1. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
(For help, go to Skills Handbook page 845.)
Simplify.
1. –(–7.2)
2. 1 – (–3)
3. –9 + (–4.5)
4. (–3.4)(–2)
5. –15 ÷ 3
6. – (– ) 2 5 3
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2. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
Solutions
1. –(–7.2) = 7.2
3. –9 + (–4.5) = –13.5
5. –15 ÷ 3 = –5
6. – (– ) = = – = –1 2 5 3
–2 + (–3)
2. 1 – (–3) = = 4
4. (–3.4)(–2) = 6.8
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3. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
A pilot uses the formula d = a a to find the number of miles d that can be seen when flying at an altitude of a miles. Which set of numbers best describes the values for each variable?
The pilot’s instrument panel will display the altitude a as a rational number in the form of a terminating decimal.
The value of d obtained by replacing a with a rational number can be either irrational or rational (consider a = 1.47).
So, the values of d are best described as real numbers.
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4. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
Graph the numbers – , , and 3.6 on a
number line. 3 4
– is between –1 and 0. 3 4
Use a calculator to find that
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5. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
Compare –9 and – Use the symbols < and >.
9 = 3, so – = –3.
Since –9 < –3, it follows that
–9 < –
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6. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
Find the opposite and the reciprocal of each number. 1 7
a. –3
b. 4
Opposite: –(–3 ) = 3 1 7
Opposite: –4
Reciprocal: = = – 7 22 1 –3 –
Reciprocal: 1 4
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7. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
Which property is illustrated?
a. (–7)(2 • 5) = (–7)(5 • 2) b. 3 • (8 + 0) = 3 • 8
The given equation is true because 2 • 5 = 5 • 2.
The given equation is true because = 8.
So, the equation uses the Commutative Property
of Multiplication.
This is an instance of the Identity Property of Addition.
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8. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
Simplify | 4 |, |–9.2|, and |3 – 8|. 1 3
4 is 4 units from 0, so | 4 | = 4 . 1 3
–9.2 is 9.2 units from 0, so |–9.2| = 9.2.
|3 – 8| = |–5| and –5 is 5 units from 0. So, |–5| = 5, and hence |3 – 8| = 5.
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9. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
Pages 8–10 Exercises
1. natural numbers, whole numbers, integers, rational numbers, real numbers
2. irrational numbers, real numbers
3. irrational numbers, real numbers
4. integers, rational numbers, real numbers
5. whole numbers, integers, rational numbers, real numbers
6. rational numbers,
real numbers
7. rational numbers,
8. irrational numbers, real numbers
9. whole numbers, rational numbers
10. natural numbers, rational numbers
11. real numbers
12.
13.
14.
15.
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10. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
16.
17. >
18. =
19. >
20. <
21. <
22. =
23. >
24. <
25. – > – , – < –
< 0.39, > 0.075
27. –2.3 < 2.1, 2.1 > –2.3
28. –5.2 < –4.8, –4.8 > –5.2
< 3.4, 3.4 > 3.04
< ,
0.4 > 0.4
31. –4 < – 4, – 4 > – 4
< 7, > 5
33. – 3 > – 5
– 5 < – 3
34. –200,
35. –3 ,
, –100 1 4 1 3 1 3 1 4 1
200 3 5 5 18
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11. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1 7 2 – 2 7
37.
38. – 3,
39. –2 ,
, –
41. 3 – ,
42. Identity Prop. Of Mult. ,
43. Dist. Prop.
44. Comm. Prop. of Add.
45. Assoc. Prop. of Mult.
46. Comm. Prop. of Mult.
47. Identity Prop. of Add.
48. Inverse Prop. of Add.
49. Inverse Prop. of Mult.
50. Dist. Prop.
51. Comm. Prop. of Mult.
52. Assoc. Prop. of Add.
55. –25
57.
58. 3 1 3 3 or 1 2 50
117 1
– 3 1 3
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12. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1 1 3
59. 3
60. –2
61–68. Answers may vary. Samples are given.
61. –5
62. –3
63. –1
64.
65. 1
66. 3
67. 4
69. natural numbers, whole numbers, integers, rational numbers, real numbers
70. irrational numbers, real numbers
71. irrational numbers, real numbers
72. rational numbers, real numbers
73. irrational numbers, real numbers
74. irrational numbers, real numbers
75. >
76. >
77. <
78. > 1 2 1 4 1 2 2 3
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13. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
79. <
80. <
81. <
82. <
83. Answers may vary. Sample: 4 is a whole
number, but is not
a whole number.
84. Answers may vary. Sample: 7 is a natural number, but –7 is not a natural number.
85. 0 is a whole number, and since –0 = 0, the opposite of 0 is a whole number.
86. Answers may vary. Sample: The integer –1 has –1 as its reciprocal, so –1 is an integer whose reciprocal is an integer.
87. Answers may vary. Sample: 2 and 2 are irrational numbers, but their product (2) is a rational number.
88. Check students’ work.
89. All except the Identity Prop. of Add. (since 0 is not in the set of natural numbers) and the inverse properties 1 4
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14. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
90. all except the Inverse Prop. of Mult. and the Inverse Prop. of Add.
91. all except the Inverse Prop. of Mult.
92. all the properties
93. Comm., Assoc., and Distr. Prop.
94. No; explanations may vary. Sample: The only pairs of integers that have a product of –12 are –1 and 12, –2 and 6, –3 and 4, –4 and 3, –6 and 2, and –12 and 1. None of these pairs has a sum of –3.
95. D
96. H
97. C
98. I
99. A
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15. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
103. –3.8
105. 0
106. –0.4
107. 7
, or 2
, or 11
,or 1
100. [2] The opposite of the reciprocal of 5 is the
opposite of , or – , and the reciprocal of the
opposite of 5 is the reciprocal of –5, or – .
[1] includes a statement or example that a reciprocal is a multiplicative inverse or that a number times its reciprocal is 1, OR a statement or example that an opposite is an additive inverse or that a number plus its opposite is 0
101. [2] 1 and –1; for a number n to be equal to its
reciprocal, n = , which means n2 = 1. So
n = 1 or –1.
[1] only includes answer 1 and –1 with no explanation 1 5 n 9 4 35 3 2
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16. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1
111. 5
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17. Properties of Real Numbers
ALGEBRA 2 LESSON 1-1 7 5
1. Graph –2.3, , and on a number line.
2. Replace each with the symbol <, >, or = to make the sentence true.
a. 
b. 
3. Fit the opposite and the reciprocal of 1 .
4. Name the property of real numbers illustrated by the equation.
(–2)( ) = (–2)( )
5. Simplify |9 – (–5)|.
< 2 9 = 5 8
–1 , 5 8
8 13
Comm. Prop. of Add. 14
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18. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
(For help, go to Skills Handbook page 845.)
Use the order of operations to simplify each expression.
1. 8 • 3 – 2 • 4
2. 8 – ÷ 3
3. 24 ÷ 12 • 4 ÷ 3
4. 3 • ÷ 4
÷ 9 –
6. ( ) ÷ 8 – (23 + 1)
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19. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
