books be positioned on a book shelf? Solution 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 Six different books can be positioned 720 ways on a book shelf. 3. How many ways can the letters ABCDEFG be arranged? Solution 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040 The letters ABCDEFG can be arranged 5040 ways.

4. On an English test, a student must write two essays. For the first essay, the student must select from topics A, B, and C. For the second essay, the student must select from topics 1, 2, 3, and 4. How many different ways can the student select the two essay topics? The task of choosing essay topics has two stages: choosing an essay from topics A, B, and C and choosing an essay from topics 1, 2, 3, and 4. The total number of ways of selecting the two essays is the product of the choices available at each of the two stages. Number of ways = 3 × 4 = 12 The student can select the two essays 12 ways. 5. How many three-letter combinations can be made from A, D and L. The task of choosing essay topics has three stages: choosing a first, second and third letter. The total number of ways of selecting the three-letter arrangements is the product of the choices available at each of the three stages. Number of ways = 3 × 3 × 3 = 27 There are 27 three-letter arrangements. 6. How many different license plates can be generated if the first 4 characters have to be letters and the last 3 characters have to be numbers? The first stage, choosing the first letter, has 26 choices. The second, third, and fourth stages also have 26 possible choices. The fifth stage, choosing the first number, has 10 choices. The sixth and seventh stages also have 10 possible choices.

7. Determine the number of distinguishable arrangements for MISSISSIPPI . Solution In the word MISSISSIPPI, there are four Is, four Ss, and two Ps. There are 11 letters altogether.

8. How many different numbers can be formed by using all of the following digits? 2, 2, 4, 4, and 5 Solution There are two 2s and two 4s. There are five digits altogether.

9. Marion has to walk five blocks east and three blocks north. How many routes can she take? Solution Marion has to walk five blocks east and three blocks north. The number of pathways is the same as the number of ways of arranging five Es and 3 Ns. There are a total of eight letters.

Marion can take 56 different routes.

10.

How many different two-digit numbers can be made from the first 5 natural numbers if repetition is allowed. Solution 5 × 5 = 25 25 two-digit numbers can be made using the first 5 natural numbers if repetition is allowed.

11.

How many different three letter combinations can be made using the vowels a, e, i, and o if repetition is allowed. Solution 4 × 4 × 4 = 64 64 three letter numbers can be made using the vowels a, e, i, and o if repetition is allowed

12.

Evaluate the expression using your understanding of factorial.

13.

Evaluate the expression using your understanding of factorial.

14.

Evaluate the expression using your understanding of factorial.

15.

Loreta has 3 different math books and 4 different geography books. Determine the number of ways all 7 books can be arranged on the shelf if  there are no restrictions Use the formula for nPn. Because all 7 books are being arranged, n = 7.

All seven books can be arranged 5040 ways. Use the formula for nPn.  the math books are on the left and the geography books are on the right . There are two stages in this problem. You must arrange the math books first, then arrange the geography books on the right. The number of ways of arranging the math books is 3!. The number of ways of arranging the geography books is 4!. Using the Fundamental Counting Principle,

There are 144 ways to arrange the math books followed by the geography books

 the books in the same subject must be kept together This problem is the same as the previous, except that either the math books or the geography books can be on the left. Therefore, you must multiply the answer to question b. by 2!.

There are 288 ways to arrange the seven books so the subjects are kept together. 16.

A father, mother, 2 boys, and 3 girls are asked to line up for a photograph. Determine the number of ways they can line up if a. there are no restrictions Use the formula for nPn. Since all 7 people must line up, n = 7.

There are 5040 ways the family can line up for a photograph. b. the parents stand together If the parents stand together, they can be considered a unit, reducing the value of n from 7 to 6. If n = 6, there are 6! arrangements. However, in each of these arrangements, the parents can switch places and still stand side by side.

There are 1440 ways the family can line up for the photograph if the parents stand together.

c. the parents do not stand together Approach this question indirectly. Recall from the answer to question a. that there were 5040 arrangements possible without any restrictions. From the answer to question b., there were 1440 arrangements if the parents stood together. Therefore, if the parents don’t stand together, the number of arrangements is

d. all the females stand together This question is similar to question b. Consider the females as a unit. There are 3 other people in addition to the females. Therefore, n = 4. When, n = 4, there are 4! possible arrangements. But, even though the 4 females are standing together, they can be arranged among themselves in 4! ways.

There are 576 different ways the family can line up if the females stand together.

17.

Evaluate 10P3 using pencil and paper and a graphing calculator.

18.

Solve 2(nP2) = 60 for n.

However, for nP2 to be defined, the value of n must be a whole number and n 2.Therefore, n = 6. 19.

Solve nP2 = 72 for n.

20.

Solve nP3 = 42n for n.

21.

How many possible letter arrangements can be formed using all of the letters of COUPLES if the consonants and vowels must alternate? There are four consonants: C, P, L, and S. There are three vowels: O, U, and E. If the consonants, c, and vowels, v, must alternate, each arrangement is of the form cvcvcvc. Arrange the consonants first. 4v3v2v1 Next, arrange the vowels. 4332211 There are 144 possible letter arrangements that can be formed.

22.

How many even, 4-digit numbers can be formed using 2, 3, 4, and 5? If numbers formed are even, they must end in a 2 or a 4. Therefore, there are only 2 choices for the last digit. 2 The remaining three digits can be positioned in any of 3! arrangements. 3212 Therefore, the number of even, 4-digit numbers that can be formed is

23.

How many different ways can 4 math books and 5 other books be arranged on a shelf if the math books are kept together?

If the math books are kept together, treat them as a single book. There are the group of math books and five other books to arrange. These six elements can be arranged in 6! ways. Next, arrange the four math books among themselves. They can be arranged in 4! ways. Using the Fundamental Counting Principle, the total number of arrangements possible is

24.

