aat solutions - ch11.pdf

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641Algebra 2

Worked-Out Solution Key

Prerequisite Skills (p. 742)

1. The probability of an event is a number from 0 to 1 that indicates the likelihood the event will occur.

2. The binomial distribution is not skewed. Instead, it is symmetric.

3.

202224 23 21 1 3

21.3 0.43422

2312 7

The numbers in increasing order: 2 Ï}

12 , 21.3, 2 3 } 4 ,

0.4, 2 }

3 , Ï

} 7

4.

2022 21 1

21.24 1.565

432 3 2

The numbers in increasing order: 2 Ï}

3 , 21.24, 6 } 5 ,

4 }

3 ,

Ï}

2 , 1.5

5. Odds: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19

Number of odds

}} Total

5 10

} 20 5 1 } 2

6. Perfect squares: 1, 4, 9, 16

Number of perfect squares

}} Total

5 4 } 20 5

1 } 5

7. Multiples of 3: 3, 6, 9, 12, 15, 18

Number of multiples of 3

}} Total

5 6 } 20 5

3 } 10

8. Factors of 50: 1, 2, 5, 10, 25, 50

Factors of 50 less than 20: 1, 2, 5, 10

Number of factors

}} Total

5 4 } 20 5

1 } 5

Lesson 11.1

11.1 Guided Practice (pp. 745–746)

1. Mean: } x 5 2 1 3 1 . . . 1 15

}} 10 5 74

} 10 5 7.4

Median: 7 1 8

} 2 5

15 } 2 5 7.5

Mode: 8

2. Range: 15 2 2 5 13

s 5 Ï}}}}

(2 2 7.4)2 1 (3 2 7.4)2 1 . . . 1 (15 2 7.4)2

}}}} 10

5 Ï}

144.4

} 10

5 3.8

3. Mean: } x 5 14 1 15 1 . . . 1 25

}} 11 5 165

} 11 5 15

Median: 15

Mode: 15

Range: 25 2 11 5 14

s 5 Ï}}}}

(14 2 15)2 1 (15 2 15)2 1 . . . 1 (25 2 15)2

}}}} 11

5 Ï}

138

} 11

ø 3.5

11.1 Exercises (pp. 747–748)

Skill Practice

1. Measures of central tendency represent the center or middle of a data set. Measures of dispersion tell you how spread out the values in a data set are.

2. The mean is the average of n numbers. The median is the middle number of n numbers when the numbers are written in increasing order. The mode is the number(s) that occurs most frequently.

3. Mean: } x 5 3 1 4 1 . . . 1 8

}} 9 5 48

} 9 ø 5.3

Median: 5

Modes: 4, 5, 6

4. Mean: } x 5 14 1 15 1 . . . 1 20

}} 7 5 120

} 7 ø 17.1

Median: 17

Mode: 20

5. Mean: } x 5 69 1 70 1 . . . 1 84

}} 10 5 745

} 10 5 74.5

Median: 73 1 74

} 2 5 73.5

Modes: 73, 78

6. Mean: } x 5 15 1 19 1 . . . 1 42

}} 11 5 296

} 11 ø 26.9

Median: 25

Mode: 19

7. B;

Median: 0.7 1 1.2

} 2 5

1.9 } 2 5 0.95

8. B;

Mean: } x 5 2 1 2 1 . . . 1 10

}} 6 5 36

} 6 5 6

9. The numbers should have been written in increasing order prior to fi nding the median. Order: 3, 5, 8, 9, 10, 10, 12The median is 9.

10. There are two modes because the numbers 9 and 12 both occur three times. The modes of the data set are 9 and 12.

11. Range: 9 2 4 5 5

Mean: } x 5 4 1 5 1 . . . 1 9

}} 8 5 51

} 8 5 6.375

s 5 Ï}}}}

(4 2 6.375)2 1 (5 2 6.375)2 1 . . . 1 (9 2 6.375)2

}}}} 8

5 Ï}

19.875

} 8 ø 1.6

12. Range: 20 2 6 5 14

Mean: } x 5 6 1 7 1 . . . 1 20

}} 9 5 90

} 9 5 10

s 5 Ï}}}}

(6 2 10)2 1 (7 2 10)2 1 . . . 1 (20 2 10)2

}}}} 9

5 Ï}

144

} 9 5 4

Chapter 11

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642Algebra 2Worked-Out Solution Key

13. Range: 6.0 2 1.3 5 4.7

Mean: } x 5 1.3 1 2.0 1 . . . 1 6.0

}} 7 5 23

} 7 ø 3.3

s 5 Ï}}}}

(1.3 2 3.3)2 1 (2.0 2 3.3)2 1 . . . 1 (6.0 2 3.3)2

}}}} 7

5 Ï}

19.67

} 7 ø 1.7

14. Range: 50 2 42 5 8

Mean: } x 5 42 1 44 1 . . . 1 50

}} 10 5 460

} 10 5 46

s 5 Ï}}}}

(42 2 46)2 1 (44 2 46)2 1 . . . 1 (50 2 46)2

}}}} 10

5 Ï}

52

} 10

ø 2.3

15. Range: 158 2 135 5 23

Mean: } x 5 135 1 136 1 . . . 1 158

}} 8 5 1165

} 8 5 145.625

s 5 Ï}}}}

(135 2 145.625)2 1 . . . 1 (158 2 145.625)2

}}}} 8

5 Ï}

519.87

} 8 ø 8.1

16. Range: 336 2 301 5 35

Mean: } x 5 301 1 308 1 . . . 1 336

}} 8 5 2535

} 8 5 316.875

s 5 Ï}}}}

(301 2 316.875)2 1 . . . 1 (336 2 316.875)2

}}}} 8

ø Ï}

800.878

} 8 ø 10.0

17. Outlier: 68

When included:

Mean: } x 5 2 1 2 1 . . . 1 68

}} 9 5 96

} 9 ø 10.7

Median: 4

Mode: 4

Range: 68 2 2 5 66

s 5 Ï}}}}

(2 2 10.7)2 1 (2 2 10.7)2 1 . . . 1 (68 2 10.7)2

}}}} 9

5 Ï}

3710.01

} 9 ø 20.3

When not included:

Mean: } x 5 2 1 2 1 . . . 1 6

}} 8 5 28

} 8 5 3.5

Median: 3 1 4

} 2 5

7 } 2 5 3.5

Mode: 4

Range: 6 2 2 5 4

s 5 Ï}}}}

(2 2 3.5)2 1 (2 2 3.5)2 1 . . . 1 (6 2 3.5)2

}}}} 8

5 Ï}

12

} 8 ø 1.2

18. Outlier: 0

When included:

Mean: } x 5 0 1 72 1 . . . 1 91

}} 8 5 562

} 8 ø 70.3

Median: 75 1 83

} 2 5

158 } 2 5 79

Mode: 83

Range: 91 2 0 5 91

s 5 Ï}}}}

(0 2 70.25)2 1 (72 2 70.25)2 1 . . . 1 (91 2 70.25)2

}}}} 8

5 Ï}

5901.5

} 8 ø 27.2

When not included:

Mean: } x 5 72 1 75 1 . . . 1 91

}} 7 5 562

} 7 ø 80.3

Median: 83

Mode: 83

Range: 91 2 72 5 19

s 5 Ï}}}}

(72 2 80.3)2 1 (75 2 80.3)2 1 . . . 1 (91 2 80.3)2

}}}} 7

5 Ï}

261.43

} 7 ø 6.1

19. Outlier: 0.7

When included:

Mean: } x 5 0.7 1 10.9 1 . . . 1 12.8

}} 6 5 60

} 6 5 10

Median: 11.6 1 11.6

} 2 5

23.2 } 2 5 11.6

Mode: 11.6

Range: 12.8 2 0.7 5 12.1

s 5 Ï}}}}

(0.7 2 10)2 1 (10.9 2 10)2 1 . . . 1 (12.8 2 10)2

}}}} 6

5 Ï}

106.02

} 6 ø 4.2

When not included:

Mean: } x 5 10.9 1 11.6 1 . . . 1 12.8

}} 5 5 59.3

} 5 ø 11.9

Median: 11.6

Mode: 11.6

Range: 12.8 2 10.9 5 1.9

s 5 Ï}}}}}

(10.9 2 11.86)2 1 (11.6 2 11.86)2 1 . . . 1 (12.8 2 11.86)2

}}}}} 5

5 Ï}

2.232

} 5 ø 0.67

Chapter 11, continued

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643Algebra 2

Worked-Out Solution Key 643Algebra 1


20. Outlier: 100

When included:

Mean: } x 5 20 1 20 1 . . . 1 100

}} 7 5 246

} 7 ø 35.1

Median: 25

Modes: 20, 25, 28

Range: 100 2 20 5 80

s 5 Ï}}}}

(20 2 35.1)2 1 (20 2 35.1)2 1 . . . 1 (100 2 35.1)2

}}}} 7

5 Ï}

4972.87

} 7 ø 26.7

When not included:

Mean: } x 5 20 1 20 1 . . . 1 28

}} 6 5 146

} 6 ø 24.3

Median: 25 1 25

} 2 5

50 } 2 5 25

Modes: 20, 25, 28

Range: 28 2 20 5 8

s 5 Ï}}}}

(20 2 24.3)2 1 (20 2 24.3)2 1 . . . 1 (28 2 24.3)2

}}}} 6

5 Ï}

65.34

} 6 5 3.3

21. Outlier: 152

When included:

Mean: } x 5 60 1 66 1 . . . 1 152

}} 10 5 788

} 10 5 78.8

Median: 71 1 72

} 2 5

143 } 2 5 71.5

Mode: 66

Range: 152 2 60 5 92

s 5 Ï}}}}

(60 2 78.8)2 1 (66 2 78.8)2 1 . . . 1 (152 2 78.8)2

}}}} 10

5 Ï}

6279.6

} 10

ø 25.1

When not included:

Mean: } x 5 60 1 66 1 . . . 1 80

}} 9 5 636

} 9 ø 70.7

Median: 71

Mode: 66

Range: 80 2 60 5 20

s 5 Ï}}}}

(60 2 70.7)2 1 (66 2 70.7)2 1 . . . 1 (80 2 70.7)2

}}}} 9

5 Ï}

326.01

} 9 ø 6.0

22. Outlier: 65

When included:

Mean: } x 5 65 1 173 1 . . . 1 199

}} 8 5 1372

} 8 5 171.5

Median: 184 1 188

} 2 5

372 } 2 5 186

Mode: no mode

Range: 199 2 65 5 134

s 5 Ï}}}}}

(65 2 171.5)2 1 (173 2 171.5)2 1 . . . 1 (199 2 171.5)2

}}}}} 8

5 Ï}

13,382

} 8 ø 40.9

When not included:

Mean: } x 5 173 1 181 1 . . . 1 199

}} 7 5 1307

} 7 ø 186.7

Median: 188

Mode: no mode

Range: 199 2 173 5 26

s 5 Ï}}}}}

(173 2 186.7)2 1 (181 2 186.7)2 1 . . . 1 (199 2 186.7)2

}}}}} 7

5 Ï}

419.43

} 7 ø 7.7

23. Sample answer:

Let n 5 9.