1. 8 • 3 – 2 • 4 = (8 • 3) – (2 • 4) = 24 – 8 = 16
2. 8 – ÷ 3 = 8 – 4 + (6 ÷ 3) = 8 – = (8 – 4) + 2 = = 6
÷ 12 • 4 ÷ 3 = (24 ÷ 12) • 4 ÷ 3 = 2 • 4 ÷ 3 = (2 • 4) ÷ 3 = 8 ÷ 3 = or 2
4. 3 • ÷ 4 = 3 • ÷ 4 + (3 • 64) + (12 ÷ 4) = = 195
÷ 9 – = ÷ 9 – = 27 + (18 ÷ 9) – = – = (27 + 2) – = 29 – = (29 – 9) + 1 = = 21
6. ( ) ÷ 8 – (23 + 1) = ( ) ÷ 8 – (8 + 1) = 64 ÷ 8 – 9 = (64 ÷ 8) – 9 = 8 – 9 = –1
Solutions 8 3 2 3
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20. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
Evaluate 7x – 3xy for x = –2 and y = 5.
7x – 3xy = 7(–2) – 3(–2) (5) Substitute –2 for x and 5 for y.
= –14 – (–30) Multiply first.
= – To subtract, add the opposite.
= 16 Add.
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21. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
Evaluate (k – 18)2 – 4k for k = 6.
(k – 18)2 – 4k = (6 – 18)2 – 4(6) Substitute 6 for k.
= (–12)2 – 4(6) Subtract within parentheses.
= 144 – 4(6) Simplify the power.
= 144 – 24 Multiply.
= 120 Subtract.
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22. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
The expression –0.08y 2 + 3y models the percent increase of Hispanic voters in a town from 1990 to In the expression, y represents the number of years since Find the approximate percent of increase of Hispanic voters by 1998.
Since 1998 – 1990 = 8, y = 8 represents the year 1998.
–0.08y2 + 3y = –0.08(8)2 + 3(8)  Substitute 8 for y 19
The number of Hispanic voters had increased by about 19%.
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23. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
Simplify by combining like terms.
2h – 3k + 7(2h – 3k)
2h – 3k + 7(2h – 3k) = 2h – 3k + 14h – 21k Distributive Property
= 2h + 14h – 3k – 21k Commutative Property
= (2 + 14)h – (3 + 21)k Distributive Property
= 16h – 24k
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24. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
Find the perimeter of this figure. Simplify the answer.
P = c d + (d – c) + d c + d c 2
= c d + d – c + d c + d c 2
= c + 4d c 2
= c + 4d 2c 2
= c + c + 4d
= 2c + 4d
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25. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
Pages 15–17 Exercises
1. –30
2. 26
3. 368
4. –16
5. –16
6. 28
7. –70
8. –12
9. 1 ft
10. 4 ft ft ft mm mm mm mm
17. $1210
18. $1331
19. $
20. $
21. 4a
22. 2s + 5
23. –9a + b
24. 6a + 3b
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26. Algebraic Expressions
ALGEBRA 2 LESSON 1-2 41 4
25. 10r + 5s
26. 6w + 5z
27. 2x2 + x
28. xy +3x
29. –0.5x
30. 6x – 5
31. 10y + x – 4
32. –y – 6x
33. –3a + 15b
34. 4g – 2
35. –3x + 4y – z
36. 4a
37. 4a
38. 5
39. 17
40. 41
41. 66
42. –1
43.
44. 10
45. –765
46. a. about 180 million voters
b. about 242 million voters; about 263 million voters
c. –0.0078y y
d. about 87 million
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27. Algebraic Expressions
ALGEBRA 2 LESSON 1-2 3 4
47. – a2 + 2b2
48.
50. y
51. 3x + 6y
52. 4x + 2y
53. –2x2 + 2y2
54. F
55. C
56. A
57. G
58. B
59. H
60. E
61. I
62. D
63. Assoc. Prop. of Add., Comm. Prop. of Add., Assoc. Prop. of Add., Identity Prop. of Mult., Distr. Prop. of Add.
64. Distr. Prop.; addition; Distr. Prop.
65. Answers may vary. Sample: x3 + x2 + x – x3 – 2x, –x2 + 2x2 + 3x – 3x – x, x2 – 5x + 4x – x5 + x5, 3x2 – x2 – x2 + 7x – 8x
5x2 2
7y 2 12 2y 15
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28. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
66. Answers may vary.
Sample:  2(b – a) + 5(b – a) = (2 + 5) (b – a) Dist. Prop.
= 7 (b – a) Addition
= 7b – 7a Dist. Prop.
67. a. 18
b. 2x2, 18
c. Properties of operations were used to simplify the original expression. Those properties convert one expression into an equivalent expression, and equivalent expressions have equal values for all replacements of their variables.
d. No; explanations may vary. Sample: Students should check each step in the simplification; they also should substitute more than one value for x in the original and simplified expressions.
68. B
69. G
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29. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
70. [2] x = 0 or x = 3 since for 6x(x – 3) = 0 either 6x or x – 3 must be equal to 0.
[1] x = 0 and x = 3 with no explanation
71. A
72. D
73. C
74. –1.5, – 2, –1.4, –0.5
75. –4.3, –|3.4|,
|–3.4|, |–4.3|
76. – , – , – ,
77. – , – , ,
78. >
79. >
80. >
81. <
82. <
83. = 5 6 3 4 3 8 1 2 1 8 1 10 1 16 1 4
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30. Algebraic Expressions
ALGEBRA 2 LESSON 1-2
1. Evaluate each expression for the given values of the variables.
a. 4p – 3q + 8; p = –6, q = –10
b. –3m2 – (m + n)2; m = 2, n = –4
2. Simplify by combining like terms.
a. 7 – 4(x + y) + 2(x – 3y)
b. 3y2 + 8 – (2y2 + 12)
c. –12(x2 + y) + 3(y + x2)
3. Find the perimeter. Simplify the answer. 14
–16
7 – 2x – 10y
y2 – 4
–9x2 – 9y
8c + 4d
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31. Simplify each expression.
Solving Equations
ALGEBRA 2 LESSON 1-3
(For help, go to Lesson 1-2.)
Simplify each expression.
1. 5x – 9x + 3
2. 2y + 7x + y – 1 x 3 y 3 2y 3
3. 10h + 12g – 8h – 4g
– y
5. (x + y) – (x – y)
6. –(3 – c) – 4(c – 1)
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32. 3. 10h + 12g – 8h – 4g = (12 – 4)g + (10 – 8)h = 8g + 2h
Solving Equations
ALGEBRA 2 LESSON 1-3
1. 5x – 9x + 3 = (5 – 9)x + 3 = –4x + 3
2. 2y + 7x + y – 1 = 7x + (2 + 1)y – 1 = 7x + 3y – 1
3. 10h + 12g – 8h – 4g = (12 – 4)g + (10 – 8)h = 8g + 2h
– y = – = =
= =
5. (x + y) – (x – y) = (x + y) + (y – x) = (x – x) + (y + y) = 0 + 2y = 2y
6. –(3 – c) – 4(c – 1) = (c – 3) – 4c – 4(–1) = c – 3 – 4c + 4 = (1 – 4)c
+ (–3 + 4) = –3c + 1
Solutions x 3 y 2y 3y
x + y + 2y – 3y
x + (1 + 2 – 3)y
x + 0y
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33. 5x + 3 = –12 Subtract 2x from each side.
Solving Equations
ALGEBRA 2 LESSON 1-3
Solve 7x + 3 = 2x – 12.
7x + 3 = 2x – 12
5x + 3 = –12 Subtract 2x from each side.
5x = –15 Subtract 3 from each side.
x = –3 Divide each side by 5.
Check: 7x + 3 = 2x – 12
7(–3) (–3) – 12
–18 = –18
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34. 4m + 36 = –3m + 12 Distributive Property
Solving Equations
ALGEBRA 2 LESSON 1-3
Solve 4(m + 9) = –3(m – 4).
4(m + 9) = –3(m – 4)
4m = –3m + 12 Distributive Property
7m = 12 Add 3m to each side.
7m = –24 Subtract 36 from each side.
m = – Divide each side by 7. 24 7
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35. A = 2 w + 2 h + 2wh Distributive Property
Solving Equations
ALGEBRA 2 LESSON 1-3
The formula for the surface area of a rectangular prism units long, w units wide, and h units high is A = 2( w + h + wh). Solve the formula for w.
A = 2( w + h + wh)
A = 2 w h + 2wh Distributive Property
A – 2 h = 2 w + 2wh Subtract 2 h from each side.
A – 2 h = ( h)w Distributive Property
= w Divide both sides by h.
A – 2 h h
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36. Solve + 8 = b for x. Find any restrictions on a and b.
Solving Equations
ALGEBRA 2 LESSON 1-3
Solve = b for x. Find any restrictions on a and b. x a
+ 8 = b x a
a( ) + a(8) = ab Multiply each side by the least common denominator (LCD), a. x a
x + 8a = ab Simplify.
x = ab – 8a Subtract 8a from each side. = /
The denominator cannot be zero, so a
1-3