How many arrangements are possible of all of the letters of CANTER if the vowels cannot be adjacent? Solution Use an indirect approach. First, determine the number of arrangements without restriction. Because there are six letters, 6! = 720 arrangements are possible. Next, determine the number of arrangements with A and E together. Treat them as a single letter, reducing the number of letters from six to five. Therefore, there are 5! = 120 arrangements of five letters. Because A and E can be interchanged, the total number of arrangements with A and E together is 5!2! = 240. Therefore, the number of arrangements possible if the vowels A and E are not adjacent is

25.

Aaron has 3 math books, 4 science books, and 5 history books. How many different ways can these books be arranged on a shelf if books of the same subject must be kept together? Solution Suppose the three math books were first, the four science books second, and the five history books in the last position. There would be 3!4!5! arrangements possible. But the three subjects can be arranged in 3! ways. Therefore, the number of arrangements possible is 3!4!5! × 3! = 103 680.

26.

If a committee has eight members. a. How many ways can the committee members be seated in a row? There are 40 320 ways to arrange the committee members.

b. How many ways can the committee select a president, vicepresident and secretary. There are 336 ways the club can choose a president, vice president, and treasurer. 27.

Determine the number of distinguishable arrangements for each of the following words. a. SASKATOON In the word SASKATOON, there are two Ss, two As, and two Os. There are nine letters altogether.

28.

How many different numbers can be formed by using all of the following digits? 2, 2, 4, 4, and 5 There are two 2s and two 4s. There are five digits altogether.

29.

A tray contains 4 different cookies. How many ways can you select and arrange 3 cookies from the tray? Order is important. Therefore, this is a permutation of 4 objects taken 3 at a time.

There are 24 ways of selecting and arranging three cookies from the tray

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6.3

PERMUTATIONS AND COMBINATIONS

If the order doesn't matter, it is a Combination If the order does matter it is a Permutation. In other words: A Permutation is an ordered Combination. There are basically two types of permutation: 1. Repetition is Allowed: such as the lock above. It could be "555". These are the easiest to calculate. When we have n things to choose from ... we have n choices each time. When choosing r of them, the permutations are: n × n × ... (r times) (In other words, there are n possibilities for the first choice, then there are n possibilities for the second choice, and so on, multiplying each time.) Which is easier to write down using an exponent of r: n × n × ... (r times) = nr Example: in the lock below, there are 10 numbers to choose from (0,1,...9) and we choose 3 of them:

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205

10 × 10 × ... (3 times) = 103 = 1,000 permutations So, the formula is simply: nr where n is the number of things to choose from, and we choose r of them (Repetition allowed, order matters) 2. No Repetition: for example the first three people in a running race. You can't be first and second. In this case, we have to reduce the number of available choices each time.

Example: what order could 16 pool balls be in? After choosing, say, number "14" we can't choose it again. So, our first choice has 16 possibilities, and our next choice has 15 possibilities, then 14, 13, etc. And the total permutations are: 16 × 15 × 14 × 13 × ... = 20,922,789,888,000 But maybe we don't want to choose them all, just 3 of them, so that is only: 16 × 15 × 14 = 3,360 In other words, there are 3,360 different ways that 3 pool balls could be arranged out of 16 balls. Without repetition our choices get reduced each time.

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The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples:   

4! = 4 × 3 × 2 × 1 = 24 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040 1! = 1

Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets us 1, but it helps simplify a lot of equations. So, when we want to select all of the billiard balls the permutations are: 16! = 20,922,789,888,000 But when we want to select just 3 we don't want to multiply after 14. How do we do that? There is a neat trick ... we divide by 13! ... 16 × 15 × 14 × 13 × 12 ... = 16 × 15 × 14 = 3,360 13 × 12 ... 16! / 13! = 16 × 15 × 14 The formula is written:

where n is the number of things to choose from, and we choose r of them (No repetition, order matters)

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207

Example: Our "order of 3 out of 16 pool balls example" is: 16! 16! 20,922,789,888,000 = = = 3,360 (16-3)! 13! 6,227,020,800 (which is just the same as: 16 × 15 × 14 = 3,360) Example: How many ways can first and second place be awarded to 10 people? 10! 10! 3,628,800 = = = 90 (10-2)! 8! 40,320 (which is just the same as: 10 × 9 = 90)

Notation Instead of writing the whole formula, people use different notations such as these:

Example: P(10,2) = 90

Two types of combinations (remember the order does not matter now):

1. Repetition is Allowed: such as coins in your pocket (5,5,5,10,10)

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2. No Repetition: such as lottery numbers (2,14,15,27,30,33) This is how lotteries work. The numbers are drawn one at a time, and if we have the lucky numbers (no matter what order) we win! The easiest way to explain it is to:  

assume that the order does matter (ie permutations), then alter it so the order does not matter. Going back to our pool ball example, let's say we just want to know which 3 pool balls are chosen, not the order. We already know that 3 out of 16 gave us 3,360 permutations. Example: let us say balls 1, 2 and 3 are chosen. These are the possibilities: Order does matter

Order doesn't matter

123 132 213 231 312 321

123

So, the permutations will have 6 times as many possibilities. In fact there is an easy way to work out how many ways "1 2 3" could be placed in order, and we have already talked about it. The answer is:

3! = 3 × 2 × 1 = 6

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209

where n is the number of things to choose from, and we choose r of them (No repetition, order doesn't matter) Notation As well as the "big parentheses", people also use these notations:

Example: So, our pool ball example (now without order) is: 16! 16! 20,922,789,888,000 = = = 560 3!(16-3)! 3!×13! 6×6,227,020,800 Or we could do it this way: 16×15×14 3360 = = 560 3×2×1 6 It is interesting to also note how this formula is nice and symmetrical:

In other words choosing 3 balls out of 16, or choosing 13 balls out of 16 have the same number of combinations. 16! 16! 16! = = = 560 3!(16-3)! 13!(16-13)! 3!×13!