} x 5 sum

} 9

10 5 sum

} 9

90 5 sum

Data set: 8, 8, 8, 8, 11, 11, 12, 12, 12

24. a. } x 5 70 1 55 1 . . . 1 74

}} 16 5 993

} 16 ø 62.1

s 5 Ï}}}

(70 2 62.1)2 1 . . . 1 (74 2 62.1)2

}}} 16

5 Ï}

5288.96

} 16

ø 18.2

x 2 } x

} s

> 3

6 2 62.1 }

18.2 >?3

56.1

} 18.2

>?3

3.08 > 3 ✓

So, 6 is an outlier.


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b. } x 5 18 1 20 1 . . . 1 26

}} 16 5 434

} 16 ø 27.1

s 5 Ï}}}

(18 2 27.1)2 1 . . . 1 (26 2 27.1)2

}}} 16

5 Ï}

5001.76

} 16

ø 17.7

x 2 } x

} s

> 3

88 2 27.1 }

17.1 >?3

60.9

} 17.7

>?3

3.4 > 3 ✓


c. } x 5 50 1 93 1 . . . 1 37

}} 15 5 1364

} 15 ø 90.9

s 5 Ï}}}

(50 2 90.9)2 1 . . . 1 (37 2 90.9)2

}}} 15

5 Ï}

16,578.95

} 15

ø 33.2

x 2 } x

} s

> 3

199 2 90.9 }}

33.2 >?3

3.26 > 3 ✓


25. } x 5 x1 1 x2 1 x3

} 3

s 5 Ï}}}

(x1 2 } x )2 1 (x2 2 } x )2 1 (x3 2 } x )2

}}} 3

5 Ï}}}}

x1

2 2 2x1 } x 1 } x 2 1 x22 2 2x2 } x 1 } x 2 1 x3

2 2 2x3 } x 1 } x 2

}}}} 3

5 Ï}}}}

3 } x 2 1 x1

2 1 x22 1 x3

2 2 2 } x (x1 1 x2 1 x3) }}}}

3

5 Ï}}}}}

1 }

3 (x1 1 x2 1 x3)2 1 x1

2 1 x22 1 x3

2 2 2 } 3 (x1 1 x2 1 x3)2

}}}}} 3

5 Ï}}}

x1

2 1 x22 1 x3

2 2 1 } 3 (x1 1 x2 1 x3)2

}}} 3

5 Ï}}

x1

2 1 x22 1 x3

2 2 3 } x 2

}} 3

5 Ï}}

x1

2 1 x22 1 x3

2

}} 3 2 } x 2

Problem Solving

26. Mean: } x 5 21 1 21 1 . . . 1 49

}} 12 5 351

} 12 5 29.25

Median: 27 1 28

} 2 5

55 } 2 5 27.5

Mode: 27

Range: 49 2 21 5 28

s 5 Ï}}}

(21 2 29.25)2 1 . . . 1 (49 2 29.25)2

}}} 12

5 Ï}

694.25

} 12

ø 7.6

27. Mean: } x 5 2 1 6 1 . . . 1 30

}} 12 5 213

} 12 5 17.75

Median: 20 1 20

} 2 5

40 } 2 5 20

Modes: 6, 20

28. Mean: } x 5 97 1 102 1 . . . 1 120

}} 10 5 1103

} 10 5 110.3

Median: 111 1 113

} 2 5

224 } 2 5 112

Mode: 114

Range: 120 2 97 5 23

s 5 Ï}}}

(97 2 110.3)2 1 . . . 1 (120 2 110.3)2

}}} 10

5 Ï}

426.1

} 10

ø 6.5

29. a. The outlier is 5.

b. When included:

Mean: } x 5 5 1 19 1 . . . 1 25

}} 10 5 202

} 10 5 20.2

Median: 21 1 23

} 2 5

44 } 2 5 22

Mode: 23

Range: 25 2 5 5 20

s 5 Ï}}}

(5 2 20.2)2 1 . . . 1 (25 2 20.2)2

}}} 10

5 Ï}

295.6

} 10

ø 5.4

When not included:

Mean: } x 5 19 1 19 1 . . . 1 25

}} 9 5 197

} 9 ø 21.9

Median: 23

Mode: 23

Range: 25 2 19 5 6

s 5 Ï}}}

(19 2 21.9)2 1 . . . 1 (25 2 21.9)2

}}} 9

5 Ï}

38.89

} 9 ø 2.1

c. Sample answer: The mean and the median increase when the outlier is removed and the range and standard deviation decrease.


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645Algebra 2


30. a. Mean: } x 5 74.36 1 79.43 1 . . . 1 86.50

}}} 12 5 987.79

} 12

ø 82.32

Median: 83.14 1 83.25

}} 2 5

166.39 } 2 ø 83.20

Mode: no mode

Range: 86.50 2 74.36 5 12.14

s 5 Ï}}}}

(74.36 2 82.32)2 1 . . . 1 (86.50 2 82.32)2

}}}} 12

5 Ï}

115.8775

} 12

ø 3.11

b. Mean: } x 5 74.26 1 74.72 1 . . . 1 82.66

}}} 8 5 630.36

} 8

ø 78.80

Median: 78.72 1 80.17

}} 2 5

158.89 } 2 ø 79.45

Mode: no mode

Range: 82.66 2 74.26 5 8.40

s 5 Ï}}}}

(74.26 2 78.8)2 1 . . . 1 (82.66 2 78.8)2

}}}} 8

5 Ï}

73.0238

} 8 ø 3.02

c. Sample answer: While the distances were more consistent in 1964 (smaller range), the average distance (mean) was greater in 2004.

31. a. (a 2 b)2 ≥ 0 because the result of squaring any real number is always greater than or equal to zero.

b. (a 2 b)2 ≥ 0 a2 2 2ab 1 b2 ≥ 0 a2 1 b2 ≥ 2ab

a2 1 b2 1 2ab ≥ 2ab 1 2ab

a2 1 2ab 1 b2 ≥ 4ab

(a 1 b)2 ≥ 4ab

c. (a 1 b)2 ≥ 4ab

Ï}

(a 1 b)2 ≥ Ï}

4ab

a 1 b ≥ 2 Ï}

ab

a 1 b

} 2 ≥ Ï

}

ab

d. (a 2 b)2 5 0

a 2 b 5 0

a 5 b

The arithmetic mean of a and b is equal to the geometric mean of a and b when a and b are equal.

Mixed Review

32. f (x) 5 x 1 709

f (3.8) 5 3.8 1 709 5 712.8

f (600) 5 600 1 709 5 1309

33. f (x) 5 x 1 11.68

f (3.8) 5 3.8 1 11.68 5 15.48

f (600) 5 600 1 11.68 5 611.68

34. f (x) 5 15.4x

f (3.8) 5 15.4(3.8) 5 58.52

f (600) 5 15.4(600) 5 9240

35. f (x) 5 200x

f (3.8) 5 200(3.8) 5 760

f (600) 5 200(600) 5 120,000

36. f (x) 5 5x 1 136

f (3.8) 5 5(3.8) 1 136 5 155

f (600) 5 5(600) 1 136 5 3136

37. f (x) 5 22x 2 450

f (3.8) 5 22(3.8) 2 450 5 2366.4

f (600) 5 22(600) 2 450 5 12,750

38. Because the vertex is on the y-axis and the co-vertex is on the x-axis, the major axis is vertical with a 5 5 and

b 5 3. An equation is x2

} 9 1

y2

} 25 5 1.

39. Because the vertex and focus are points on a horizontal line, the major axis is horizontal with a 5 7 and c 5 3.

c2 5 a2 2 b2

9 5 49 2 b2

b2 5 40

An equation is x2

} 49

1 y2

} 40 5 1.

40. Because the co-vertex is on the y-axis and the focus is on the x-axis, the major axis is horizontal with b 5 Ï

}

11 and c 5 2.

c2 5 a2 2 b2

4 5 a2 2 11

15 5 a2

An equation is x2

} 15

1 y2

} 11 5 1.

41. R 5 V

} I 5 120

} 0.8 5 150 ohms

42. Total dogs 5 3 1 5 1 4 1 3 5 15

The number of distinguishable permutations is

15! }}

3! p 5! p 4! p 3! 5 12,612,600 ways.

Graphing Calculator Activity 11.1 (p. 750)

1. } x ø 48.8

Med 5 48

Range 5 maxX 2 minX 5 58 2 40 5 18

sx ø 5.3

2. } x 5 3.3

Med 5 2.9

Range 5 maxX 2 minX 5 6 2 1.3 5 4.7

sx ø 1.6


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3. } x ø 87.4

Med 5 88

Range 5 maxX 2 minX 5 98 2 78 5 20

sx ø 6.4

4. } x ø 18.0

Med 5 17.75

Range 5 maxX 2 minX 5 24.8 2 12.7 5 12.1

sx ø 3.8

5. } x ø 106.9

Med 5 107

Range 5 maxX 2 minX 5 112 2 101 5 11

sx ø 3.4

6. } x ø 272

Med 5 275

Range 5 maxX 2 minX 5 298 2 252 5 46

sx ø 14.4

7. } x ø 54.0

Med 5 53

Range 5 maxX 2 minX 5 71 2 36 5 35

sx ø 8.4

Lesson 11.2

11.2 Guided Practice (p. 752)

1. Mean: 408 1 300 5 708

Median: 408 1 300 5 708

Mode: 410 1 300 5 710

Range: 33

Standard deviation: 9.3

2. Mean: 1.09(17.53) ø 19.11

Median: 1.09(17.39) ø 18.96

Mode: 1.09(17.35) ø 18.91

Range: 1.09(1.32) ø 1.44

Standard deviation: 1.09(0.37) ø 0.40

11.2 Exercises (pp. 753–755)

Skill Practice

1. Multiplying each value in a data set by a constant is an example of a transformation of the data.

2. The mean, median, and mode increase by the value of the constant, but the range and standard deviation remain the same.