37. Relate: 2 • width + length = perimeter Define: Let w = the width.
Solving Equations
ALGEBRA 2 LESSON 1-3
Adrian will use part of a garage wall as one of the long sides of a rectangular rabbit pen. He wants the pen to be 3 times as long as it is wide. He plans to use 68 ft of fencing. Find the dimensions of the pen.
Relate: 2 • width + length = perimeter
Define: Let w = the width.
Then 3w = the length.
Write: 2 w + 3w = 68
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38. The width is 13 ft and the length is 40 ft.
Solving Equations
ALGEBRA 2 LESSON 1-3
(continued)
5w = 68 Add.
w = 13 Divide each side by 5. 3 5
3w = 40 Find the length. 4 5
Check: Is the answer reasonable? Since the dimensions are about 14 ft by 41 ft and = 69, the perimeter is about 69 ft. The answer is reasonable. 3 5
The width is ft and the length is ft. 4
1-3

39. Relate: Perimeter equals the sum of the lengths of the four sides.
Solving Equations
ALGEBRA 2 LESSON 1-3
The sides of a quadrilateral are in the ratio 1 : 2 : 3 : 6. The perimeter is 138 cm. Find the lengths of the sides.
Relate: Perimeter equals the sum of the lengths of the four sides.
Define: Let x = the length of the shortest side.
Then 2x = the length of the second side.
Then 3x = the length of the third side.
Then 6x = the length of the fourth side.
1-3