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NAME:______________________________ GRADE /SECTION:_____________________

SCORE:_____________ DATE:_______________

PRACTICE SET 6.3 1. Determining the number of ways a selection from a set can be made when the order does matter. A tray contains 4 different cookies. How many ways can you select 3 cookies from the tray if order doesn’t matter?

2. Determining the number of ways a selection from a set can be made when the order does matter. A tray contains 4 different cookies. How are the answers to questions a. and b. related?

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211

3. How many different 3-member committees can be selected from a class of 20 students?

4. A cookie jar contains 5 cookies. How many ways can a child choose all 5 cookies if order doesn’t matter?

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5. A cookie jar contains 5 cookies. A child is offered a cookie, but refuses. How many ways can this be done?

6. A Pure Mathematics 30 class has 10 boys and 12 girls. The teacher wants to form a committee of 3 students to plan the year-end picnic. Determine the number of committees possible if a. There are no restrictions

6.3

b. There are no boys on the committee

permutations and combinations

213

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c. There must be at least one boy on the committee.

7. How many diagonals are there in a regular octagon?

6.3

permutations and combinations

8. How many diagonals are in a regular polygon with 20 sides?

9. How many triangles can be formed by joining the vertices of a regular polygon with 20 sides?

10.

If 3(nC2) = 30 solve for n. Remember that for nCr, n r.

215

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11.

If nC3 = 3(nP2) solve for n. Remember that for nCr, n r.

12.

All the possible two-number permutations and combinations from {1, 2, 3, 4, 5}.

6.3

13.

permutations and combinations

All the possible three-number permutations and combinations from {1, 2, 3, 4, 5}.

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14.

There are six colors: red, white, blue, green, yellow, and black. Find the number of two-color combinations.

15.

There are six colors: red, white, blue, green, yellow, and black. Find the number of three-color combinations. There are six colors: red, white, blue, green, yellow, and black.

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219

16.

How many 13 card hands can be made from a deck of 52 cards?

17.

Bridge hands have 13 card hands. How many bridge hands can be made with 4 aces?

18.

How many bridge hands can be made with no aces?

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19.

How many 5 card hands can be made from a deck of 52 cards?

20.

How many 5 card hands with only red cards can be made from a deck of 52 cards?

6.3

permutations and combinations

21.

How many 5 card hands can be made of the same suit?

22.

How many distinct 5 card hands can be made from a deck of 52 cards with: a. 3 hearts and 2 spades?

221

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b. only 1 queen and 3 kings

c. 3 red cards and 2 black cards?

23.

How many distinct 5 card hands can be made from a deck of 52 cards with: a. no fives?

6.3

b. 4 aces?

c. only 3 face cards?

permutations and combinations

223

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 6.3 1. Determining the number of ways a selection from a set can be made when the order does matter. A tray contains 4 different cookies. How many ways can you select 3 cookies from the tray if order doesn’t matter? Solution Let the set {A, B, C, D} represent the cookies on the tray. The three-cookie combinations are {A, B, C}, {A, B, D}, {A, C, D}, and {B, C, D}.There are 4 ways of selecting three cookies from the tray if order does not matter. 2. Determining the number of ways a selection from a set can be made when the order does matter. A tray contains 4 different cookies. How are the answers to questions a. and b. related? Solution The symbol used to represent the number of combinations is 4C3, which is read “4 choose 3.” This implies, “From 4, choose 3.” Here, 4C3 = 4. The answers to questions a. and b. were 24 and 4 respectively. Notice that 4P3 = 4C3 × 6 or 4C3 × 3!.There are 6 or 3! times as many permutations as combinations because there are 3! ways of arranging the three cookies once they are chosen. The permutation nPr represents two tasks: selecting r objects from a set of n objects first, then arranging the r objects selected. Using the Fundamental Counting Principle,

3. How many different 3-member committees can be selected from a class of 20 students? Solution Order is not important; so, determine the number of combinations.

There are 1140 different committees that can be selected. 4. A cookie jar contains 5 cookies. How many ways can a child choose all 5 cookies if order doesn’t matter? Solution Because the order doesn’t matter, determine the number of combinations.

If order is not important, the child can choose all five cookies only 1 way!

5. A cookie jar contains 5 cookies. A child is offered a cookie, but refuses. How many ways can this be done? Solution

There is only 1 way the child can reject the cookie. This is done simply by saying, “No.” 6. A Pure Mathematics 30 class has 10 boys and 12 girls. The teacher wants to form a committee of 3 students to plan the year-end picnic. Determine the number of committees possible if a. There are no restrictions Solution

Without restrictions, there are 1540 committees possible.

b. There are no boys on the committee Solution Because there are no boys on the committee, you must select committee members from the 12 girls.

There are 220 possible committees of only girls.

c. There must be at least one boy on the committee Solution Using an Indirect Approach From the answer to question a., you know there are 1540 committees possible when there are no restrictions. Some of these committees will be all boys, others will be all girls, and the rest will be boys and girls. From the answer to question b., you know there are 220 committees that consist of only girls.

There are 1320 possible committees with at least one boy. 7. How many diagonals are there in a regular octagon? Solution A regular octagon has 8 sides and 8 vertices. Choose any two vertices and join them. The line segment formed is either a side or a diagonal. The total number of ways of joining the 8 vertices is 8C2.

The number of diagonals of any regular polygon with n sides can be determined using the following equation: Number of diagonals = nC2 - n

8. How many diagonals are in a regular polygon with 20 sides?

9. How many triangles can be formed by joining the vertices of a regular polygon with 20 sides?

10.

If 3(nC2) = 30 solve for n. Remember that for nCr, n r. Solution

Because nC2 is only defined when n 2, where n is a whole number, n = 5. Remember that for nCr, n r.

11.

If nC3 = 3(nP2) solve for n. Remember that for nCr, n r.

12.

All the possible two-number permutations and combinations from {1, 2, 3, 4, 5}. The number of permutations of two numbers from {1, 2, 3, 4, 5} with no repetitions is

Each of the combinations in this question can be arranged in two ways.