3.Original data set

Adding 6 to data values

Mean 18 18 1 6 5 24

Median 17 17 1 6 5 23

Mode 17 17 1 6 5 23

Range 9 9

Std. dev. 3.0 3.0

4. Original data set


Mean 39.3 39.3 1 18 5 57.3

Median 39 39 1 18 5 57

Mode 42 42 1 18 5 60

Range 17 17

Std. dev. 5.1 5.1



Mean 78 78 1 17 5 95

Median 77 77 1 17 5 94

Mode 77 77 1 17 5 94

Range 9 9

Std. dev. 2.8 2.8



Mean 202.2 202.2 1 155 5 357.2

Median 207.5 207.5 1 155 5 362.5

Mode 211 211 1 155 5 366

Range 38 38

Std. dev. 13.0 13.0



Mean 56 56 2 21 5 35

Median 53 53 2 21 5 32

Mode 53 53 2 21 5 32

Range 21 21

Std. dev. 7.0 7.0



Mean 285.7 285.7 2 45 5 240.7

Median 280.5 280.5 2 45 5 235.5

Mode 279 279 2 45 5 234

Range 23 23

Std. dev. 9.0 9.0

9. The standard deviation does not change when adding a constant. So, the new standard deviation is 10.


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647Algebra 2


10.Original data set

Multiplying datavalues by 3

Mean 26.6 3(26.6) 5 79.8

Median 26 3(26) 5 78

Mode 23 3(23) 5 69

Range 15 3(15) 5 45

Std. dev. 4.9 3(4.9) 5 14.7



Mean 61.9 4(61.9) 5 247.6

Median 62 4(62) 5 248

Mode 58 4(58) 5 232

Range 9 4(9) 5 36

Std. dev. 3.4 4(3.4) 5 13.6


Multiplying datavalues by 1.5

Mean 34.3 1.5(34.3) 5 51.5

Median 35 1.5(35) 5 52.5

Mode 35 1.5(35) 5 52.5

Range 12 1.5(12) 5 18

Std. dev. 4.0 1.5(4.0) 5 6.0



Mean 98.2 2.5(98.2) 5 245.5

Median 100.5 2.5(100.5) 5 251.3

Mode 102 2.5(102) 5 255

Range 19 2.5(19) 5 47.5

Std. dev. 6.6 2.5(6.6) 5 16.5



Mean 122.7 0.5(122.7) ø 61.4

Median 125.5 0.5(125.5) ø 62.8

Mode 130 0.5(130) 5 65

Range 24 0.5(24) 5 12

Std. dev. 8.9 0.5(8.9) ø 4.5



Mean 228.7 0.9(228.7) ø 205.8

Median 226.5 0.9(226.5) ø 203.9

Mode 222 0.9(222) 5 199.8

Range 38 0.9(38) 5 34.2

Std. dev. 12.0 0.9(12.0) 5 10.8

16. D; new range 5 21 p 3 5 63

17. ax1 1 ax2 1 . . . 1 axn

}} n 5

a(x1 1 x2 1 . . . 1 xn) }} n 5 a } x

Problem Solving

18. a-b. Salaries

without bonusSalaries

with bonus

Mean 37.2 37.2 1 1.2 5 38.4

Median 38 38 1 1.2 5 39.2

Mode 28.5 28.5 1 1.2 5 29.7

Range 19.5 19.5

Std. dev. 6.9 6.9

19. a-b. Heights

without stiltsHeights

with stilts

Mean 70.8 70.8 1 28 5 98.8

Median 72 72 1 28 5 100

Mode 72 72 1 28 5 100

Range 8 8

Std. dev. 2.4 2.4

20. B; mean: 68 1 10 5 78

standard deviation: 15

21. a-b. Distancesin meters

Distancesin feet

Mean 6.84 3.28(6.84) ø 22.44

Median 6.89 3.28(6.89) ø 22.60

Mode (none) (none)

Range 1.16 3.28(1.16) ø 3.80

Std. dev. 0.324 3.28(0.324) ø 1.06


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22.a-b. Weights with

equipmentWeights without

equipment

Mean 280.6 280.6 2 115 5 165.6

Median 280 280 2 115 5 165

Mode 280 280 2 115 5 165

Range 30 30

Std. dev. 9.2 9.2

A constant of 2115 is added to the original mean, median, and mode to determine the values. The range and standard deviation remain the same.

23. a. Mean: 75.8

Median: about 75.4

Mode: 74.5

Range: 9.9

Standard deviation: about 3.0

b. C1 5 5 } 9 (74.5 2 32) ø 23.6

C2 5 5 } 9 (81.9 2 32) ø 27.7

C3 5 5 } 9 (72.5 2 32) ø 22.5

C4 5 5 } 9 (73.4 2 32) ø 23

C5 5 5 } 9 (78.4 2 32) ø 25.8

C6 5 5 } 9 (72.6 2 32) ø 22.6

C7 5 5 } 9 (76.8 2 32) ø 24.9

C8 5 5 } 9 (74.5 2 32) ø 23.6

C9 5 5 } 9 (77.6 2 32) ø 25.3

C10 5 5 } 9 (72.0 2 32) ø 22.2

C11 5 5 } 9 (79.2 2 32) ø 26.2

C12 5 5 } 9 (76.2 2 32) ø 24.6

c. Mean: about 24.3

Median: 24.1

Mode: 23.6

Range: 5.5


d. When the temperatures are converted from Fahrenheit to Celsius, the measures of central tendency also undergo the same addition and multiplication transformations

1 adding 232 and multiplying the result by 5 }

9 2 . The

measures of dispersion only undergo the multiplication transformation.

24. Rainfall in

centimetersRainfall in

inches

Mean 10 0.3937(10) ø 3.9

Median 10.05 0.3937(10.05) ø 4.0

Mode 9.9 0.3937(9.9) ø 3.9

Range 3.1 0.3937(3.1) ø 1.2

Std. dev. 1.0 0.3937(1.0) ø 3.9

25. mean salary 5 sum of salaries

}} total employees

49,500 5 sum of salaries

}} 100

$4,950,000 5 sum of salaries

(15 supervisors)($5000) 5 $75,000

4,950,000 1 75,000 5 5,025,000

new mean 5 5,025,000

} 100 5 $50,250

Because all of the supervisors’ salaries are greater than the median before and after the raise, the median stays the same ($42,000).

Mixed Review

26. x2 2 11x 1 24 5 0

(x 2 8)(x 2 3) 5 0

x 2 8 5 0 or x 2 3 5 0

x 5 8 or x 5 3

27. 3x2 2 13x 1 8 5 0

x 5 2(213) 6 Ï

}}

(213)2 2 4(3)(8) }}}

2(3)

5 13 6 Ï

}

169 2 96 }} 6 5

13 6 Ï}

73 } 6

28. Ï}

4x 1 6 5 Ï}

7x 2 15

1 Ï}

4x 1 6 2 2 5 1 Ï

}

7x 2 15 2 2

4x 1 6 5 7x 2 15

21 5 3x

7 5 x

29. 2x 5 17

log22x 5 log217

x 5 log217 5 log 17

} log 2

ø 4.1

30. P(k 5 4) 5 30C4(0.5)4(1 2 0.5)26

5 30!

} 4!26! (0.5)4(0.5)26

ø 0.0000255

31. P(k 5 17) 5 30C17(0.5)17(1 2 0.5)13

5 30! } 17!13! (0.5)17(0.5)13

ø 0.112


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649Algebra 2


32. P(k 5 21) 5 30C21(0.5)21(1 2 0.5)9

5 30!

} 21!9! (0.5)21(0.5)9

ø 0.0133

33. P(k 5 7) 5 30C7(0.5)7(1 2 0.5)23

5 30!

} 7!23! (0.5)7(0.5)23

ø 0.00190

34. P(k 5 14) 5 30C14(0.5)14(1 2 0.5)16

5 30! } 14!16! (0.5)14(0.5)16

ø 0.135

35. P(k 5 11) 5 30C11(0.5)11(1 2 0.5)19

5 30! } 11!19! (0.5)11(0.5)19

ø 0.0509

36. P(k 5 27) 5 30C27(0.5)27(1 2 0.5)3

5 30!