40. Check: Is the answer reasonable? Since 12 + 23 + 35 + 69 = 139,
Solving Equations
ALGEBRA 2 LESSON 1-3
(continued)
Write: 138 = x + 2x + 3x + 6x
138 = 12x Combine like terms.
11.5 = x
2x = 2(11.5)  3x = 3(11.5)  6x = 6(11.5)   Find the length of = = = 69 each side.
Check:  Is the answer reasonable? Since = 139,
the answer is reasonable.
The lengths of the sides are 11.5 cm, 23 cm, 34.5 cm, and 69 cm.
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41. Relate: distance first plane travels = distance second plane travels.
Solving Equations
ALGEBRA 2 LESSON 1-3
A plane takes off from an airport and flies east at a speed of 350 mi/h. Thirty-five minutes later, a second plane takes off from the same airport and flies east at a higher altitude at a speed of 400 mi/h. How long does it take the second plane to overtake the first plane?
Relate: distance first plane travels = distance second plane travels.
Define: Let t = the time in hours for the second plane.
Then t = the time in hours for the first plane. 35 60
Write: 400t = 350 (t ) 7 12
400t = 350t + Distributive Property
1225 6
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42. Solving Equations (continued) 50t = Solve for t.
ALGEBRA 2 LESSON 1-3
(continued)
50t = Solve for t.
1225 6
t = h or about 4 h 5 min 1 12
Check:  Is the answer reasonable? In 4 h, the second plane travels 1600 mi. In 4 h, the first plane travels about 1600 mi. The answer is reasonable. 2 3
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43. Solving Equations Pages 21–24 Exercises 1. 23 2. 8 3. 4. –5 5. 6. – 7.
ALGEBRA 2 LESSON 1-3
Pages 21–24 Exercises
1. 23
2. 8 3.
4. –5 5.
6. – 7. 8.
9. 8
10. –6
11. 2
12. 0
13. –
14. –6
15. 2
16. 17 2 7 1 9 3
17. h =
18. g =
19. w =
20. r =
21. r =
22. h =
23. x = , a –b
24. x = , b c 2A b 2s t2 V lh I pt S
2 h
r 2 c
a + b
c – b = /
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44. Solving Equations 25. x = a(c – b) or ac – ab, a 0
ALGEBRA 2 LESSON 1-3 2 3 2 3 2 3 27 5 2 5
25. x = a(c – b) or ac – ab, a 0
26. x = a(b + 5) or ab + 5a, a 0
27. x = 2(m + n) + 2 or 2m + 2n + 2
28. x = – 1
29. 4 h
mi/h, 600 mi/h
31. width = 4.5 cm, length = 7.5 cm
in., 5 in., 6 in.
cm, 11 cm, cm, 16.5 cm
cm, 10 cm, cm
35. a. x + (x + 1) + (x + 2) = 90; 29, 30, 31
b. (x – 1) + x + (x + 1) = 90; 29, 30, 31
, or 1
37. 3
, or 5
39.
40. 30
, or
42. R =
43. r2 =
44. h =
45. v = = / 3 2 = / 79 80
r1r2
r1 + r2 5g 2
Rr1
r1 – R
S – 2 r 2
2 r 46 39 7 39
h + 5t2 t
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45. Solving Equations 46. h = 47. b2 = – b1 48. 40°, 140° 49. 34°, 56°
ALGEBRA 2 LESSON 1-3
2(v – s2) s
10c a
46. h =
47. b2 = – b1
48. 40°, 140°
49. 34°, 56°
50. about mi m
52. $360; $746.40
53. 43, 45, 47, 49
54. 34, 36, 38, 40
55. x = ab – b2 – a, b 0
56. x = , b d
57. x = , a c
58. x = , a b
59. x = , b c
60. x = , 3at cd
61. x = , 5bp aq
62. x = , a, b, d 0 = /
63. x = , a 0
64. x = a, m 0,
x a
65. a. t =
b. about 40.9°F
c. C = (F – 32)
d. about 4.9°C = / 2A h
c – a
b – d
a – c m = / = / = /
b + d
c – a = /
s – 1055
1.1
3a – b – 8
a – b = /
3b + 2c – 5
b – c = / 5 9
2ab – 2c
3at – cd = /
4a – 3bc
aq – 5bp = / cb
2da = /
1-3

46. b. 80 legs; 20 legs; 2 legs; 10 cows
Solving Equations
ALGEBRA 2 LESSON 1-3
66. a. 10 cows, 30 chickens; Sample equation: 4c + 2(40 – c) = 100, where c is the number of cows.
b. 80 legs; 20 legs; 2 legs; 10 cows
c. Answers may vary. Sample: In all, a repair shop has 11 bicycles and tricycles to repair. These have a total of 26 wheels. How many bicycles and how many tricycles are there? 7 bicycles, 4 tricycles
67. a. If you solve ax – b = c for x, you
get x = Since b and c are
integers, b + c is an integer. But
a is a nonzero integer. So
is the quotient of two integers and hence, by the definition of a
rational number, is a
rational number.
b + c a
b + c a
b + c a
1-3

47. b. Solutions are rational when 0, a 0,
Solving Equations
ALGEBRA 2 LESSON 1-3
b. Solutions are rational when , a 0,
and is a perfect square (a whole number
perfect square or, in simplest form, a fraction whose numerator and denominator are whole number perfect squares).
68. about ft
70. 20 °
cm2
c – b a
73. –7
74. – , or –5
75. –20
76. – , or –5
77. 7x2 – 2x
78. ab – 6a
79. 2y – 7x
80. 11x – 6
81. –6x – 5
82. –2r – 5s 16 3 1 11 2 = /
67. (continued)
> –
1-3

48. 5. Find three consecutive odd integers whose sum is 111. 3
Solving Equations
ALGEBRA 2 LESSON 1-3
1. Solve 16x – 15 = –5x + 48.
2. Solve 5(1 – 3m) = 30 – 2(4m + 7).
3. Solve s = for b.
4. Mrs. Chern drove at a rate of 45 mi/h from her home to her sister’s house. She spent 1.5 hours having lunch with her sister. She then drove back home at a rate of 55 mi/h. The entire trip, including lunch, took 4 hours. How far does Mrs. Chern live from her sister?
5. Find three consecutive odd integers whose sum is 111. 3 – 11 7
a + b + c 2
b = 2s – a – c
61 mi 7 8
35, 37, 39
1-3

49. State whether each inequality is true or false.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
(For help, go to Lessons 1-1 and 1-3.)
State whether each inequality is true or false.
1. 5 < 12
2. 5 < –12
–12
> –
< –
< –
> –
Solve each equation.
7. 3x + 3 = 2x – 3
8. 5x = 9(x – 8) + 12
1-4

50. Solving Inequalities Solutions 1. 5 < 12, true 3. 5 12, false
ALGEBRA 2 LESSON 1-4
Solutions
1. 5 < 12, true
, false
, true
7. 3x + 3 = 2x – 3
3x – 2x = –3 – 3 x = –6
2. 5 < –12, false
–12, false
, true
8. 5x = 9(x – 8) + 12
5x = 9x – –4x = –60 x = 15
< –
> –
> –
< –
1-4

51. Solve –2x < 3(x – 5). Graph the solution.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
Solve –2x < 3(x – 5). Graph the solution.
–2x < 3(x – 5)
–2x < 3x – 15 Distributive Property
–5x < –15 Subtract 3x from both sides.
x > 3 Divide each side by –5 and reverse the inequality.
1-4

52. Solve 7x 7(2 + x). Graph the solution. > –
Solving Inequalities
ALGEBRA 2 LESSON 1-4
Solve 7x 7(2 + x). Graph the solution.
> –
7x 7(2 + x)
> –
7x x Distributive Property
> –
Subtract 7x from both sides.
> –
The last inequality is always false, so 7x 7(2 + x) is always false. It has no solution.
> –
1-4