The second list has double the number of pairs compared to the first list. Notice that 2 = 2!

13.

All the possible three-number permutations and combinations from {1, 2, 3, 4, 5}. The number of permutations of two numbers from {1, 2, 3, 4, 5} with no repetitions is 5P3 = 5 × 4 × 3 = 60

There are 3! or 6 times as many permutations as combinations. Order is important with permutations, but not with combinations. Notice that 6 = 3! 5C3 = 5P3 / 3! = 60 / 6 = 10

14.

There are six colours: red, white, blue, green, yellow, and black. Find the number of two-colour combinations. There are six colours: red, white, blue, green, yellow, and black.

There are 15 two-color combinations 15.

There are six colors: red, white, blue, green, yellow, and black. Find the number of three-color combinations. There are six colors: red, white, blue, green, yellow, and black. n = 6 and r = 3

There are 20 three-color combinations.

16.

How many 13 card hands can be made from a deck of 52 cards?

17. Bridge hands have 13 card hands. How many bridge hands can be made with 4 aces?

18. How many bridge hands can be made with no aces?

There are approximately 1.93 × 1011 different hands with no aces.

19.

How many 5 card hands can be made from a deck of 52 cards?

There are 2 598 960 different five-card hands. There are 52 ÷ 2 = 26 red cards in a deck of cards. 20.

How many 5 card hands with only red cards can be made from a deck of 52 cards?

There are 65 780 different five-card hands containing only red cards

21.

How many 5 card hands can be made of the same suit? Solution There are 4 suits in a standard deck of cards; therefore, there are 52 ÷ 4 = 13 cards in each suit. You need to determine the number of different hands containing all cards of the same suit among the 4 different suits.

There are 5148 different five-card hands containing all cards of the same suit. 22.

How many distinct 5 card hands can be made from a deck of 52 cards with: a. 3 hearts and 2 spades? Solution We must use the fundamental counting principle and combinations: 13C3 × 13C2 = 286 × 78 = 22308 There are 22 308 different five-card hands.

b. only 1 queen and 3 kings Solution We must use the fundamental counting principle and combinations: 4C1 × 4C3 × 48C1= 4 × 4 × 48 = 768 There are 768 different five-card hands containing 1 queen and 3 kings c. 3 red cards and 2 black cards? Solution There are 52 ÷ 2 = 26 red cards in a deck of cards. There are 52 ÷ 2 = 26 black cards in a deck of cards. We must use the fundamental counting principle and combinations: 26C3 × 26C2 = × = There are different five-card hands containing 3 red and 2 black cards. 23.

How many distinct 5 card hands can be made from a deck of 52 cards with: a. no fives?

We must use the fundamental counting principle and combinations: 4C0 × 48C5 = 1 × 1712304 = 1712304 There 1 712 304 are different five-card hands with no fives

b. 4 aces? Solution We must use the fundamental counting principle and combinations: 4 C4

× 48C1 = 1 × 48 = 48

There are 48 different five-card hands with 4 aces.

c. only 3 face cards?

Solution We must use the fundamental counting principle and combinations: There are 4(3) = 12 face cards in a deck of cards. 12C3

× 40C2 = 220 × 780 = 171600

There are 171 600 different five-card hands containing only 3 face card

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Chapter 7 PROBABILITY OF COMPOUND EVENTS 7.1 7.2 7.3 7.4

CARDINALITY OF A UNION OF TWO SETS UNION AND INTERSECTION OF EVENTS PROBABILITY OF A UNION OF TWO SETS PROBABILITIES OF A UNION AND INTERSECTION OF EVENTS

Specific Objectives: at the end of the chapter the students should be able to;  Find the cardinality of a union of two sets A and B,  Recognize events, union events and intersection of events,  Define the probability of a union of two events using the definition of the probability of an event e,  Find the probability P(A B),  Define events that are independent,  Find the probability P(A . B)  Solve problems involving probabilities of a union and intersection of events.

Pascal, Blaise [paskal](1623--62) Mathematician, physicist, theologian, and man-of-letters, born in Clermont-Ferrand, C France. He invented a calculating machine (1647), and later the barometer, the hydraulic press, and the syringe. Until 1654 he spent his time between mathematics and the social round in Paris, but a mystical experience that year led him to join his sister, who was a member of the Jansenist convent at Port-Royal, where he defended Jansenism against the Jesuits in Lettres provinciales (1656--7). Fragments jotted down for a case book of Christian truth were discovered after his death and published as the Pensées (1669, Thoughts).

7.1

7.1

cardinality of a union of two sets

225

CARDINALITY OF A UNION OF TWO SETS

The cardinality of a finite set is defined as simply the number of elements in the set. We will say that the cardinality of an infinite set is infinity (written as ∞). However, as the study of infinite sets is more complicated, in discussing cardinality we restrict our attention to finite sets. We use the notation |A| to refer to the cardinality of a set A; that is, |A| = number of elements in A . First suppose that two sets A and B are disjoint combining the elements of A and B into one set, and because A and B have no number of elements in A and the number of elements in B; thus , if A and B are disjoint. However, if A and B are not disjoint then the situation is not as simple, as illustrated in the next example. Example: The art club at Gusa Regional Science High School has 10 members, the music club has 12 members, and 5 students belong to both of these clubs. How many students will be invited to the joint art club - music club Christmas party? If the art and music clubs were disjoint, then the answer would be 10 + 12 = 22. However, the 5 students who are members of both clubs are counted twice in this addition; thus we must subtract 5 from the total, to arrive at 10 + 12 − 5 = 17 as the number of students in the union of the art and music clubs. The preceding example demonstrates the general formula for the cardinality of the union of two sets A and B (when A and B might intersect), |A

B| = |A| + |B| − |A ∩ B|.