} 27!3! (0.5)27(0.5)3

ø 0.00000378

37. P(k 5 18) 5 30C18(0.5)18(1 2 0.5)12

5 30! } 18!12! (0.5)18(0.5)12

ø 0.0806

38. 20C5 1 20C6 1 20C7 1 20C8 1 20C9 1 20C10 1

20C11 1 20C12 1 20C13 1 20C14 1 20C15 1 20C16 1

20C17 1 20C18 1 20C19 1 20C20

5 15,504 1 38,760 1 77,520 1 125,970 1 167,960 1 184,756 1 167,960 1 125,970 1 77,520 1 38,760 1 15,504 1 4845 1 1140 1 190 1 20 1 1

5 1,042,380 combinations

Quiz 11.1–11.2 (p. 755)

1. Mean: 5 1 5 1 . . . 1 16

}} 9 ø 9.7

Median: 9

Modes: 5 and 11

Range: 16 2 5 5 11

Std. dev.: Ï}}}

(5 2 9.7)2 1 . . . 1 (16 2 9.7)2

}}} 9 ø 3.7

2. Mean: 14 1 15 1 . . . 1 24

}} 9 ø 18.9

Median: 19

Mode: 19

Range: 24 2 14 5 10

Std. dev.: Ï}}}

(14 2 18.9)2 1 . . . 1 (24 2 18.9)2

}}} 9 ø 3.2

3. Mean: 43 1 44 1 . . . 1 56

}} 8 5 49.25

Median: 47.5

Mode: 56

Range: 56 2 43 5 13

Std. dev.: Ï}}}

(43 2 49.25)2 1 . . . 1 (56 2 49.25)2

}}} 8 ø 5.2

4. Mean: 67 1 68 1 . . . 1 74

}} 9 5 71

Median: 71

Mode: 73

Range: 74 2 67 5 7

Std. dev.: Ï}}}

(67 2 71)2 1 . . . 1 (74 2 71)2

}}} 9 ø 2.3

5. Mean: 145 1 150 1 . . . 1 181

}} 7 ø 161.1

Median: 159

Mode: no mode

Range: 181 2 145 5 36

Std. dev.: Î}}}

(145 2 161.1)2 1 . . . 1 (181 2 161.1)2

}}} 7 ø 11.4

6. Mean: 231 1 232 1 . . . 1 261

}} 8 5 246

Median: 246

Mode: 246

Range: 261 2 231 5 30

Std. dev.: Î}}}

(231 2 246)2 1 . . . 1 (261 2 246)2

}}} 8 ø 10.1

7. Original prices($)

Saleprices($)

Mean 231 (0.8)(231) 5 184.80

Median 230 (0.8)(230) 5 184

Mode 320 (0.8)(320) 5 256

Range 230 (0.8)(230) 5 184

Std. dev. 84.40 (0.8)(84.4) 5 67.52

Mixed Review of Problem Solving (p. 756)

1. a. Original prices($)

Saleprices($)

Mean 85.50 (0.75)(85.5) ø 64.13

Median 80 (0.75)(80) 5 60

Mode 75, 80(0.75)(75) 5 56.25

(0.75)(80) 5 60

Range 50 (0.75)(50) 5 37.50

Std. dev. 15.5 (0.75)(15.51) ø 11.63


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b. Mean: 85.50 2 20 5 65.50

Median: 80 2 20 5 60

Modes: 75 2 20 5 55

80 2 20 5 60

Range: 50

Standard deviation: ø 15.51

Sample answer: The 25% discount has a lower mean, range, and standard deviation. Both discounts have the same median and have one mode in common.

2. a. Mean: 75.5

Median: 76.2

Range: 15


b. Mean: about 47.7

Median: 48.1

Range: 47.4


c. Sample answer: Buffalo, New York, has a much wider range of temperatures than Miami, Florida.

3. } x 5 sum }}

number of values

20 5 sum

} 5

100 5 sum

Sample answer: 3, 14, 20, 26, 37Add 10 to each data value to change the mean from 20 to 30: 13, 24, 30, 36, 47

4. a. Teammate 1 Teammate 2

Mean 11.92 12.06

Median 11.3 11.9

Mode 11.3 11.8

Range 2.3 0.7

Std. dev. 0.89 0.29

b. Teammate 2 has the more consistent times because the range and standard deviation are smaller.

5. a. Mean: $310.36

Median: $250

Mode: $200

Range: $655

Standard deviation: $189.06

b. The data values $670 and $850 are outliers because all the other values are in or very close to the 200s.

c. Mean: $235.42

Median: $240

Mode: $200

Range: $95

Standard deviation: $32.24

d. The range and standard deviation are most affected by the outliers. The mode is least affected by the outliers.

6. Mean fi rst round: 84 strokes Mean last round: 84 2 2 5 82 strokes

Lesson 11.3


1. P(x ≤ } x ) 5 0.5

2. P(x ≥ } x ) 5 0.5

3. P( } x ≤ x ≤ } x 1 2s) 5 0.34 1 0.135 5 0.475

4. P( } x 2 s ≤ x ≤ } x ) 5 0.34

5. P(x ≤ } x 2 3s) 5 0.0015

6. P(x ≥ } x 1 s) 5 0.135 1 0.0235 1 0.0015 5 0.16

7. 34% 1 13.5% 5 47.5%

47.5% of the women have readings between 172 and 200.

8. z 5 x 2 } x

} s

5 90 2 73

} 14.1 ø 1.2

Using the standard normal table, P(x ≤ 90) ø P(z ≤ 1.2) 5 0.8849.

9. A z-score of 0 indicates that the z-score and the mean are the same. Therefore, the area under the normal curve is divided into two equal parts with the mean and the z-score being equal to 0.5.

11.3 Exercises (pp. 760–762)

Skill Practice

1. A normal curve is a bell-shaped curve that is symmetric about the mean.

2. To fi nd P(z ≤ 1.4) using the table, fi nd the value where row 1 and column .4 intersect.

3. P(x ≤ } x 2 s) 5 0.135 1 0.0235 1 0.0015 5 0.16

4. P(x ≥ } x 1 2s) 5 0.0235 1 0.0015 5 0.025

5. P(x ≤ } x 1 s) 5 0.34 1 0.34 1 0.135 1 0.0235 1 0.0015

5 0.84

6. P(x ≥ } x 2 s) 5 0.34 1 0.34 1 0.135 1 0.0235 1 0.0015

5 0.84

7. P( } x 2 s ≤ x ≤ } x 1 s) 5 0.34 1 0.34 5 0.68

8. P( } x 2 3s ≤ x ≤ } x ) 5 0.0235 1 0.135 1 0.34 5 0.4985

9. 13.5% 1 2.35% 1 0.15% 5 16%

10. 2.35% 1 2.35% 5 4.7%

11–16.

21 25 29 33 37 41 45

11. P(29 ≤ x ≤ 37) 5 0.34 1 0.34 5 0.68

12. P(33 ≤ x ≤ 45) 5 0.34 1 0.135 1 0.0235 5 0.4985

13. P(21 ≤ x ≤ 41)

5 0.0235 1 0.135 1 0.34 1 0.34 1 0.135 5 0.9735


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651Algebra 2


14. P(x ≥ 25) 5 1 2 P(x ≤ 25)

5 1 2 (0.0235 1 0.0015)

5 0.975

15. P(x ≥ 29) 5 1 2 P(x ≤ 29)

5 1 2 (0.135 1 0.0235 1 0.0015)

5 0.84

16. P(x ≤ 37) 5 1 2 P(x ≥ 37)

5 1 2 (0.135 1 0.0235 1 0.0015)

5 0.84

17. C;

69 74 79 84 89 94 99

P(74 ≤ x ≤ 94) 5 0.135 1 0.34 1 0.34 1 0.135 5 0.95

18. B;

42 45 48 51 54 57 60

P(x ≤ 48) 5 0.135 1 0.0235 1 0.0015 5 0.16

19. z 5 x 2 } x

} s

5 68 2 64

} 7 ø 0.6

P(x ≤ 68) ø P(z ≤ 0.6) 5 0.7257

20. z 5 x 2 } x

} s

5 80 2 64

} 7 ø 2.3

P(x ≤ 80) ø P(z ≤ 2.3) 5 0.9893

21. z 5 x 2 } x

} s

5 45 2 64

} 7 ø 22.7

P(x ≤ 45) ø P(z ≤ 22.7) 5 0.0035

22. z 5 x 2 } x

} s

5 54 2 64

} 7 ø 21.4

P(x ≤ 54) ø P(z ≤ 21.4) 5 0.0808

23. z 5 x 2 } x

} s

5 64 2 64

} 7 ø 0

P(x ≤ 64) ø P(z ≤ 0) 5 0.5

24. z 5 x 2 } x

} s

5 59 2 64

} 7 ø 20.7

P(x > 59) 5 1 2 P(x ≤ 59)

ø 1 2 P(z ≤ 20.7)

5 1 2 0.2420

5 0.758

25. z 5 x 2 } x

} s

5 75 2 64

} 7 ø 1.6

P(x > 75) 5 1 2 P(x ≤ 75)

ø 1 2 P(z ≤ 1.6)

5 1 2 0.9452

5 0.0548

26. z 5 x 2 } x

} s

5 60 2 64

} 7 ø 20.6

z 5 x 2 } x

} s

5 75 2 64

} 7 ø 1.6

P(60 < x ≤ 75) ø P(20.6 < z ≤ 1.6)

5 P(z ≤ 1.6) 2 P(z ≤ 20.6)

5 0.9452 2 0.2743

5 0.6709

27. z 5 x 2 } x

} s

5 45 2 64

} 7 ø 22.7

z 5 x 2 } x

} s

5 65 2 64

} 7 ø 0.1

P(45 < x ≤ 65) ø P(22.7 < z ≤ 0.1)

5 P(z ≤ 0.1) 2 P(z ≤ 22.7)

5 0.5398 2 0.0035

5 0.5363

28. 0.9192 corresponds to a z-score of 1.4.

z 5 k 2 } x

} s

1.4 5 k 2 80

} 10

14 5 k 2 80

94 5 k

29. The table was interpreted incorrectly. The value 0.2119 is the probability that z is less than or equal to 20.8. To fi nd P(z ≥ 20.8), you must subtract this from 1.

P(z ≥ 20.8) 5 1 2 P(z ≤ 20.8) 5 1 2 0.2119 5 0.7881

30. a. Sample answer:

b. Sample answer: As the standard deviation decreases, the normal curve gets steeper.

Problem Solving

31.

3.4 3.8 4.2 4.6 5.0 5.4 5.8

P(x ≥ 5) 5 1 2 P(x ≤ 5)

5 1 2 (0.0015 1 0.0235 1 0.135 1 0.34 1 0.34)

5 1 2 0.84 5 0.16

The probability that a randomly selected housefl y has a wing length of at least 5 millimeters is 0.16.


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32. a.

3 4 5 6 7 8 9

P(x ≤ 8) 5 0.0015 1 0.0235 1 0.135 1 0.34 1 0.34 1 0.135

5 0.975

The probability that the fi re department takes at most 8 minutes is 0.975.

b.

3 4 5 6 7 8 9

P(4 ≤ x ≤ 7) 5 0.135 1 0.34 1 0.34 5 0.815

The probability that the fi re department takes between 4 and 7 minutes is 0.815.

