View
49
Download
0
Category
Tags:
Preview:
DESCRIPTION
1. –(–7.2). 2. 1 – (–3). 3. –9 + (–4.5). 4. (–3.4)(–2). 2 5. 3 5. 6. – + ( – ). 5. –15 ÷ 3. Properties of Real Numbers. ALGEBRA 2 LESSON 1-1. (For help, go to Skills Handbook page 845.). Simplify. 1-1. Solutions. 1. –(–7.2) = 7.2 3. –9 + (–4.5) = –13.5 - PowerPoint PPT Presentation
Citation preview
1-1
ALGEBRA 2 LESSON 1-1
Simplify.
1. –(–7.2) 2. 1 – (–3)
3. –9 + (–4.5) 4. (–3.4)(–2)
(For help, go to Skills Handbook page 845.)
Properties of Real Numbers
5. –15 ÷ 3 6. – + (– )25
35
ALGEBRA 2 LESSON 1-1Properties of Real Numbers
Solutions
1. –(–7.2) = 7.2
3. –9 + (–4.5) = –13.5
5. –15 ÷ 3 = –5
6. – + (– ) = = – = –125
35
–2 + (–3)5
55
2. 1 – (–3) = 1 + 3 = 4
4. (–3.4)(–2) = 6.8
1-1
A pilot uses the formula d = a2 + 7290a 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?
ALGEBRA 2 LESSON 1-1Properties of Real Numbers
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.
1-1
Graph the numbers – , 7 , and 3.6 on anumber line.
ALGEBRA 2 LESSON 1-1Properties of Real Numbers
34
1-1
Use a calculator to find that 7 2.65.
– is between –1 and 0.34
Compare –9 and – 9. Use the symbols < and >.
ALGEBRA 2 LESSON 1-1Properties of Real Numbers
Since –9 < –3, it follows that
9 = 3, so – 9 = –3.
–9 < – 9.
1-1
Find the opposite and the reciprocal of each number.
ALGEBRA 2 LESSON 1-1Properties of Real Numbers
a. –317
Opposite: –4
1-1
Opposite: –(–3 ) = 317
17
Reciprocal: = = – 7 2222
7
1
–3 17
1
–Reciprocal: 1
4
b. 4
Which property is illustrated?
ALGEBRA 2 LESSON 1-1Properties of Real Numbers
a. (–7)(2 • 5) = (–7)(5 • 2) b. 3 • (8 + 0) = 3 • 8
The given equation is true because 2 • 5 = 5 • 2.
So, the equation uses the Commutative Propertyof Multiplication.
The given equation is true because 8 + 0 = 8.
This is an instance of the Identity Property of Addition.
1-1
Simplify | 4 |, |–9.2|, and |3 – 8|.
ALGEBRA 2 LESSON 1-1Properties of Real Numbers
13
–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.
1-1
4 is 4 units from 0, so | 4 | = 4 .13
13
13
13
Properties of Real NumbersALGEBRA 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, real numbers
8. irrational numbers, real numbers
9. whole numbers, rational numbers
10. natural numbers, rational numbers
11. real numbers
12.
13.
14.
15.
1-1
Properties of Real NumbersALGEBRA 2 LESSON 1-1
16.
17. >
18. =
19. >
20. <
21. <
22. =
23. >
24. <
25. – > – , – < –26. 0.075 < 0.39,
0.39 > 0.075
27. –2.3 < 2.1, 2.1 > –2.3
28. –5.2 < –4.8, –4.8 > –5.2
29. 3.04 < 3.4, 3.4 > 3.04
14
13
13
14
30. 0.4 < 0.4, 0.4 > 0.4
31. –4 < – 4,– 4 > – 4
32. 5 < 7, 7 > 5
33. – 3 > – 5– 5 < – 3
34. –200,
35. –3 ,
36. 0.01, –100
1200
35
518
1-1
Properties of Real NumbersALGEBRA 2 LESSON 1-1
37.
38. – 3,
39. –2 ,
40. 2.34, –
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.
53. 10.3
54. 0.06
55. –25
56. 1.6
57.