53. Define: Let x = sales (in dollars).
Solving Inequalities
ALGEBRA 2 LESSON 1-4
A real estate agent earns a salary of $2000 per month plus 4% of the sales. What must the sales be if the salesperson is to have a monthly income of at least $5000?
Relate: $ % of sales $5000
> –
Define: Let x = sales (in dollars).
Write: x
> –
0.04x Subtract 2000 from each side.
> –
x 75,000 Divide each side by 0.04.
> –
The sales must be greater than or equal to $75,000.
1-4

54. Graph the solution of 2x – 1 3x and x > 4x – 9. < –
Solving Inequalities
ALGEBRA 2 LESSON 1-4
Graph the solution of 2x – x and x > 4x – 9.
< –
2x – x and x > 4x – 9
< –
–1 x > 3x
< –
– x and 3 > x
< –
This compound inequality can be written as –1 x < 3.
< –
1-4

55. Graph the solution of 3x + 9 < –3 or –2x + 1 < 5.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
Graph the solution of 3x + 9 < –3 or –2x + 1 < 5.
3x + 9 < –3 or –2x + 1 < 5
3x < – –2x < 4
x < –4 or x > –2
1-4

56. Relate: minimum length final length maximum length
Solving Inequalities
ALGEBRA 2 LESSON 1-4
A strip of wood is to be 17 cm long with a tolerance of ± 0.15 cm. How much should be trimmed from a strip 18 cm long to allow it to meet specifications?
Relate: minimum length final length maximum length
Define:  Let x = number of centimeters to remove.
< –
Write: 17 – – x
< –
– x Simplify.
< –
– – x –0.85 Subtract 18.
< –
x Multiply by –1.
> –
At least 0.85 cm and no more than 1.15 cm should be trimmed off to meet specifications.
1-4

57. 10. All real numbers are solutions.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
Pages 29–31 Exercises 1 2
1. x –
2. k > –9
3. a > 11
4. t
< –
5. y –14
6. y –6
7. x 8
8. m < 10
< –
9. n > 8
10. All real numbers are solutions.
11. All real numbers are solutions.
< –
< –
< –
1-4

58. 12. All real numbers are solutions.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
12. All real numbers are solutions.
13. no solutions
14. The width is less than 11.5 in., and the length is 3 in. greater than the width.
15. The longest side is less than 21 cm.
16. The smaller number is an integer greater than or equal to 8.
or more chips
18. –5 < x < 2
19. –4 x 2
20. –4 x < 6
< –
< –
< –
1-4

59. 23. All real numbers are solutions.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
21. –5 < x 6
22. x < 4 or x > 12
23. All real numbers are solutions.
24. x –8
25. x –3 or x 9
< –
26. between about 36.4 lb and 45.5 lb flour
27. between 4 and 5 days
28. between 0.26 cm and 0.30 cm
29. z 6
30. y 1 2 1 2
> –
> – 3 13
< –
< –
> –
1-4

60. 34. All real numbers are solutions. > –
Solving Inequalities
ALGEBRA 2 LESSON 1-4
31. x –48
32. x < 42
33. no solutions
34. All real numbers are solutions.
> –
35. Answers may vary. Sample: Mario has a coin collection that consists of dimes and nickels. There are half as many dimes as nickels. There are no more than 60 coins in the collection. Describe the collection.
36. 2 < AB < 6
37. between $204,000 and $254,000
38. a. 0 makes y true, but it does
not make (y – 16)
y + 2 true.
b. y –20
39. Dist. Prop.; arithmetic; Sub. Prop. of Inequality; Mult. Prop. of Inequality
< – 1 2
> –
< –
1-4

61. 45. All real numbers are solutions.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
40. Mult. Prop. of Inequality; Dist. Prop.; Add. Prop. of Inequality; Subt. Prop. of Inequality; Div. Prop. of Inequality
41. –1 < x < 8
42. 3 < y < 6
43. no solutions
44. –7 z < 4
45. All real numbers are solutions.
46. b < 2 or b > 5
47. All real numbers are solutions. 2 5
< –
1-4

62. 49. All real numbers are solutions.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
48. no solutions
49. All real numbers are solutions.
50. no solutions
51. Answers may vary. Sample: 2x – 7 –11
52. Answers may vary. Sample: –3x + 1 > 4
53. Answers may vary. Sample: –9 < 5x + 1 < 6
54. Answers may vary. Sample: 2x or –3x –
55. a. no
b. yes; values of a that are 8 or greater
c. (a) yes; values of a that are less than 8, (b) no
56. C
57. B
58. H
59. D
60. [2] The maximum number of songs can be recorded when the songs are short, so the maximum number of songs is
= 30
songs; the minimum number of songs can be recorded when the songs are long, so the minimum number of songs is
= 18 songs.
90 min
3 min per song
5 min per song
< –
> –
1-4

63. [1] provides 30 and 18 but not the explanation
Solving Inequalities
ALGEBRA 2 LESSON 1-4
[1] provides 30 and 18 but not the explanation
61. [4] Answer includes all of the following six parts of the explanation (or equivalent statements), which consist of solving each compound inequality, graphing each compound inequality, and explaining why the choices of or and and result in solutions of all real numbers and no real numbers.
x + 5 > 0 x – 3 < 0
x > –5 x < 3
The solutions to the two simple inequalities overlap. By writing or in the box, the solution is any value on either of the two arrows, or all real numbers.
x + 5 < 0 x + 5 > 0
x < –5 x > –5
The solutions to the two simple inequalities do not overlap. By writing and in the box, the solution is any point on both arrows, or no real numbers.
1-4

64. [3] omits one or two parts of the six parts of the explanation
Solving Inequalities
ALGEBRA 2 LESSON 1-4
[3] omits one or two parts of the six parts of the explanation
[2] omits three or four of the six parts of the explanation
[1] includes at least one of the six parts of the explanation
62. 4
63. no solution
64.
65. –20
66. 7a + 5
67. –2x + 14y
68.
– 0.1k
b + 12 12 9 10
1-4