33. a. z 5 x 2 } x

} s

5 19.4 2 20

} 0.25 5 22.4

z 5 x 2 } x

} s

5 20.4 2 20

} 0.25 5 1.6

b. P(x ≤ 19.4) ø P(z ≤ 22.4)

5 0.0082

c. P(19.4 ≤ x ≤ 20.4) ø P(22.4 ≤ z ≤ 1.6)

5 P(z ≤ 1.6) 2 P(z ≤ 22.4)

5 0.9452 2 0.0082

5 0.937

34. a. A height of 16 inches is 2 standard deviations above the mean, so P(x ≤ 16)

5 0.0015 1 0.0235 1 0.135 1 0.34 1 0.34 1 0.135

5 0.975, and P(x ≥ 16) 5 1 2 P(x ≤ 16) 5 0.025.

About 2.5% of the plants are taller than 16 inches.

b. z 5 x 2 } x

} s

5 13 2 12

} 2 ø 0.5

P(x ≤ 13) ø P(z ≤ 0.5) 5 0.6915

About 69.15% of the plants are at most 13 inches.

c. z 5 x 2 } x

} s

5 7 2 12

} 2 5 22.5

z 5 x 2 } x

} s

5 14 2 12

} 2 5 1

P(7 ≤ x ≤ 14) ø P(22.5 ≤ z ≤ 1)

5 P(z ≤ 1) 2 P(z ≤ 22.5)

5 0.8413 2 0.0062

5 0.8351

About 83.51% of the plants are between 7 inches and 14 inches.

d. z 5 x 2 } x

} s

5 9 2 12

} 2 5 21.5

z 5 x 2 } x

} s

5 15 2 12

} 2 5 1.5

P(x ≤ 9) or P(x ≥ 12) ø P(z ≤ 21.5) or P(z ≥ 1.5)

5 P(z ≤ 21.5) 1 P(z ≥ 1.5)

5 0.0668 1 [1 2 P(z ≤ 1.5)] 5 0.0668 1 (1 2 0.9332)

5 0.1336

About 13.36% of the plants are at least 3 inches taller than or shorter than the mean height.

35. a. z 5 x 2 } x

} s

5 30 2 20

} 4.2 ø 2.4

b. z 5 x 2 } x

} s

5 610 2 500

} 90 ø 1.2

c. Lisa scored better because her test score is 2.4 standard deviations above the mean, while Ann’s is 1.2 standard deviations above the mean.

36. a. z 5 x 2 } x

} s

5 72 2 69

} 2.75 ø 1.1

P(x > 72) 5 1 2 P(x ≤ 72)

ø 1 2 P(z ≤ 1.1)

5 1 2 0.8643

5 0.1357

The probability that all 3 men are more than 6 feet tall is about (0.1357)3 ø 0.0025.

b. z 5 x 2 } x

} s

5 65 2 69

} 2.75 ø 21.5

z 5 x 2 } x

} s

5 75 2 69

} 2.75 ø 2.2

P(65 ≤ x ≤ 75) ø P(21.5 ≤ z ≤ 2.2)

5 P(z ≤ 2.2) 2 P(z ≤ 21.5)

5 0.9861 2 0.0668

5 0.9193

The probability that all 5 men are between 65 and 75 inches tall is about (0.9193)5 ø 0.6566.

Mixed Review

37. 4x3 1 5x2 1 (23x) 1 7

2 4 5 23 7

8 26 46

4 13 23 53

f (2) 5 53

38. 23x3 1 (25x2) 1 0x 1 10

4 23 25 0 10

212 268 2272

23 217 268 2262

f (4) 5 2262

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653Algebra 2


39. 2x4 1 x3 1 (22x2) 1 x 1 0

2 2 1 22 1 0

4 10 16 34

2 5 8 17 34

f (2) 5 34

40. x4 1 0x3 1 (24x2) 1 3x 1 (26)

23 1 0 24 3 26

23 9 215 36

1 23 5 212 30

f (23) 5 30

41. f (x) 1 g(x) 5 4x 2 9 1 6x2

5 6x2 1 4x 2 9

The domain is all real numbers.

42. f (x) 2 g(x) 5 4x 2 9 2 6x2

5 26x2 1 4x 2 9


43. g(x) 2 f (x) 5 6x2 2 (4x 2 9)

5 6x2 2 4x 1 9


44. f (x) p g (x) 5 (4x 2 9)(6x2) 5 24x3 2 54x2


45. f (x)

} g(x)

5 4x 2 9

} 6x2

The domain consists of all real numbers except x 5 0.

46. g(x)

} f (x)

5 6x2

} 4x 2 9

The domain consists of all real numbers except x 5 9 } 4 .

47. f (g(x)) 5 f (6x2) 5 4(6x2) 2 9 5 24x2 2 9


48. g(f (x)) 5 g(4x 2 9) 5 6(4x 2 9)2

5 6(4x 2 9)(4x 2 9)

5 6(16x2 2 72x 1 81) 5 96x2 2 432x 1 486


49. Ï} x 1 3 5 11

Ï} x 5 8

x 5 64

50. Ï}

4x 1 5 5 9

4x 1 5 5 81

4x 5 76

x 5 19

51. Ï}

3x 2 3 } 8 5 0

Ï}

3x 5 3 } 8

3x 5 9 } 64

x 5 3 } 64

52. Ï}

5(x 1 3) 5 10

5(x 1 3) 5 100

5x 1 15 5 100

5x 5 85

x 5 17

53. Ï}

x 2 3 5 x 2 5

x 2 3 5 (x 2 5)2

x 2 3 5 x2 2 10x 1 25

0 5 x2 2 11x 1 28

0 5 (x 2 7)(x 2 4)

x 5 7 or x 5 4

Check x 5 7: Check x 5 4:

Ï}

7 2 3 0 7 2 5 Ï}

4 2 3 0 4 2 5

Ï}

4 0 2 Ï}

1 0 21

2 5 2 ✓ 1 Þ 21

The only solution is 7.

54. Ï}

9x 2 2 5 x 1 2

9x 2 2 5 (x 1 2)2

9x 2 2 5 x2 1 4x 1 4

0 5 x2 2 5x 1 6

0 5 (x 2 3)(x 2 2)

x 5 3 or x 5 2

Check x 5 3: Check x 5 2:

Ï}

9(3) 2 2 0 3 1 2 Ï}

9(2) 2 2 0 2 1 2

Ï}

25 0 5 Ï}

16 0 4

5 5 5 ✓ 4 5 4 ✓

The solutions are 2 and 3.

11.3 Extension (p. 765)

1. } x 5 np 5 24(0.4) 5 9.6

s 5 Ï}

np(1 2 p) 5 Ï}}

24(0.4)(1 2 0.4) 5 2.4

2. } x 5 np 5 40(0.6) 5 24

s 5 Ï}

np(1 2 p) 5 Ï}}

40(0.6)(1 2 0.6) ø 3.1

3. } x 5 np 5 46(0.3) 5 13.8

s 5 Ï}

np(1 2 p) 5 Ï}}

46(0.3)(1 2 0.3) ø 3.1

4. } x 5 np 5 55(0.15) 5 8.25

s 5 Ï}

np(1 2 p) 5 Ï}}

55(0.15)(1 2 0.15) ø 2.6


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5. } x 5 np 5 36(0.7) 5 25.2

s 5 Ï}

np(1 2 p) 5 Ï}}

36(0.7)(1 2 0.7) ø 2.7

6. } x 5 np 5 66(0.2) 5 13.2

s 5 Ï}

np(1 2 p) 5 Ï}}

66(0.2)(1 2 0.2) ø 3.2

7. } x 5 np 5 110(0.08) 5 8.8

s 5 Ï}

np(1 2 p) 5 Ï}}

110(0.08)(1 2 0.08) ø 2.8

8. } x 5 np 5 125(0.35) 5 43.75

s 5 Ï}

np(1 2 p) 5 Ï}}

125(0.35)(1 2 0.35) ø 5.3

9. } x 5 np 5 140(0.75) 5 105

s 5 Ï}

np(1 2 p) 5 Ï}}

140(0.75)(1 2 0.75) ø 5.1

10–12. } x 5 np 5 460(0.04) 5 18.4

s 5 Ï}

np(1 2 p) 5 Ï}

18.4(0.96) ø 4.2

10. P(x ≤ 15) ø P 1 z ≤ 15 2 18.4 } 4.2

2 ø P(z ≤ 20.8)

5 0.2119

11. P(x ≥ 12) ø 1 2 P 1 z ≤ 12 2 18.4 } 4.2

2

ø 1 2 P(z ≤ 21.5)

5 1 2 0.0668 5 0.9332

12. P(6 ≤ x ≤ 18) ø P 1 z ≤ 18 2 18.4 } 4.2

2 2 P 1 z ≤ 6 2 18.4 } 4.2

2 ø P(z ≤ 20.1) 2 P(z ≤ 23.0)

5 0.4602 2 0.0013 5 0.4589

13–15. } x 5 np 5 1221(0.09) ø 110

s 5 Ï}

np(1 2 p) 5 Ï}

110(0.91) ø 10

80 90 100 110 120 130 140

13. P(x ≥ 140) ø 0.0015

14. P(x ≤ 100) ø 0.0015 1 0.0235 1 0.135 5 0.16

15. P(80 ≤ x ≤ 130)

ø 0.0235 1 0.135 1 0.34 1 0.34 1 0.135 5 0.9735

16–18. } x 5 np 5 192(0.25) 5 48

s 5 Ï}

np(1 2 p) 5 Ï}

48(0.75) 5 6

30 36 42 48 54 60 66

16. P(x ≥ 42) 5 1 2 P(x ≤ 42)

ø 1 2 (0.0015 1 0.0235 1 0.135)

5 1 2 0.16

5 0.84

17. P(x ≤ 66) 5 1 2 P(x ≥ 66) ø 1 2 0.0015 5 0.9985

18. P(36 ≤ x ≤ 60) ø 0.135 1 0.34 1 0.34 1 0.135 5 0.95

19. Hypothesis: 85% of people are generally happy. In your survey, 19 out of 26 people, or about 73%, are happy.

n 5 26, p 5 0.85

} x 5 np 5 26(0.85) 5 22.1

s 5 Ï}

np(1 2 p) 5 Ï}}

26(0.85)(0.15) ø 1.82

P(x ≤ 19) ø P 1 z ≤ 19 2 22.1 } 1.82 2 ø P(z ≤ 21.7)

5 0.0446

Because 0.0446 < 0.05, you should reject the hypothesis.