58. 3
72 , – 2
7
1 3
3 3or
12
50117
1 – 3 1
3
1-1
Properties of Real NumbersALGEBRA 2 LESSON 1-1
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
68. 4.8
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. >
1214
12
23
13
1-1
Properties of Real NumbersALGEBRA 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
14
1-1
Properties of Real NumbersALGEBRA 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
1-1
Properties of Real NumbersALGEBRA 2 LESSON 1-1
102. 1.9
103. –3.8
104. 27
105. 0
106. –0.4
107. 7
108. , or 2
109. , or 11
110. ,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
15
15
15
1n
94
14
353
23
32
12
1-1
Properties of Real NumbersALGEBRA 2 LESSON 1-1
111. 5
112. 38
113. 45
1-1
Properties of Real NumbersALGEBRA 2 LESSON 1-1
1. Graph –2.3, 12, and on a number line.
2. Replace each with the symbol <, >, or = to make the sentence true.
a. 2.03 2.8
b. 1 1.2
3. Fit the opposite and the reciprocal of 1 .
4. Name the property of real numbers illustrated by the equation.
(–2)(3 + ) = (–2)( + 3)
5. Simplify |9 – (–5)|.
29
75
<
=
58
Comm. Prop. of Add.
14
1-1
–1 , 58
8 13
1-2
ALGEBRA 2 LESSON 1-2
Use the order of operations to simplify each expression.
1. 8 • 3 – 2 • 4 2. 8 – 4 + 6 ÷ 3
3. 24 ÷ 12 • 4 ÷ 3 4. 3 • 82 + 12 ÷ 4
5. 27 + 18 ÷ 9 – 32 + 1 6. (40 + 24) ÷ 8 – (23 + 1)
(For help, go to Skills Handbook page 845.)
Algebraic Expressions
Solutions
ALGEBRA 2 LESSON 1-2Algebraic Expressions
1. 8 • 3 – 2 • 4 = (8 • 3) – (2 • 4) = 24 – 8 = 16
2. 8 – 4 + 6 ÷ 3 = 8 – 4 + (6 ÷ 3) = 8 – 4 + 2 = (8 – 4) + 2 = 4 + 2 = 6
3. 24 ÷ 12 • 4 ÷ 3 = (24 ÷ 12) • 4 ÷ 3 = 2 • 4 ÷ 3 = (2 • 4) ÷ 3 = 8 ÷ 3 =
or 2
4. 3 • 82 + 12 ÷ 4 = 3 • 64 + 12 ÷ 4 + (3 • 64) + (12 ÷ 4) = 192 + 3 = 195
5. 27 + 18 ÷ 9 – 32 + 1 = 27 + 18 ÷ 9 – 9 + 1 = 27 + (18 ÷ 9) – 9 + 1 = 27 + 2 – 9 + 1 = (27 + 2) – 9 + 1 = 29 – 9 + 1 = (29 – 9) + 1 = 20 + 1 = 21
6. (40 + 24) ÷ 8 – (23 + 1) = (40 + 24) ÷ 8 – (8 + 1) = 64 ÷ 8 – 9 = (64 ÷ 8) – 9 = 8 – 9 = –1
83
23
1-2
ALGEBRA 2 LESSON 1-2Algebraic Expressions
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.
= –14 + 30 To subtract, add the opposite.
= 16 Add.
1-2
Algebraic Expressions
Evaluate (k – 18)2 – 4k for k = 6.
ALGEBRA 2 LESSON 1-2
(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.
1-2
Algebraic ExpressionsALGEBRA 2 LESSON 1-2
The expression –0.08y 2 + 3y models the percent increase of Hispanic voters in a town from 1990 to 2000. In the expression, y represents the number of years since 1990. 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%.
1-2
Algebraic ExpressionsALGEBRA 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
1-2
Algebraic ExpressionsALGEBRA 2 LESSON 1-2
Find the perimeter of this figure. Simplify the answer.
= 2c + 4d
= c + c + 4d
1-2
P = c + + d + (d – c) + d + + c + dc2
c2
= c + + d + d – c + d + + c + d c2
c2
= + + c + 4dc2
c2
= + c + 4d2c2
Algebraic ExpressionsALGEBRA 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
11. 64 ft
12. 1600 ft
13. 0.013 mm
14. 0.032 mm
15. 0.4 mm
16. 1.4 mm
17. $1210
18. $1331
19. $1464.10
20. $1610.51
21. 4a
22. 2s + 5
23. –9a + b
24. 6a + 3b
1-2
Algebraic ExpressionsALGEBRA 2 LESSON 1-2
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. –0.0078y2 + 1.265y +
65.27
d. about 87 million
414
1-2
Algebraic ExpressionsALGEBRA 2 LESSON 1-2
47. – a2 + 2b2
48.