65. 1. Solve –2(x – 3) 4. Graph the solution.
Solving Inequalities
ALGEBRA 2 LESSON 1-4
1. Solve –2(x – 3) Graph the solution.
2. Solve –5(4 – x) < 5x. Graph the solution.
3. Graph the solution of 3x and –2x
4. A copper wire is to have a length of 16 cm with a tolerance of ±0.02 cm. How much must be trimmed from a wire that is 18 cm long for it to meet specifications?
> – –
<
x 1
all real numbers
> –
> –
at least 1.98 cm and no more than 2.02 cm
1-4

66. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
(For help, go to Lessons 1-3 and 1-4.)
Solve each equation.
1. 5(x – 6) = 40
2. 5b = 2(3b – 8)
3. 2y + 6y = 15 – 2y + 8
Solve each inequality.
4. 4x + 8 > 20
5. 3a – 2 a + 6
> –
6. 4(t – 1) < 3t + 5
1-5

67. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
Solutions
1. 5(x – 6) = b = 2(3b – 8)
= 5b = 6b – 16
x – 6 = 8 –b = –16
x = 14 b = 16
5(x – 6) 5 40 5
3. 2y + 6y = 15 – 2y x + 8 > 20
2y + 6y + 2y = x > 12
10y = 23 x > 3
y = 2.3
5. 3a – 2 a (t – 1) < 3t + 5
3a – a t – 4 < 3t + 5
2a t – 3t < 5 + 4
a 4 t < 9
> –
> –
> –
> –
1-5

68. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
Solve |15 – 3x| = 6.
|15 – 3x| = 6
15 – 3x = 6 or 15 – 3x = –6 The value of 15 – 3x can be 6 or –6 since |6| and |–6| both equal 6.
–3x = – –3x = –21 Subtract 15 from each side of both equations.
x = 3 or x = 7 Divide each side of both equations by –3.
Check: |15 – 3x| = 6 |15 – 3x| = 6
|15 – 3(3)| 6 |15 – 3(7)| 6
|6| 6 |–6| 6
6 = = 6
1-5

69. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
Solve 4 – 2|x + 9| = –5.
4 – 2|x + 9| = –5
–2|x + 9| = –9 Add –4 to each side.
|x + 9| = Divide each side by –2. 9 2
x + 9 = or x + 9 = – Rewrite as two equations. 9 2
x = –4.5 or x = –13.5 Subtract 9 from each side of both equations.
Check: 4 – 2 |x + 9| = –5 4 – 2|x + 9| = –5
4 – 2 |– | –5 4 – 2 |– | –5
4 – 2 |4.5| –5 4 – 2 |–4.5| –5
4 – 2 (4.5) –5 4 – 2 (4.5) –5
–5 = –5 –5 = –5
1-5

70. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
Solve |3x – 4| = –4x – 1.
|3x – 4| = –4x – 1
3x – 4 = –4x – 1 or 3x – 4 = –(–4x –1) Rewrite as two equations.
7x – 4 = – 1 3x – 4 = 4x + 1 Solve each equation.
7x = 3 –x = 5
x = or x = –5 3 7
1-5

71. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
(continued)
Check: |3x – 4| = –4x – 1 |3x – 4| = –4x – 1
|3( ) – 4| –4( ) – 1 |3(–5) – 4| (–4(–5) –1) 3 7 3 7
|– | – |–19| 19 19 7 19 7
– 19 = 19 19 7 19 7 = /
The only solution is – is an extraneous solution. 3 7
1-5

72. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
Solve |2x – 5| > 3. Graph the solution.
|2x – 5| > 3
2x – 5 < –3 or 2x – 5 > 3 Rewrite as a compound inequality.
2x < 2 2x > 8 Solve for x.
x < 1 or x > 4
1-5

73. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
Solve –2|x + 1| –3. Graph the solution.
> –
–2|x + 1| –3
> –
–2|x + 1| –8 Isolate the absolute value expression. Subtract 5 from each side.
> –
|x + 1| 4 Divide each side by –2 and reverse the inequality.
< –
–4 x Rewrite as a compound inequality.
< –
–5 x 3 Solve for x.
< –
1-5

74. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
The area A in square inches of a square photo is required to satisfy A Write this requirement as an absolute value inequality.
< –
< –
Find the tolerance.
8.9 – 8.5 2 =
0.4
= 0.2
Find the average of the maximum and minimum values. 2 =
17.4
= 8.7
Write an inequality.
– A –
< –
Rewrite as an absolute value inequality.
|A – 8.7|
< –
1-5

75. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
Pages 36–38 Exercises
1. –6, 6
2. –8, 8
3. –6, 12
4. 3, –
5. no solution
6. no solution
7. –18, 10
8. –7, 17
9. –7, 15
10. –
11.
12.
13. –4, 8
14. –1,
15. 1
16. x < –12 or x > 6
17. x –3 or x
18. y –9 or y 15
19. All real numbers are solutions. 3 2 2 3
< –
> – 5 3 3 2
< –
> – 3 2
1-5

76. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
20. x –3 or x 4
21. z < –4 or z > 4
22. 0 < y < 18
23. –2 < y < 3
< –
> –
24. –2 < x < 6
25. no solution
26. – w
27. – t
28. |h – 1.4|
29. |k – 50.5|
30. |C – 27.5| 1 2 1 2
< –
< – 11 15 17 15
< –
< – 2 3 1 3
< –
< –
< –
1-5

77. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5 5 2
31. |b – 52.2|
32. |m – 1250|
33. |d – |
34. no solutions
35. – , –1
36. – ,
37. –
38. no solutions
< –
39.
40.
41. –1, –3
42. –
43. 2
44. –6 x 8
45. x –8 or x 5
46. All real numbers are solutions.
47. t < – or t > 2
48. All real numbers are solutions.
49. x < – or x > 11 8
< –
< – 3 2 71 36 3 2 2 3
< –
< – 14 3 16 3 1 3 1 2 3 2
< –
> –
1-5

78. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
50. x –8.4 or x
51. – x
52. –5 < x < 11
53. x < –32 or x > 22
< –
> –
54. Region 1
0 s , |s – 0.05| ;
Region 2
s 1, |s – 0.55| ;
Region 3
1 s 3, |s – 2| 1;
Region 4
3 s 6, |s – 4.5| ;
55. |C – 28.75| ; C
56. The graph of | x | < a (where a > 0) is the set of all points on the number line that lie between the points for a and –a. The graph of | x | > a has two parts: the left part consists of the points to the left of the point for –a, and the right part consists of the points to the right of the point for a.
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
1-5

79. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
ab + d c
–ab + d c
57. Answers may vary. Sample: |x – 1| 0; | x | < –5
58. |x – 36.8| , x
59. |x – 9.55| , x
60. x is in inches: |x – 3600| 4, x
61. – , , ,
64. (–6 x –5) or (5 x 6)
65. (x –6) or (–5 < x < 5) or (x 6)
66. x
67. D
68. F
69. C
70. G
71. [2] |x – 3| 5
–5 x –
–2 x 8
–1, 0, 1, 2, 3, 4, 5, 6, 7, 8
[1] only includes –2 x and does not show work
> –
ac + d ab
ac – d ab
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
< –
> –
< –
< –
b + c a
b + c a
< –
< – 5 2
> –
1-5

80. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
72. [4] 3 |2x – 4| + 5 < 41
3 |2x – 4| < 36 Subtr. Prop. of Ineq.
|2x – 4| < 12 Div. Prop. of Ineq.
–12 < 2x – 4 < 12 Def. of absolute value
–8 < 2x < 16 Add. Prop. of Ineq.
–4 < x < 8 Div. Prop. of Ineq.
[3] appropriate methods, but with one computational error
[2] does not include steps
[1] only includes final answer of –4 < x < 8, with no steps or justification of steps
73. y < 6
74. s > –
75. a 4
76. x 7 5
> –
< –
1-5

81. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5 11 14
77. x
78. t 7
79. 36
80. –
81. Inverse Prop. of Add.
82. Closure Prop. Of Mult.
83. Comm. Prop. of Mult.
> –
< – 4 15
1-5

82. Absolute Value Equations and Inequalities
ALGEBRA 2 LESSON 1-5
1. Solve |3x + 1| = 4.
2. Solve |–2x + 3| + 7 > 9. Graph the solution.
3. Solve |4x – 12| Graph the solution.
4. A machinist is to drill a hole with diameter D inches, that satisfies D Express the specification using an absolute value inequality.
– , 1 5 3
x < or x > 1 2 5
< –
1 x 5 –
<
< –
< –
|D – 0.175| –
<
1-5

83. Write each number as a percent.
Probability
ALGEBRA 2 LESSON 1-6
(For help, go to Skills Handbook page 842.)
Write each number as a percent. 3 8 5 6 1.
2. 1 4. 1
400
6. 3
1-6

84. Probability 1. = 3 ÷ 8 = 0.375 = 0.375(100%) = 37.5% Solutions
ALGEBRA 2 LESSON 1-6
= 3 ÷ 8 = = 0.375(100%) = 37.5%
= = 11 ÷ 6 = 1.83 = 1.83(100%) = %
= (100%) = 0.43%
= 1 ÷ 400 = = (100%) = 0.25%
= 1.04(100%) = 104%
6. 3 = 3(100%) = 300%
Solutions 3 8 5 6 11 6 1 3 1
400
1-6

85. Probability
ALGEBRA 2 LESSON 1-6
A player hit the bull’s eye on a circular dartboard 8 times out of 50. Find the experimental probability that the player hits the bull’s eye.
P(bull’s eye) = = 0.16, or 16% 8 50
1-6

86. Count to see how many times heads came up exactly twice.
Probability
ALGEBRA 2 LESSON 1-6
Describe a simulation you could use that involves flipping a coin to find the experimental probability of guessing exactly 2 answers out of 6 correctly on a true-false quiz.
Getting heads with a flip of a coin has the same probability as guessing the correct answer to a question on a true-false test.
So, let heads represent a correct answer and tails represent an incorrect answer. To simulate guessing the answers for a six-question true-false test, flip a coin six times. Record the number of heads.
Repeat 100 times.
Count to see how many times heads came up exactly twice.
Divide this number by 100.
The result is the experimental probability that the simulation gives for guessing 2 correct answers out of 6.
1-6

87. To roll a multiple of 3 with a number cube, you must roll 3 or 6.
Probability
ALGEBRA 2 LESSON 1-6
Find the theoretical probability of rolling a multiple of 3 with a number cube.
To roll a multiple of 3 with a number cube, you must roll 3 or 6. 2 6
6 equally likely outcomes are in the sample space.
2 outcomes result in a multiple of 3. 1 3 =
1-6

88. Brown is a dominant eye color for human beings.
Probability
ALGEBRA 2 LESSON 1-6
Brown is a dominant eye color for human beings.
If a father and mother each carry a gene for brown eyes and a gene for blue eyes, what is the probability of their having a child with blue eyes?
B b
B BB Bb
b Bb bb
Gene from Father
Gene from Mother
Let B represent the dominant gene for brown eyes. Let b represent the recessive gene for blue eyes.
The sample space contains four equally likely outcomes {BB, Bb, Bb, bb}. 1 4
The outcome bb is the only one for which a child will have blue eyes. So, P(blue eyes) = . 1 4
The theoretical probability that the child will have blue eyes is , or 25%.
1-6

89. Probability
ALGEBRA 2 LESSON 1-6
For the dartboard in Example 5, find the probability that a dart that lands at random on the dartboard hits the outer ring.
P(outer ring) =
area of outer ring
area of circle with radius 4r =
(area of circle with radius 4r) – (area of circle with radius 3r) =
(4r)2 – (3r)2
(4r2) =
16 r 2 – 9 r 2
16 r 2 =
16 r 2
7 r 2 = 7 16
The theoretical probability of hitting the outer ring is , or about 44%. 7 16
1-6

90. Probability Pages 42–45 Exercises 1. or about 47%; or about 53%
ALGEBRA 2 LESSON 1-6
Pages 42–45 Exercises
161
340
179
340
or about 47%; or about 53%
2. the number 1: or about 15.7%; the number 2: or about 16.4%; the number 3: or about 16.8%; the number 4: or about 16.4%; the number 5: or about 17.5%; the number 6: or about 17.2%
3. Answers may vary. Sample: Generate random numbers between 0 and 1 using a graphing calculator. In each random number, examine the first five digits. Let even digits represent correct answers and odd digits incorrect answers. If there are two or more even digits, make a tally mark for that number. Do this 100 times. Find the total number of tally marks. This, as a percent, gives the experimental probability. The simulated probability should be about 70%. 21
134 11 67 45
268 11 67 45
268 23
134
1-6