20. Hypothesis: 30% of graduating seniors will buy a class ring. In your survey, 4 out of 15 seniors, or about 27%, are planning to buy a class ring.

n 5 15, p 5 0.3

} x 5 np 5 15(0.3) 5 4.5

s 5 Ï}

np(1 2 p) 5 Ï}

15(0.3)(0.7) ø 1.77

P(x ≤ 4) ø P 1 z ≤ 4 2 4.5 } 1.77 2 ø P(z ≤ 20.3)

5 0.3821

Because 0.3821 > 0.05, you should not rejectthe hypothesis.

21. Hypothesis: 1% of a manufacturer’s computers will fail to operate. At a small business, 2 out of 40 computers, or 5%, fail to operate.

n 5 40, p 5 0.01

} x 5 np 5 40(0.01) 5 0.4

s 5 Ï}

np(1 2 p) 5 Ï}}

40(0.01)(0.99) ø 0.63

P(x ≥ 2) 5 1 2 P(x ≤ 2)

ø 1 2 P 1 z ≤ 2 2 0.4 } 0.63

2 5 1 2 P(z ≤ 2.5)

5 1 2 0.9938

5 0.0062


22. Hypothesis: 80% of people prefer the new apple juice. In a taste test, 12 out of 20, or 60%, prefer the new apple juice.

n 5 20, p 5 0.8

} x 5 np 5 20(0.8) 5 16

s 5 Ï}

np(1 2 p) 5 Ï}

20(0.8)(0.2) ø 1.79

P(x ≤ 12) ø P 1 z ≤ 12 2 16 } 1.79

2 5 P(z ≤ 22.2) 5 0.0139



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655Algebra 2


Lesson 11.4


1. The computer science teacher selected students that are easily accessible. So, the sample is a convenience sample. The sample is biased because students in a computer science class are more likely to visit the school’s website.

2. Sample answer:

Place each student’s name on a small piece of paper in a hat and draw 40 names.

3. Margin of error 5 6 1 }

Ï}

n

5 6 1 }

Ï}

1202

ø 6 0.029

ø 6 2.9%

Margin of error 5 6 1 }

Ï}

n

6 0.02 5 6 1 }

Ï}

n

0.0004 5 1 } n

n 5 2500 people

11.4 Exercises (pp. 769–771)

Skill Practice

1. A sample for which each member of a population has an equal chance of being selected is a random sample.

2. An unbiased sample is representative of the population being sampled, while a biased sample over or under represents part of a population.

3. Because every tenth customer is surveyed, the sample is systematic. Because the sample is representative of the population, it is unbiased.

4. Because the sample is easy to reach, it is a convenience sample. Because people who do not own dogs were not surveyed, the sample is biased.

5. Because each student has an equal chance of being selected, the sample is an unbiased random sample.


Ï}

n

5 6 1 }

Ï}

260

ø 60.062

ø 66.2%


Ï}

n

5 6 1 }

Ï}

1000

ø 60.032

ø 63.2%


Ï}

n

5 6 1 }

Ï}

750

ø 60.037

ø 63.7%


Ï}

n

5 6 1 }

Ï}

6400

ø 60.013

ø 61.3%


Ï}

n

5 6 1 }

Ï}

3275

ø 60.017

ø 61.7%


Ï}

n

5 6 1 }

Ï}

525

ø 60.044

ø 64.4%


Ï}

n

5 6 1 }

Ï}

2024

ø 60.022

ø 62.2%


Ï}

n

5 6 1 }

Ï}

10,000

5 60.01

5 61.0%

14. B; Margin of error 5 6 1 }

Ï}

n

5 6 1 }

Ï}

2000

ø 60.022

ø 62.2%


Ï}

n

60.03 5 6 1 }

Ï}

n

0.0009 5 1 } n

n ø 1111 people


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Ï}

n

60.08 5 6 1 }

Ï}

n

0.0064 5 1 } n

n ø 156 people


Ï}

n

60.1 5 6 1 }

Ï}

n

0.01 5 1 } n

n 5 100 people


Ï}

n

60.042 5 6 1 }

Ï}

n

0.001764 5 1 } n

n ø 567 people


Ï}

n

60.056 5 6 1 }

Ï}

n

0.003136 5 1 } n

n ø 319 people


Ï}

n

60.015 5 6 1 }

Ï}

n

0.000225 5 1 } n

n ø 4444 people


Ï}

n

60.065 5 6 1 }

Ï}

n

0.004225 5 1 } n

n ø 237 people


Ï}

n

60.025 5 6 1 }

Ï}

n

0.000625 5 1 } n

n 5 1600 people

23. D; Margin of error 5 6 1 }

Ï}

n

60.02 5 6 1 }

Ï}

n

0.0004 5 1 } n

n 5 2500 people

24. In the calculation, 4% should be used instead of 13%.

60.04 5 6 1 }

Ï}

n

0.0016 5 1 } n

n 5 625

25. 52.3% 1 61.7%

}} 2 5

114% } 2 5 57%

57% 2 52.3% 5 4.7%

60.047 5 6 1 }

Ï}

n

0.002209 5 1 } n

n ø 453 people

26. 6E 5 6 1 }

Ï}

n

E2 5 1 } n

n 5 1 }

E2

To cut the margin of error in half:

n2 5 1 }

1 1 } 2 E 2 2

5 1 }

1 }

4 E2

5 4 1 1 } E2 2 5 4n

The new sample size must be 4 times as large.

Problem Solving

27. a. Margin of error 5 6 1 }

Ï}

n

5 6 1 }

Ï}

439

ø 0.048

ø 64.8%

b. 14% 2 4.8% 5 9.2%

14% 1 4.8% 5 18.8%

It is likely that between 9.2% and 18.8% of all U.S. teenagers worked during their summer vacation.

28. Sample answer: Assign each student a number from1 to 1225 and use a random number generator to generate 250 unique numbers from the given set of numbers.

29. Sample answer: It is not reasonable to assume that Kosta is going to win the election, because the margin of error is 65%. If the margin of error works in favor of Murdock and against Kosta, Kosta will have 49% (54% 2 5%) and Murdock will have 51% (46% 1 5%).

30. a. 0.23 5 181

} n

n ø 787 students

b. Margin of error 5 6 1 }

Ï}

n

5 6 1 }

Ï}

787

ø 60.036

ø 63.6%

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657Algebra 2



c. 23% 1 3.6% 5 26.6%

23% 2 3.6% 5 19.4%

It is likely that between 19.4% and 26.6% of all students would say that math is their favorite subject.

31. a. Candidate A: 235

} 500

5 47%

Candidate B: 100% 2 47% 5 53%


Ï}

n

5 6 1 }

Ï}

500

ø 60.045

ø 64.5%

c. Candidate A: 47% 6 4.5% 5 42.5% to 51.5%

Candidate B: 53% 6 4.5% 5 48.5% to 57.5%

d. Because the intervals overlap between 48.5% and 51.5%, you cannot be confident that canidate B won.

If the intervals do not overlap:

265 1 x

} 500

2 0.045 > 235 2 x

} 500

1 0.045

0.53 1 x } 500 > 0.47 2

x } 500 1 0.09

20.03 > 2 2x

} 500

215 > 22x

7.5 < x

The number of voters for candidate B must be 265 1 8 5 273 in order for you to be confident of her victory.

32. 52% 2 50% 5 2%

For the interval to contain only values greater than or equal to 50%, the margin of error must be less than 2%.

0.02 > 1 }

Ï} n

0.0004 > 1 }

n

n > 2500 people

So, at least 2501 people would have to be surveyed for you to be confi dent that cola X is truly preferred by more than half the population.

Mixed Review

33. y 5 2 2 } 3 x 1 4

x

y

1

1

(0, 4)

(3, 2)

34. y 5 x2 2 5x 2 24

x 5 2b

} 2a 5 2 (25)

} 2(1)

5 2.5

y 5 1 5 } 2 2 2 2 5 1 5 }

2 2 2 24 5 230.25

Vertex: (2.5, 230.25)

Axis of symmetry: x 5 2.5

5

x

y

2

35. y 5 1 } 4 (x 2 3)2 1 2

Vertex: (h, k) 5 (3, 2)

Axis of symmetry: x 5 3

1

x

y

21

36. f (x) 5 (x 1 1)(x2 2 3x 1 3) x-intercept: 21

6

x

y

1

37. g(x) 5 4 p 2x

Plot (0, 4), (21, 2), and (1, 8).

x

y

4

2

38. h(x) 5 22(0.25)x

Plot (0, 22), (21, 28), and 1 1, 2 1 } 2 2 .

x

y2

2

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39. f (1) f (2) f (3) f (4) f (5) f (6)

29 210 27 0 11 26

21 3 7 11 15

4 4 4 4

a(1)2 1 b(1) 1 c 5 29 → a 1 b 1 c 5 29

a(2)2 1 b(2) 1 c 5 210 → 4a 1 2b 1 c 5 210

a(3)2 1 b(3) 1 c 5 27 → 9a 1 3b 1 c 5 27

Using a calculator to solve the system gives a 5 2, b 5 27, and c 5 24.

A polynomial function that fi ts the data isf (x) 5 2x2 2 7x 2 4.

11.4 Extension (p. 773)

1. Sample answer: This is a leading question. Respondents may think a “no” response means they are not supporters of city growth. Better question: Do you think the city should invest in building a new baseball stadium?

2. Sample answer: This is a leading question. A person may want to be part of the majority. Better question: Do you favor a tax cut?

3. Sample answer: Many patients may answer untruthfully because their dentist is asking the question. The correction is to have the question asked by someone not involved in dental health.

4. Sample answer: The question encourages the respondent to agree. Better question: Would you like to see the old town hall renovated?

5. Sample answer: The question assumes the respondent is familiar with the Carter case. Correction: First give the facts of the Carter case and then ask the question.

6. Sample answer: The question assumes the respondent is familiar with the city council candidates’ platforms. Correction: First state each platform and then ask the question.

7. Sample answer: The fl aw is that Algebra 2 students are the experimental group and the Algebra 1 students are the control group. The experimental and control groups should both be Algebra 2 students.