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Any set S is disjoint from its complement ~S, while the union of S and ~S is the universal set U. Consequently, if we add the number of elements in S to the number not in S, we get the total number of elements in U - that is, |S| + |~S| = |U| . Example: In Althea’s refrigerator there are 9 eggs, of which 3 are rotten. How many of the eggs are good? The universe U, consisting of all the eggs, has 9 members, while the subset R of rotten eggs has 3 members. The good eggs make up the complement, ~R. Therefore, |R| + |~R| = |U| , |~R| = |U| − |R| =9−3 =6 There are 6 good eggs. Example: Among the voters at a neighborhood board meeting there are 10 females, 23 Democrats, and 7 female Democrats. How many voters are either female or Democrat? First we work the problem with our formula. Let F be the set of females at the meeting and D the set of Democrats. We are given that |F| = 10, |D| = 23, and |F ∩ D| = 7. The number of voters who are female or Democrat then is |F

D| = |F| + |D| − |F ∩ D| = 10 + 23 − 7 = 26 .

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227

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 7.1 1. Of the 17 meals that Flax-n-Soy can sell, 13 come with flax and 15 come with soy. If he chooses a meal at random, what is the probability that Health man will receive a meal containing both flax and soy?

2. A is the set of all odd numbers less than 10 and B, the set of all prime numbers less than 20. If C is the union of sets A and B, which of the following sets represent the set C?

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3. Find the union between two sets. Let A = {1 orange, 1 pineapple, 1 banana, 1 apple} and B = { 1 spoon, 1 knife, 1 fork}

4. Find the union of A and B

5. Find the union of A and B . A = { x / x is a number bigger than 4 and smaller than 8}, B = { x / x is a positive number smaller than 7}

7.1

cardinality of a union of two sets

6. Find the union of A and B, A = { x / x is a country in Asia}, B = { x / x is a country in Africa}

7. Find the union between A and B.

8. Find A B C by putting all the elements of A, B, and C together , then make a Venn Euler Diagram A = { 1, 2, 4, 6}, B = { a, b, c,} and C = A = {#, %, &, * , $ }

229

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9. Given U = {1, 2, 3, 4, 5, 6, 7, 8, 10} X = {1, 2, 6, 7} and Y = {1, 3, 4, 5, 8} Find X

10.

Y and draw a Venn diagram to illustrate X

Y.

Given U = {1, 2, 3, 4, 5, 6, 7, 8, 10} X = {1, 6, 9} and Y = {1, 3, 5, 6, 8, 9} Find X

Y and draw a Venn diagram to illustrate X

Y.

NAME:______________________________

SCORE:_____________

GRADE /SECTION:_____________________

DATE:_______________

PRACTICE SET 7.2 1. Of the 17 meals that Flax-n-Soy can sell, 13 come with flax and 15 come with soy. If he chooses a meal at random, what is the probability that Health man will receive a meal containing both flax and soy? Apply the principle of inclusion-exclusion to find the number of meals that come with both flax and soy, denoted with x:

Solve for x:

The probability is the ratio between the number of favorable events and the total number of events:

2. A is the set of all odd numbers less than 10 and B, the set of all prime numbers less than 20. If C is the union of sets A and B, which of the following sets represent the set C? Step 1: Set of all odd numbers less than 10 = {1, 3, 5, 7, 9} Step 2: Set of all prime numbers less than 20 = {2, 3, 5, 7, 11, 13, 17, 19} Step 3: Union of these two sets = {1, 3, 5, 7, 9} {2, 3, 5, 7, 11, 13, 17, 19} Step 4: = {1, 2, 3, 5, 7, 9, 11, 13, 17, and 19} Step 5: So, {1, 2, 3, 5, 7, 9, 11, 13, 17, and 19} is the union of the given two sets.

3. Find the union between two sets. Let A = {1 orange, 1 pinapple, 1 banana, 1 apple} and B = { 1 spoon, 1 knife, 1 fork} A

B = {1 orange, 1 pineapple, 1 banana, 1 apple, 1 spoon, 1 knife, 1fork}

4. Find the union of A and B

A = { 1, 2, 4, 6} and B = { 4, a, b, c, d, f} A

B = { 1, 2, 4, 6, 4, a, b, c, d, f} = { 1, 2, 4, 6, a, b, c, d, f }

5. Find the union of A and B .A = { x / x is a number bigger than 4 and smaller than 8} B = { x / x is a positive number smaller than 7} A = { 5, 6, 7} and B = { 1, 2, 3, 4, 5, 6} A È B = { 1, 2, 3, 4, 5, 6, 7} Or A È B = { x / x is a number bigger than 0 and smaller than 8}

6. Find the union of A and B, A = { x / x is a country in Asia}, B = { x / x is a country in Africa} A B = {x / x is a country in Asia and Africa} = { all countries in Asia and Africa} Technically, using the definition, we should have said A country in either Asia or Africa or both countries}

B = {x / x is a

7. Find the union between A and B.

A = {#, %, &, * , $ } B={} A

B = {#, %, &,* , $}

8. Find A B C by putting all the elements of A, B, and C together , then make a Venn Euler Diagram A = { 1, 2, 4, 6}, B = { a, b, c,} and C = A = {#, %, &, * , $ } A B C = { 1, 2, 4, 6, a, b, c,#, %, &, * , $ }

9. Given U = {1, 2, 3, 4, 5, 6, 7, 8, 10} X = {1, 2, 6, 7} and Y = {1, 3, 4, 5, 8} Find X

Y and draw a Venn diagram to illustrate X

Y.

Solution: X

Y = {1, 2, 3, 4, 5, 6, 7, 8} ←1 is written only once.

If X ⊂ Y then X 10.

Y = Y.

Given U = {1, 2, 3, 4, 5, 6, 7, 8, 10} X = {1, 6, 9} and Y = {1, 3, 5, 6, 8, 9} Find X

Y and draw a Venn diagram to illustrate X

Solution: X

Y = {1, 3, 5, 6, 8, 9}

Y.