49. +
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
34
5x2
27y 2
122y15
1-2
Algebraic ExpressionsALGEBRA 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
1-2
Algebraic ExpressionsALGEBRA 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. >
56
34
38
12
18
110
116
14
80. >
81. <
82. <
83. =
1-2
Algebraic ExpressionsALGEBRA 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
1-2
1-3
ALGEBRA 2 LESSON 1-3
Simplify each expression.
1. 5x – 9x + 3 2. 2y + 7x + y – 1
3. 10h + 12g – 8h – 4g 4. + + – y
5. (x + y) – (x – y) 6. –(3 – c) – 4(c – 1)
(For help, go to Lesson 1-2.)
Solving Equations
y3
2y3
x3
ALGEBRA 2 LESSON 1-3Solving Equations
Solutions
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
4. + + – 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
1-3
x3
y3
2y3
x3
y3
2y3
3y3
x + y + 2y – 3y3
x + (1 + 2 – 3)y3
x + 0y3
x3
Solving EquationsALGEBRA 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.
1-3
Check: 7x + 3 = 2x – 12
7(–3) + 3 2(–3) – 12
–18 = –18
Solving EquationsALGEBRA 2 LESSON 1-3
Solve 4(m + 9) = –3(m – 4).
4(m + 9) = –3(m – 4)
4m + 36 = –3m + 12 Distributive Property
7m + 36 = 12 Add 3m to each side.
7m = –24 Subtract 36 from each side.
1-3
m = – Divide each side by 7.247
Solving EquationsALGEBRA 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 + 2 h + 2wh Distributive Property
A – 2 h = 2 w + 2wh Subtract 2 h from each side.
A – 2 h = (2 + 2h)w Distributive Property
1-3
= w Divide both sides by 2 + 2h.A – 2 h2 + 2h
Solving EquationsALGEBRA 2 LESSON 1-3
Solve + 8 = b for x. Find any restrictions on a and b.xa
x + 8a = ab Simplify.
1-3
+ 8
= b
xa
a( ) + a(8) = ab Multiply each side by the least common denominator (LCD), a.
xa
x = ab – 8a Subtract 8a from each side.
=/The denominator cannot be zero, so a 0.
Solving EquationsALGEBRA 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
1-3
Solving EquationsALGEBRA 2 LESSON 1-3
(continued)
5w = 68 Add.
Check: Is the answer reasonable? Since the dimensions are about 14 ft by 41 ft and 14 + 14 + 41 = 69, the perimeter is about 69 ft. The answer is reasonable.
1-3
w = 13 Divide each side by 5.35
3w = 40 Find the length.45
35The width is 13 ft and the length is 40 ft.4
5
Solving EquationsALGEBRA 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
Solving EquationsALGEBRA 2 LESSON 1-3
(continued)
Write: 138= x + 2x + 3x + 6x138 = 12xCombine like terms.11.5 = x2x = 2(11.5) 3x = 3(11.5) 6x = 6(11.5) Find the length of
= 23 = 34.5 = 69 each side.
Check: Is the answer reasonable? Since 12 + 23 + 35 + 69 = 139, the answer is reasonable.
The lengths of the sides are 11.5 cm, 23 cm, 34.5 cm, and 69 cm.
1-3
Solving EquationsALGEBRA 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?
1-3
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.3560
Write: 400t = 350 (t + ) 7 12
400t = 350t + Distributive Property 12256
Solving EquationsALGEBRA 2 LESSON 1-3
(continued)
50t = Solve for t.12256
1-3
t = 4 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.
23
Solving EquationsALGEBRA 2 LESSON 1-3
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.
172
72
19
17732
23
12
17. h =
18. g =
19. w =
20. r =
21. r =
22. h =
23. x = , a –b
24. x = , b c
2Ab
2st2
VlhI
ptS
2 hV r
2
ca + b
cc – b
=/
=/
Solving EquationsALGEBRA 2 LESSON 1-3
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 or2m + 2n + 2
28. x = – 1
29. 4 h
30. 300 mi/h, 600 mi/h
31. width = 4.5 cm, length = 7.5 cm
32. 4 in., 5 in., 6 in.
33. 11 cm, 11 cm, 16.5 cm, 16.5 cm
34. 7.5 cm, 10 cm, 12.5 cm
35. a. x + (x + 1) + (x + 2) = 90; 29, 30, 31
b. (x – 1) + x + (x + 1) = 90; 29, 30, 31
36. , or 1
37. 3
38. , or 5
39.