Lesson 11.5

Investigating Algebra Activity 11.5 (p. 774)

1. The time to complete the activity is decreasing. This pattern makes sense because your partner is familiar with the words on each card and their alphabetical order.

2. Answers will vary.

3. Answers will vary.

4. Answers will vary; While the data is different, the pattern should be similar because the new partner would become more familiar with the words on the cards.


1. The shape of the scatter plot appears to be quadratic. Using the quadratic regression feature, a model isy 5 0.0106x2 1 0.116x 1 21.6.

2. The shape of the scatter plot appears to be linear. Using the linear regression feature, a model is y 5 12.2x 1 30.3.

3. f (x) 5 20.00793x2 1 0.727x 1 13.8

f(70) 5 20.00793(70)2 1 0.727(70) 1 13.8 ø 25.8

The average fuel effi ciency of a car traveling 70 miles per hour is about 25.8 miles per gallon.

4. The shape of the scatter plot appears to be quadratic. Using the quadratic regression feature, a model isy 5 20.000451x2 1 0.404x 2 23.1.

5. The shape of the scatter plot appears to be cubic. Using the cubic regression feature, a model is y 5 0.868x3 1 4.00x2 1 0.902x 2 6.41.

11.5 Exercises (pp. 778–780)

Skill Practice

1. A function of the form y 5 abx is an exponential function.

2. Sample answer: Plot the points in a coordinate plane. If the points appear to lie on a straight line, then a linear model is appropriate. If the points appear to lie on a parabola, then a quadratic model is appropriate.

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659Algebra 2



3. The shape of the scatter plot appears to be quadratic. Using the quadratic regression feature, a model is y 5 20.381x2 1 1.12x 1 15.7.

4. The shape of the scatter plot appears to be linear. Using the linear regression feature, a model is y 5 3.68x 1 23.1.

5. D; The points are level at fi rst and then rapidly increase. This suggests an exponential model.

6. B; After plotting the points, you can see that an exponential model, y 5 126(0.931)x, best fi ts the data.

7. Error: the x and the value of b have been interchanged. The correct model is y 5 9.71(1.55)x.

8. Sample answer:

x 1 3 5 7 8

y 8 2 1 6 10

9. x inches 1 1 foot }

12 inches 2 5

x } 12 feet

y 5 5 1 x } 12

2 2.3

Problem Solving

10. The points are level at fi rst and then rapidly increase. This suggests an exponential model.

y 5 67.4(1.07)x

11. The points increase, then seem to level off briefl y, then increase again. This suggests a cubic model.

y 5 0.00211x3 2 0.0766x2 1 1.26x 2 0.0664

12. The points lie approximately on a line. This suggests a linear model. y 5 0.847x 1 13.7, where x is the number of years since 1975.

13. a.

A quadratic function best models the data.

b. y 5 22.97x2 1 40.4x 2 85.9

c.

d. No, because at 1:00 P.M., the function predicts a negative number of customers.

14. From the scatter plot, you can see that a quadratic model best fi ts the data.

f (x) 5 20.00000357x2 1 0.0365x 2 19.8

Using the maximum feature of the graphing calculator, you can fi nd the maximum of the function. An engine speed of about 5112 rpm maximizes the car’s engine power.

15. The asymptote represents the minimum production cost.The production cost appears to approach $25(horizontal asymptote).

y 5 abx 1 25 or y 2 25 5 abx

Using a graphing calculator, a model for the data is y 5 189.01(0.999)x 1 25.

Mixed Review

16. 6(5) 2 2 5 30 2 2 5 28

17. 8 2 3(26) 5 8 1 18 5 26

18. 43 2 1 5 42 5 16

19. 8(8 2 2) 5 8(6) 5 48

20. 5(4) 2 7 1 2(4) 5 20 2 7 1 8 5 21

21. 6(29) 2 (29 1 5) 5 254 2 (24) 5 254 1 4 5 250

22. log3 5 2 log3 8 5 log3 5 }

8

23. 2 ln 4 1 ln 3 5 ln 42 1 ln 3

5 ln 16 1 ln 3

5 ln(16 p 3)

5 ln 48

24. 2 log x 2 4 log y 5 log x2 2 log y4

5 log x2

} y4

25. 7 log4 x 1 5 log4 y 5 log4 x7 1 log4 y5

5 log4(x7 p y5)

5 log4 x7y5

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26. log3 2 1 1 } 2 log3 y 5 log3 2 1 log3 y1/2

5 log3(2 p y1/2)

5 log3 2 Ï}

y

27. 1 }

4 log5 81 2

1 } 4 log5 4 5 log5 811/4 2 log5 41/4

5 log5 3 2 log5 Ï}

2

5 log5 3 }

Ï}

2

5 log5 3 Ï

}

2 }

2

28.

56 mi

35 mi

c

352 1 562 5 c2

4361 5 c2

66.04 ø c Your house is about 66.04 miles away from the radio

station, which is farther than the signal can reach.

Quiz 11.3–11.5 (p. 780)

1–3.

29 35 41 47 53 59 65

1. P(35 ≤ x ≤ 65)

5 0.135 1 0.34 1 0.34 1 0.135 1 0.0235 5 0.9735

2. P(x ≥ 41) 5 1 2 P(x ≤ 41)

5 1 2 (0.135 1 0.0235 1 0.0015)

5 0.84

3. P(x ≤ 29) 5 0.0015


Ï}

n

60.03 5 6 1 }

Ï}

n

0.0009 5 1 } n

n ø 1111 people


Ï}

n

60.07 5 6 1 }

Ï}

n

0.0049 5 1 } n

n ø 204 people


Ï}

n

60.045 5 6 1 }

Ï}

n

0.002025 5 1 } n

n ø 494 people


Ï}

n

60.008 5 6 1 }

Ï}

n

0.000064 5 1 } n

n ø 15,625 people

8. From the scatter plot, you can see that the points lie approximately on a line. This suggests a linear model. y 5 0.164x 2 20.11

Problem Solving Workshop 11.5 (p. 781)

1. The temperature of the soup appears to decay to 748F, so the temperature of the room is about 748F.

y 5 abx 1 74

y 5 61.3(0.962)x 1 74

2. The temperature of the water appears to warm up to about 198C, so the temperature of the room isabout 198C.

y 5 abx 2 19

y 5 24.88(1.004)x 2 19

Mixed Review of Problem Solving (p. 782)

1. a. From a scatter plot, you can see that a quadratic model best fi ts the data.

y 5 0.0179x2 2 0.313x 1 1.70

b. y ø 0.0179(20)2 2 0.313(20) 1 1.70

5 7.16 2 6.26 1 1.70 5 2.6

A Maine land locked salmon that is 20 inches long weighs about 2.6 pounds.

2. a. Margin of error 5 6 1 }

Ï}

n

5 6 1 }

Ï}

1022

ø 60.031

ø 63.1%

b. 73% 2 3.1% 5 69.9%

73% 1 3.1% 5 76.1%

An interval is 69.9% to 76.1%.

3. z 5 x 2 } x

} s

5 220 2 200

} 20 5 1

P(x ≥ 220) 5 1 2 P(x ≤ 220)

ø 1 2 P(z ≤ 1)

5 1 2 0.84 5 0.16 5 16%

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661Algebra 2



4. Sample answer:

x 23 22 0 1 3 5

y 1 1.2 2 3 7 16

5. a. From a scatter plot, you can see that a quadratic model best fi ts the data.

y 5 20.044x2 1 1.63x 2 3.17

b. y 5 20.044(17)2 1 1.63(17) 2 3.17

5 11.824 ø 12

At age 17 his shoe size will be about size 12.

c. After a certain age, a person’s feet will stop growing and his shoe size will remain constant.

6. 68% represents one standard deviation on either side of the mean.

10.5 6 0.75 5 9.75 to 11.25

So, between 9.75 ounces and 11.25 ounces represents 68% of the amounts dispensed.

7. 69 minutes is two standard deviations to the right of the mean, 45 minutes.

P( } x ≤ x ≤ } x 1 2s) 5 P(45 ≤ x ≤ 69)

5 0.34 1 0.135 5 0.475

The probability that a shopper will spend between 45 and 69 minutes in the store is about 0.475.

8. a. 15

} 100

n 5 315

n 5 2100 students


Ï}

n

5 6 1 }

Ï}

2100

ø 60.022

ø 62.2%

c. 15% 1 2.2% 5 17.2%

15% 2 2.2% 5 12.8%

It is likely that between 12.8% and 17.2% of all students would prefer to have gym class during the last period of the day.

9. The station selected people that are easily accessible. So, the sample is a convenience sample. The sample is biased because people in a sports stadium are more likely to watch sporting events on TV than people in general.

Chapter 11 Review (pp. 784–786)

1. Standard deviation is a measure of dispersion that describes the typical difference between a value in a data set and the mean.

2. When every value in a data set is multiplied by the same constant, the mean, median, mode, range, and standard deviation are also multiplied by the constant.

3. The z-score for an x-value from a normal distribution represents the number of standard deviations the x -value lies above or below the mean.