7.2

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231

UNION AND INTERSECTION OF EVENTS

When two events (call them "A" and "B") are mutually exclusive it is impossible for them to happen together: P(A and B) = 0 "The probability of A and B together equals 0 (impossible)" But the probability of A or B is the sum of the individual probabilities: P(A or B) = P(A) + P(B) "The probability of A or B equals the probability of A plus the probability of B" Example: A Deck of Cards In a Deck of 52 Cards:  

the probability of a King is 1/13, so P(King)=1/13 the probability of an Ace is also 1/13, so P(Ace)=1/13

When we combine those two Events:  

The probability of a card being a King and an Ace is 0 (Impossible) The probability of a card being a King or an Ace is (1/13) + (1/13) = 2/13

Which is written like this: P(King and Ace) = 0 P(King or Ace) = (1/13) + (1/13) = 2/13

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Special Notation Instead of "and" you will often see the symbol ∩ (which is the "Intersection" symbol used in Venn Diagrams)instead of "or" you will often see the symbol (the "Union" symbol)

Example: Scoring Goals If the probability of: scoring no goals (Event "A") is 20% scoring exactly 1 goal (Event "B") is 15% Then: The probability of scoring no goals and 1 goal is 0 (Impossible) The probability of scoring no goals or 1 goal is 20% + 15% = 35% Which is written: P(A ∩ B) = 0 P(A

B) = 20% + 15% = 35%

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233

"Or has more ... than And" Also

is like a cup which holds more than ∩

As a formula this is: P(A or B) = P(A) + P(B) − P(A and B) "The probability of A or B equals the probability of A plus the probability of B minus the probability of A and B" Here is the same formula, but using P(A

and ∩:

B) = P(A) + P(B) − P(A ∩ B) Example: 16 people study French, 21 study Spanish and there are 30 altogether. Work out the probabilities! Let's say b is how many study both languages:  

people studying French Only must be 16-b people studying Spanish Only must be 21-b

And we get:

And we know there are 30 people, so: (16−b) + b + (21−b) = 30 37 − b = 30 b=7 And we can put in the correct numbers:

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So we know all this now:      

P(French) = 16/30 P(Spanish) = 21/30 P(French Only) = 9/30 P(Spanish Only) = 14/30 P(French or Spanish) = 30/30 = 1 P(French and Spanish) = 7/30

Lastly, let's check with our formula: P(A or B) = P(A) + P(B) − P(A and B) Put the values in: 30/30 = 16/30 + 21/30 − 7/30 Probability Let's look at the probabilities of Mutually Exclusive events. But first, a definition: Number of ways it can happen Probability of an event happening = Total number of outcomes Example: there are 4 Kings in a deck of 52 cards. What is the probability of picking a King? Number of ways it can happen: 4 (there are 4 Kings) Total number of outcomes: 52 (there are 52 cards in total) 4 1 So the probability = = 52 13

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235

Rule of Subtraction In a previous lesson, we learned two important properties of probability:  

The probability of an event ranges from 0 to 1. The sum of probabilities of all possible events equals 1.

The rule of subtraction follows directly from these properties. Rule of subtraction the probability that event A will occur is equal to 1 minus the probability that event A will not occur. P(A) = 1 - P(A') Example: The probability that Bill will graduate from college is 0.80. What is the probability that Bill will not graduate from college? Based on the rule of subtraction, the probability that Bill will not graduate is 1.00 - 0.80 or 0.20. Rule of Multiplication The rule of multiplication applies to the situation when we want to know the probability of the intersection of two events; that is, we want to know the probability that two events (Event A and Event B) both occur. Rule of multiplication the probability that Events A and B both occur is equal to the probability that Event A occurs times the probability that Event B occurs, given that A has occurred. P(A ∩ B) = P(A) P(B|A) Example: An urn contains 6 red marbles and 4 black marbles. Two marbles are drawn without replacement from the urn. What is the probability that both of the marbles are black?

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Solution: Let A = the event that the first marble is black; and let B = the event that the second marble is black. We know the following:  

In the beginning, there are 10 marbles in the urn, 4 of which are black. Therefore, P(A) = 4/10. After the first selection, there are 9 marbles in the urn, 3 of which are black. Therefore, P(B|A) = 3/9. Therefore, based on the rule of multiplication: P(A ∩ B) = P(A) P(B|A) P(A ∩ B) = (4/10) (3/9) = 12/90 = 2/15

Rule of Addition The rule of addition applies to the following situation. We have two events, and we want to know the probability that either event occurs. Rule of addition the probability that Event A or Event B occurs is equal to the probability that Event A occurs plus the probability that Event B occurs minus the probability that both Events A and B occur. P(A B) = P(A) + P(B) - P(A ∩ B)) Note: Invoking the fact that P(A ∩ B) = P( A )P( B | A ), the Addition Rule can also be expressed as P(A B) = P(A) + P(B) - P(A)P( B | A ) Example: A student goes to the library. The probability that she checks out (a) a work of fiction is 0.40, (b) a work of non-fiction is 0.30, and (c) both fiction and non-fiction is 0.20. What is the probability that the student checks out a work of fiction, non-fiction, or both? Solution: P(F N) = P(F) + P(N) - P(F ∩ N) P(F N) = 0.40 + 0.30 - 0.20 = 0.50

7.2

union and intersection of events

237

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 7.2 1. An urn contains 6 red marbles and 4 black marbles. Two marbles are drawn with replacement from the urn. What is the probability that both of the marbles are black?

2. A card is drawn randomly from a deck of ordinary playing cards. You win 10 if the card is a spade or an ace. What is the probability that you will win the game?

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3. Of the customers who bought items at a store on a particular day, \(53\) of them bought clocks and \(47\) of them bought watering cans. What is the largest possible number of people that bought either a clock or a watering can on that day?

4. Two events connected with the experiment of rolling a single die are E: “the number rolled is even” and T: “the number rolled is greater than two.” Find the complement of each.

5. Find the probability that at least one heads will appear in five tosses of a fair coin.

7.2

union and intersection of events

239

5. In the experiment of rolling a single die, find the intersection E ∩ T of the events E: “the number rolled is even” and T: “the number rolled is greater than two.”