91. Probability
ALGEBRA 2 LESSON 1-6
4. Answers may vary. Sample: Toss 5 coins. Keep a tally of the times 3 or more heads are tossed. (A head represents a correct answer.) Do this 100 times. The total number of tally marks, as a percent, gives the experimental probability. The simulated probability should be about 40%.
5. Answers may vary. Sample: Generate 100 random numbers with a calculator. Record the first five digits of each number. Let 0 and 1 represent correct answers and the other digits incorrect answers. Tally the recorded numbers with exactly one digit that represents a correct answer. Tally the recorded numbers with exactly two digits that represent correct answers. Tally the recorded numbers with exactly three digits that represent correct answers. The tally totals, as a percent, give the experimental probabilities. They should be about 40%, 20%, and 5%, respectively.
1-6

92. Probability 6. , or 30% 7. , or 50% 8. , or 80% 9. , or 80%
ALGEBRA 2 LESSON 1-6
, or 30%
7. , or 50%
8. , or 80%
9. , or 80%
, or 38.4%
, or 15.2%
, or 82.4%
, or 56%
, or 61.6% 3 10 1 2 4 5 48
125 19
103 14 25 77 1 2
15. {Gg, Gg, gg, gg}; , or 50%
16. {Gg, Gg, Gg, Gg}; 1, or 100%
, or 6.25%
, or 37.5%
, or 25%
, or 75%
21. a. 1
b. 0 1 16 3 8 4
22. Answers may vary. Sample: Let the digits 1–6 correspond to good eggs, and 7–9 correspond to bad eggs. Ignoring the digit 0, start in the first row of the table. Circle groups of three digits, 20 groups in all. Tally the circled groups that do not have a 7, 8, or 9. The experimental probability of getting 3 cartons with only unbroken eggs is
, or . 5 20 1 4
1-6

93. Probability
ALGEBRA 2 LESSON 1-6
36.
37.
38. a. (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
b. 36 outcomes c. d. 4 9 1 36 6
23. Answers may vary. Sample: Let odd digits represent heads and even digits represent tails. Use the first 50 digits of the table. The experimental probability of heads
is .
, or 78.9%
, or 35.4%
, or 29.3% 1 2
116
147 52 43
, or 21.1%
28.
29.
30.
31. 0
32.
33. 1
34.
35. 31
147 1 6 2 3 4 9
1-6

94. 41. if there are any restrictions on the last digit of a ZIP code
Probability
ALGEBRA 2 LESSON 1-6 a
a + b %
40. a. ;
b. Answers may vary. Sample: Variables such as injuries make probability a poor predictor.
41. if there are any restrictions on the last digit of a ZIP code
or 6.7%
43. a.
b. a to b – a or
c. A game where the probability of
winning is ;
when the odds of
winning are , the
probability of winning is only .
44. Check students’ work.
45. a. about
b. about
46. B
47. H
48. [2] (11, 12, 13, 21, 22, 23, 31, 32, 33); all the pairs are equally likely because the three regions have the same area, and the outcome of the first spin does not affect the outcome of the second spin.
[1] answers one of the two parts a
b – a 1 4 3 4 1 2 1 2 1 3 1 15 1 3 2 3
1-6

95. no. of favorable outcomes
Probability
ALGEBRA 2 LESSON 1-6 1 9
49. [2] ; there is one favorable outcome (11) out of 9 possible outcomes.
[1] no explanation included
50. [4] a. The favorable outcomes for an even sum are 11, 13, 22, 31, and 33.
b. The probability of an even sum is
c. The favorable outcomes for a sum that is not prime are 13, 22, 31, and 33.
d. The probability of a sum that is not prime is .
e.  > , so an even sum is more likely than a sum that is not prime.
[3] omits one of the five parts of the answer
[2] omits two OR three of the five parts of the answer
[1] omits four of the five parts of the answer
no. of favorable outcomes
total no. of outcomes 5 9
= . 4 9 5 9 4 9
1-6

96. Probability
ALGEBRA 2 LESSON 1-6 12 5 24 5
51. No; not all odds are 50–50. In this case they are, which means 1 out every 250 people will have defective chromosomes.
52. –12, 6
53. – , 5
54. –13, 6
55. –5, 9
56. –13, 10
57. – ,
58. x –4 or x 4
59. –2 x 6
60. x < 1 or x > 2
61. –5 < x < 1
62. x –8 or x 2
63. – x –3
< –
> –
< –
< – 1 3 5 3
< –
> –
< –
< –
1-6

97. Probability
ALGEBRA 2 LESSON 1-6
1. A bowler rolled the ball 35 times and got 5 strikes. What is the experimental probability that the bowler gets a strike?
2. Describe a simulation you could use that involves flipping a coin to find the experimental probability of guessing at least four correct answers on a five-question true or false quiz.
3. What is the theoretical probability of rolling a sum less than 5 using two number cubes?
4. Segments parallel to the sides are used to divide a square board 3 ft on each side into 9 equal-size smaller squares. If the board is in a level position and a grain of rice lands on the board at a random point, what is the probability that it lands on a corner section? 1 7
Answers may vary. Sample: Let heads represent a correct answer and tails represent an incorrect answer. Flip the coin five times. Record the number of heads. Repeat 100 times. Divide the number of times you got 4 or 5 heads by 100. 1 6 4 9
1-6

98. 1. Check students’ answers.
Tools of Algebra
ALGEBRA 2 CHAPTER 1
1. Check students’ answers.
2. Dist. Prop., Assoc. Prop. of Add., Identity Prop. of Mult., Dist. Prop., arithmetic
3. 29
4. 84
5. 15
6. 2a2 + a
7. –3x + 5y
8. 2a – 4b
9. 11x – 27
10. 7
11.
12.
13. 5
14. 45 16 3 8
15. 4
16. x = , a b
17. x = 3c2, c 0
18. x = a(b – 1) + 5, a 0
19. x = , a b
20. r =
21. $138
22. 9 2a
a – b
c – 2 s
2 h = /
1-A

99. Tools of Algebra 23. x –4 24. x > 7 25. t >
ALGEBRA 2 CHAPTER 1 4 3
23. x –4
24. x > 7
25. t >
26. x < –16 or x >
> –
27. 1 < d < 2
28. –1,
29. –
p 7
31. w < – or w > 0
32.
33. 0
34.
35. 1
36. about 40% 2 5 3 2 3 5 1 2 5 3 2 5
< –
< –
1-A