4. Mean: } x 5 35 1 36 1 . . . 48

}} 8 ø 40.1

Median: 38 1 41

} 2 5 39.5

Mode: 36

Range: 48 2 35 5 13

Standard deviation:

s 5 Ï}}}}

(35 2 40.1)2 1 (36 2 40.1)2 1 . . . 1 (48 2 40.1)2

}}}} 8

ø 4.4

5. Mean: } x 5 75 1 76 1 . . . 1 92

}} 8 ø 84.1

Median: 85 1 88

} 2 5 86.5

Mode: 88

Range: 92 2 75 5 17

Standard deviation:

s 5 Ï}}}}

(75 2 84.1)2 1 (76 2 84.1)2 1 . . . 1 (92 2 84.1)2

}}}} 8

ø 6.2

6. Mean: } x 5 76 1 85 1 . . . 1 102

}} 8 5 90.1

Median: 91

Mode: 91

Range: 102 2 76 5 26

Standard deviation:

s 5 Ï}}}}

(76 2 90.1)2 1 (85 2 90.1)2 1 . . . 1 (102 2 90.1)2

}}}} 8

ø 7.3

7. Mean: } x 5 103 1 115 1 . . . 1 155

}} 7 ø 129.7

Median: 130

Mode: 140

Range: 155 2 103 5 52

Standard deviation:

s 5 Ï}}}}}

(103 2 129.7)2 1 (115 2 129.7)2 1 . . . 1 (155 2 129.7)2

}}}}} 7

ø 16.1

8. Order: 0.97, 1.04, 1.13, 1.13, 1.13, 1.14, 1.26, 1.30, 1.47, 1.47, 1.59

Median: 1.14

Mean: 0.97 1 1.04 1 . . . 1 1.59

}} 11

ø 1.24

Standard deviation:

s 5 Ï}}}}}

(0.97 2 1.24)2 1 (1.04 2 1.24)2 1 . . . 1 (1.59 2 1.24)2

}}}}} 11

ø 0.19

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9.Originaldata set

Adding 27to data values

Mean 39.8 39.8 2 7 5 32.8

Median 38 38 2 7 5 31

Mode 37 37 2 7 5 30

Range 14 14

Std. dev. 4.6 4.6

10.Originaldata set


Mean 72.4 72.4(1.2) 5 86.88

Median 74 74(1.2) 5 88.8

Mode 66 66(1.2) 5 79.2

Range 20 20(1.2) 5 24

Std. dev. 6.7 6.7(1.2) 5 8.04

11.Rainfall in

millimetersRainfall in

inches

Mean 37.95 37.95(0.03937) ø 1.49

Median 35.35 35.35(0.03937) ø 1.39

Mode 59.8 59.8(0.03937) ø 2.35

Range 62.6 62.6(0.03937) ø 2.46

Std. dev. 20.8 20.8(0.03937) ø 0.82

12. z 5 x 2 } x

} s

5 89 2 95

} 7 ø 20.9

P(x ≤ 89) ø P(z ≤ 20.9) 5 0.1841

13. z 5 x 2 } x

} s

5 84 2 95

} 7 ø 21.6

P(x ≤ 84) ø P(z ≤ 21.6) 5 0.0548

14. z 5 x 2 } x

} s

5 91 2 95

} 7 ø 20.6

z 5 x 2 } x

} s

5 100 2 95

} 7 ø 0.7

P(91 < x ≤ 100) ø P(20.6 < z ≤ 0.7)

5 0.7580 2 0.2743 5 0.4837

15. z 5 x 2 } x

} s

5 50 2 95

} 7 ø 26.4

P(x ≤ 50) ø P(z ≤ 26.42) ø 0

16. z 5 x 2 } x

} s

5 1002 95

} 7 ø 0.7

P(x > 100) ø P(z > 0.7)

5 1 2 P(z ≤ 0.7)

5 1 2 0.758

5 0.242

17. z 5 x 2 } x

} s

5 50 2 95

} 7 ø 26.4

z 5 x 2 } x

} s

5 80 2 95

} 7 ø 22.1

P(50 < x ≤ 80) ø P(26.4 < z ≤ 22.1)

5 0.0179 2 0

5 0.0179


Ï}

n

5 6 1 }

Ï}

300

ø 60.058

5 65.8%


Ï}

n

5 6 1 }

Ï}

2500

5 60.02

5 62%


Ï}

n

5 6 1 }

Ï}

800

ø 60.035

5 63.5%


Ï}

n

5 6 1 }

Ï}

4900

ø 60.014

5 61.4%


Ï}

n

5 6 1 }

Ï}

517

ø 60.044

5 64.4%

23. The scatter plot appears to be linear. Using the linear regression feature, a model is y 5 23.79x 1 28.3.

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663Algebra 2



Chapter 11 Test (p. 787)

1.Originaldata set


Mean 41.75 41.75(3) 5 125.25

Median 41.5 41.5(3) 5 124.5

Mode 41 41(3) 5 123

Range 13 13(3) 5 39

Std. dev. 3.80 3.80(3) 5 11.4

2.Originaldata set

Adding 14 todata values

Mean 19 19 1 14 5 33

Median 18.5 18.5 1 14 5 32.5

Mode 21 21 1 14 5 35

Range 10 10

Std. dev. 3.04 3.04

3.Originaldata set


Mean 101.5 101.5(4.5) ø 456.8

Median 100.5 100.5(4.5) ø 452.3

Mode 92 92(4.5) 5 414

Range 24 24(4.5) 5 108

Std. dev. 8.6 8.6(4.5) 5 38.7

4–6.

57 62 67 72 77 82 87

4. P(67 ≤ x ≤ 77) 5 0.34 1 0.34 5 0.68

5. P(57 ≤ x ≤ 72) 5 0.5 2 0.0015 5 0.4985

6. P(x ≥ 62) 5 1 2 P(x ≤ 62)

5 1 2 (0.0015 1 0.0235)

5 0.975


Ï}

n

5 6 1 }

Ï}

340

ø 60.054

5 65.4%


Ï}

n

5 6 1 }

Ï}

8125

ø 60.011

5 61.1%


Ï}

n

5 6 1 }

Ï}

931

ø 60.033

5 63.3%


Ï}

n

5 6 1 }

Ï}

1560

ø 60.025

5 62.5%

11. a. Mean: } x 5 3 1 3 1 . . . 1 48

}} 15 ø 13.7

Median: 13

Modes: 3, 4, 13, 14, 17, 19

Range: 48 2 3 5 45

Standard deviation:

s 5 Ï}}}}

(3 2 13.7)2 1 (3 2 13.7)2 1 . . . 1 (48 2 13.7)2

}}}} 15

ø 10.7

b. Mean: } x 5 2 1 3 1 . . . 1 41

}} 15 5 14

Median: 11

Mode: 17

Range: 41 2 2 5 39

Standard deviation:

s 5 Ï}}}}

(2 2 14)2 1 (3 2 14)2 1 . . . 1 (41 2 14)2

}}}} 15

ø 10.1

c. Sample answer: The data is very similar except the margins of victory for the AFC are slightly more spread out.

12. a. z 5 x 2 } x

} s

5 55 2 50

} 10 5 0.5

b. z 5 x 2 } x

} s

5 70 2 50

} 10 5 2.0

c. z 5 x 2 } x

} s

5 40 2 50

} 10 5 21.0

d. z 5 x 2 } x

} s

5 47 2 50

} 10 5 20.3

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Ï}

n

5 6 1 }

Ï}

1600

5 60.025

5 62.5%

61% 1 2.5% 5 63.5%

61% 2 2.5% 5 58.5%

It is likely that between 58.5% and 63.5% of all U.S. adults have purchased a product online.

14. From a scatter plot, you can see that an exponential model best fi ts the data. y 5 48.9(0.967)x

Standardized Test Preparation (p. 789)

1. Sample answer: You can eliminate choice D because a probability of 0.99 is very close to 1, which is the total area under the normal curve. Values between 4.06 and 4.14 do not represent the whole normal curve.

2. Sample answer: You can eliminate choice C because the mode increases by the constant. The range and standard deviation do not change.

3. Sample answer: You can eliminate choice D because the points do not strictly increase or decrease; rather they lie in a U-shape (quadratic function).

Standardized Test Practice (pp. 790–791)

1. B; order: 94, 94, 97, 101, 109, 110, 113, 114, 136, 166

Median: 109 1 110

} 2 5 109.5 min

2. B;

s 5 Ï}}}}}

(94 2 113.4)2 1 (94 2 113.4)2 1 . . . 1 (166 2 113.4)2

}}}}} 10

ø 21.2 min

3. B;

55.9

58.6

61.3

64.0

66.7

69.4

72.1

P(58.6 ≤ x ≤ 66.7) 5 0.135 1 0.34 1 0.34 5 0.815

(0.815)4 ø 0.44 5 44%

4. C; 34% 1 34% 1 13.5% 1 2.35% 5 83.85%

5. A; z 5 x 2 } x

} s

5 88 2 79.3

} 7.5 5 1.16

6. B; Mean: } x 5 384 1 480 1 . . . 1 768

}} 5 5 569.6

Median: 576

Mode: no mode

Range: 768 2 384 5 384

7. C; Margin of error 5 6 1 }

Ï}

n

5 6 1 }

Ï}

1015

5 60.031

ø 63.1%

8. A; From a scatter plot, you can see that the points lie approximately on a line, suggesting a linear model.

9. D; The mean changes from 17.5 to 36.4.

The median changes from 15 to 20.

The mode stays 0.

The range changes from 45 to 150.

The range is most affected.

10. z 5 x 2 } x

} s

5 80 2 77

} 4 5 0.75

11. The standard deviation does not change. It is 12.2.

12. } x 5 1 1 3 1 . . . 1 25

}} 14 5 189

} 14 5 13.5

13. 13.5% 1 34% 1 34% 1 13.5% 5 95%


Ï}

n

6 0.02 5 6 1 }

Ï}

n

0.0004 5 1 } n

n 5 2500

About 2500 adults were surveyed. You do not know for certain that more than 50% of adults follow football because of the margin of error. It is possible that 51% 2 2%, or 49% could follow football.

15. The sample is a biased convenience sample because the sample represents only students in his address book who use the Internet. Students who do not use the Internet or have access to email were not included in the survey.

16. a.

San Antonio(in.)

Chicago(mm)

Mean 2.4 76.875

Median 2.35 75.65

Mode 1.8 no mode

Range 1.9 61.6

Std. dev. 0.62 18.79

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665Algebra 2



b. Chicago

(mm)Chicago

(in.)

Mean 76.875 76.875(0.03937) ø 3.03

Median 75.65 75.65((0.0397) ø 2.98

Mode no mode no mode

Range 61.6 61.6(0.03937) ø 2.43

Std. dev. 18.79 18.79(0.03937) ø 0.74

Sample answer: Chicago gets more rain on averagethan San Antonio.

17. a.

From the scatter plot, you can see that a linear function best fi ts the data.

b. y 5 0.986x 1 46.6

c.

d. The model predicts that the number of voters in 2004 was y 5 0.986(64) 1 46.6 5 109.704 million. The equation’s prediction is low.

n2ws-1100.indd 665 6/27/06 11:30:00 AM

aat solutions - ch11.pdf

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b mean

b median

middle number of n numbers

number of factors

increasing order

number of odds

number of multiples

average of n numbers