6. A single die is rolled. a. Suppose the die is fair. Find the probability that the number rolled is both even and greater than two.

b. Suppose the die has been “loaded” so that P(1)=1∕12, P(6)=3∕12, and the remaining four outcomes are equally likely with one another. Now find the probability that the number rolled is both even and greater than two.

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7. In the experiment of rolling a single die, find three choices for an

event A so that the events A and E: “the number rolled is even” are mutually exclusive.

8. In the experiment of rolling a single die, find the union of the events E: “the number rolled is even” and T: “the number rolled is greater than two.”

9. A two-child family is selected at random. Let B denote the event that at least one child is a boy, let D denote the event that the genders of the two children differ, and let M denote the event that the genders of the two children match. Find B D and B

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union and intersection of events

241

10. Two fair dice are thrown. Find the probabilities of the following events: a. both dice show a four

b. at least one die shows a four

11. A tutoring service specializes in preparing adults for high school equivalence tests. Among all the students seeking help from the service, 63% need help in mathematics, 34% need help in English, and 27% need help in both mathematics and English. What is the percentage of students who need help in either mathematics or English?

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12. Volunteers for a disaster relief effort were classified according to both specialty (C: construction, E: education, M: medicine) and language ability (S: speaks a single language fluently, T: speaks two or more languages fluently). The results are shown in the following two-way classification table: Language Ability Specialty S T C 12 1 E 4 3 M 6 2

13. A volunteer is selected at random, meaning that each one has an equal chance of being chosen. Find the probability that: Language Ability Specialty Total S T C 12 1 13 E 4 3 7 M 6 2 8 Total 22 6 28 a. his specialty is medicine and he speaks two or more languages;

NAME:______________________________

SCORE:_____________

GRADE/SECTION:_____________________

DATE:_______________

PRACTICE SET 7.2 1. An urn contains 6 red marbles and 4 black marbles. Two marbles are drawn with replacement from the urn. What is the probability that both of the marbles are black? Solution The correct answer is A. Let A = the event that the first marble is black; and let B = the event that the second marble is black. We know the following: In the beginning, there are 10 marbles in the urn, 4 of which are black. Therefore, P(A) = 4/10. After the first selection, we replace the selected marble; so there are still 10 marbles in the urn, 4 of which are black. Therefore, P(B|A) = 4/10. Therefore, based on the rule of multiplication: P(A ∩ B) = P(A) P(B|A) P(A ∩ B) = (4/10)*(4/10) = 16/100 = 0.16 2. A card is drawn randomly from a deck of ordinary playing cards. You win 10 if the card is a spade or an ace. What is the probability that you will win the game? Solution The correct answer is C. Let S = the event that the card is a spade; and let A = the event that the card is an ace. We know the following: There are 52 cards in the deck. There are 13 spades, so P(S) = 13/52. There are 4 aces, so P(A) = 4/52. There is 1 ace that is also a spade, so P(S ∩ A) = 1/52. Therefore, based on the rule of addition: P(S A) = P(S) + P(A) - P(S ∩ A) P(S A) = 13/52 + 4/52 - 1/52 = 16/52 = 4/13

3. Of the customers who bought items at a store on a particular day, \(53\) of them bought clocks and \(47\) of them bought watering cans. What is the largest possible number of people that bought either a clock or a watering can on that day? Solution: Let \(W\) be the event that a person bought a watering can, and let \(C\) be the event that a person bought a clock. Then: [W \cup C] = [W] + [C] - [W \cap C] To maximize the size of the union, the size of the intersection must be minimized. There are no further restrictions on the intersection found in the context of the problem, so just let it be \(0\), giving [W \cup C] = 53 + 47 = 100 So up to 100 people could have bought clocks or watering cans at the store that day. The next two examples lack context in order to emphasize the algebraic manipulations. 4. Two events connected with the experiment of rolling a single die are E: “the number rolled is even” and T: “the number rolled is greater than two.” Find the complement of each. Solution: In the sample space S={1,2,3,4,5,6} the corresponding sets of outcomes are E={2,4,6} and T={3,4,5,6}. The complements are Ec={1,3,5} and Tc={1,2}. 5. Find the probability that at least one heads will appear in five tosses of a fair coin. Solution: Identify outcomes by lists of five hs and ts, such as tthtt and hhttt. Although it is tedious to list them all, it is not difficult to count them. third toss, hence 4×2=8 outcomes for three tosses. Similarly, there are 8×2=16 outcomes for four tosses and finally 16×2=32 outcomes for five tosses. Let O denote the event “at least one heads.” There are many ways to obtain at least one heads, but only one way to fail to do so: all tails. Thus although it is difficult to list all the outcomes that form O, it is easy to write Oc={ttttt}. Since there are 32 equally likely outcomes, each has probability 1/32, so P(Oc)=1∕32, hence P(O)=1−1∕32≈0.97 or about a 97% chance.

6. In the experiment of rolling a single die, find the intersection E ∩ T of the events E: “the number rolled is even” and T: “the number rolled is greater than two.” Solution: The sample space is S={1,2,3,4,5,6}. Since the outcomes that are common to E={2,4,6} and T={3,4,5,6} are 4 and 6, E∩T={4,6}. In words the intersection is described by “the number rolled is even and is greater than two.” The only numbers between one and six that are both even and greater than two are four and six, corresponding to E ∩ T given above. 7. A single die is rolled. a. Suppose the die is fair. Find the probability that the number rolled is both even and greater than two. In both cases the sample space is S={1,2,3,4,5,6} and the event in question is the intersection E∩T={4,6} of the previous example. Since the die is fair, all outcomes are equally likely, so by counting we have P(E∩T)=2∕6. b. Suppose the die has been “loaded” so that P(1)=1∕12, P(6)=3∕12, and the remaining four outcomes are equally likely with one another. Now find the probability that the number rolled is both even and greater than two. Solution: The information on the probabilities of the six outcomes that we have so far is OutcomeProbablity11122p3p4p5p6312 since P(1)+P(6)=4∕12=1∕3 and the probabilities of all six outcomes add up to 1, P(2)+P(3)+P(4)+P(5)=1−13=23 Thus 4p=2∕3, so p=1∕6. In particular P(4)=1∕6. Therefore P(E∩T)=P(4)+P(6)=16+312=512

8. In the experiment of rolling a single die, find three choices for an

event A so that the events A and E: “the number rolled is even” are mutually exclusive. Solution: Since E={2,4,6} and we want A to have no elements in common with E, any event that does not contain any even number will do. Three choices are {1,3,5} (the complement Ec, the odds), {1,3}, and {5}. 9. In the experiment of rolling a single die, find the union of the events E: “the number rolled is even” and T: “the number rolled is greater than two.” Solution: Since the outcomes that are in either E={2,4,6} or T={3,4,5,6} (or both) are 2, 3, 4, 5, and 6, E T={2,3,4,5,6}. Note that an outcome such as 4 that is in both sets is still listed only once (although strictly speaking it is not incorrect to list it twice). In words the union is described by “the number rolled is even or is greater than two.” Every number between one and six except the number one is either even or is greater than two, corresponding to E T given above. 10. A two-child family is selected at random. Let B denote the event that at least one child is a boy, let D denote the event that the genders of the two children differ, and let M denote the event that the genders of the two children match. Find B D and B M. A sample space for this experiment is S={bb,bg,gb,gg}, where the first letter denotes the gender of the firstborn child and the second letter denotes the gender of the second child. The events B, D, and M are B={bb,bg,gb} D={bg,gb} M={bb,gg}

11. Two fair dice are thrown. Find the probabilities of the following events: c. both dice show a four There are 36 equally likely outcomes, of which exactly one corresponds to two fours, so the probability of a pair of fours is 1/36. d. at least one die shows a four 112131415161122232425262132333435363142434445464152535455565 162636465666 From the table we can see that there are 11 pairs that correspond to the event in question: the six pairs in the fourth row (the green die shows a four) plus the additional five pairs other than the pair 44, already counted, in the fourth column (the red die is four), so the answer is 11/36. To see how the formula gives the same number, let AG denote the event that the green die is a four and let AR denote the event that the red die is a four. Then clearly by counting we get P(AG)=6∕36 and P(AR)=6∕36. Since AG∩AR={44}, P(AG∩AR)=1∕36; this is the computation in part (a), of course. Thus by the Additive Rule of Probability, P(AG AR)=P(AG)+P(AR)−P(AG−AR)=636+636−136=1136 12. A tutoring service specializes in preparing adults for high school equivalence tests. Among all the students seeking help from the service, 63% need help in mathematics, 34% need help in English, and 27% need help in both mathematics and English. What is the percentage of students who need help in either mathematics or English? Imagine selecting a student at random, that is, in such a way that every student has the same chance of being selected. Let M denote the event “the student needs help in mathematics” and let E denote the event “the student needs help in English.” The information given is that P(M)=0.63, P(E)=0.34, and P(M∩E)=0.27. The Additive Rule of Probability gives P(M E)=P(M)+P(E)−P(M∩E)=0.63+0.34−0.27=0.70

13. Volunteers for a disaster relief effort were classified according to both specialty (C: construction, E: education, M: medicine) and language ability (S: speaks a single language fluently, T: speaks two or more languages fluently). The results are shown in the following two-way classification table: Language Ability Specialty S T C 12 1 E 4 3 M 6 2 The first row of numbers means that 12 volunteers whose specialty is construction speak a single language fluently, and 1 volunteer whose specialty is construction speaks at least two languages fluently. Similarly for the other two rows. 14. A volunteer is selected at random, meaning that each one has an equal chance of being chosen. Find the probability that: Language Ability Specialty Total S T C 12 1 13 E 4 3 7 M 6 2 8 Total 22 6 28

b. his specialty is medicine and he speaks two or more languages;

The probability sought is P(M∩T). The table shows that there are 2 such people, out of 28 in all, hence P(M∩T)=2∕28≈0.07 or about a 7% chance.

7.3

7.3

probability of union of two sets

243

PROBABILITY OF A UNION OF TWO SETS

Probability of the union of two events: Let us find the probability of the union of two arbitrary events A and B. One might think P(A B)=P(A)+P(B); however, each of P(A) and P(B) counts the probability of A∩B. We thus have to subtract this probability from P(A)+P(B) to obtain the correct formula: P(A B)=P(A)+P(B)−P(A∩B). Another way to derive the formula is to note that A B is the disjoint union of A/B, B/A, and A∩B. Now, since A is the disjoint union of A∩B and A/B, it follows that P(A)=P(A∩B)+P(A/B); whence P(A/B)=P(A)−P(A∩B). In a similar manner, one shows that P(B/A)=P(B)−P(B∩A). It follows that P(A B)=P(A/B)+P(B/A)+P(A∩B) =(P(A)−P(A∩B))+(P(B)−P(B∩A))+P(A∩B) =P(A)+P(B)−P(A∩B). Probability of the union of three events: For three arbitrary events A, B, and C, to find P(A B C), we first start with the guess P(A B C)=P(A)+P(B)+P(C).(1) Now 1. counts each of P(A∩B), P(A∩C), and P(B∩C) twice. To make up for this, we have to subtract P(A∩B)+P(A∩C)+P(B∩B).(2) But P(A)+P(B)+P(C) counts P(A∩B∩C) thrice, and in

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2. Subtracted it thrice. So, add P(A)+P(B)+P(C) back in again. P(A B C)=P(A)+P(B)+P(C)−P(A∩B)−P(A∩C)−P(B∩C)+P(A∩B∩C). Looking at the Venn diagram may be helpful here:

More generally, for the events A1, A2, …, An, we have the inclusion-exclusion principle: P(⋃i=1nAi)=∑i≤nP(Ai)−∑i1