40. 30
41. , or 0.9875
42. R =
43. r2 =
44. h =
45. v =
23
23
23
4639
739
275
25
32
7980
r1r2
r1 + r2
Rr1
r1 – R
h + 5t2
t
1-3
5g2
=/
=/
S – 2 r 2
2 r
Solving EquationsALGEBRA 2 LESSON 1-3
1-3
46. h =
47. b2 = – b1
48. 40°, 140°
49. 34°, 56°
50. about 169.4 mi
51. 2.98 m
52. $360; $746.4053. 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 = + 6, 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
2(v – s2)s
2Ah
c – ab – d
b + dc – a
3a – b – 8a – b
3b + 2c – 5b – c
2ab – 2c3at – cd
4a – 3bcaq – 5bp
cb2da
10ca
a – cm
59
s – 10551.1
=/
=/
=/
=/
=/
=/
=/
=/
=/
=/
=/
Solving EquationsALGEBRA 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 + ca
b + ca
b + ca
1-3
Solving EquationsALGEBRA 2 LESSON 1-3
1-3
b. Solutions are rational when 0, 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 269.4 ft
69. 18.5
70. 20
71. 19.38°
72. 560 cm2
c – bac – 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
163
13
112
12
=/67. (continued)
>–
Solving EquationsALGEBRA 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.
a + b + c2
3
b = 2s – a – c
35, 37, 39
1-3
– 117
61 mi78
1-4
ALGEBRA 2 LESSON 1-4
State whether each inequality is true or false.
1. 5 < 12 2. 5 < –12
3. 5 12 4. 5 –12
5. 5 5 6. 5 5
(For help, go to Lessons 1-1 and 1-3.)
Solving Inequalities
Solve each equation.
7. 3x + 3 = 2x – 3 8. 5x = 9(x – 8) + 12
>– <–
<– >–
2. 5 < –12, false
4. 5 –12, false
6. 5 5, true
8. 5x = 9(x – 8) + 125x = 9x – 72 + 12–4x = –60x = 15
Solutions
Solving Inequalities
1. 5 < 12, true
3. 5 12, false
5. 5 5, true
7. 3x + 3 = 2x – 33x – 2x = –3 – 3x = –6
ALGEBRA 2 LESSON 1-4
1-4
>–<–
<–>–
Solving InequalitiesALGEBRA 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
Solving InequalitiesALGEBRA 2 LESSON 1-4
Solve 7x 7(2 + x). Graph the solution.
1-4
7x 7(2 + x)>–
7x 14 + 7x Distributive Property>–
0 14 Subtract 7x from both sides.>–
The last inequality is always false, so 7x 7(2 + x) is always false. It has no solution.
>–
>–
Solving InequalitiesALGEBRA 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?
Define: Let x = sales (in dollars).
The sales must be greater than or equal to $75,000.
1-4
Relate: $2000 + 4% of sales $5000>–
Write: 2000 + 0.04x 5000>–
0.04x 3000 Subtract 2000 from each side.>–x 75,000 Divide each side by 0.04.>–
Solving InequalitiesALGEBRA 2 LESSON 1-4
Graph the solution of 2x – 1 3x and x > 4x – 9.
1-4
<–
2x – 1 3x and x > 4x – 9<–
–1 x 9 > 3x<–
–1 x and 3 > x<–
This compound inequality can be written as –1 x < 3.<–
Solving InequalitiesALGEBRA 2 LESSON 1-4
Graph the solution of 3x + 9 < –3 or –2x + 1 < 5.
3x + 9 < –3 or –2x + 1 < 5
3x < –12 –2x < 4
x < –4 or x > –2
1-4
Solving Inequalities
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?
ALGEBRA 2 LESSON 1-4
At least 0.85 cm and no more than 1.15 cm should be trimmed off to meet specifications.
1-4
Relate: minimum length final length maximum length
Define: Let x = number of centimeters to remove.
<– <–
Write: 17 – 0.15 18 – x 17 + 0.15<– <–
16.85 18 – x 17.15 Simplify.<– <–
–1.15 – x –0.85 Subtract 18.<– <–
1.15 x 0.85 Multiply by –1.>– >–
5. y –14
6. y –6
7. x 8
8. m < 10
Solving InequalitiesALGEBRA 2 LESSON 1-4
1. x –
2. k > –9
3. a > 11
4. t 11
1-4
Pages 29–31 Exercises
9. n > 8
10. All real numbers are solutions.
11. All real numbers are solutions.
12
<–
<–
<–
<–
<–
Solving InequalitiesALGEBRA 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.
17. 4546 or more chips
18. –5 < x < 2
19. –4 x 2
20. –4 x < 6
1-4
<– <–
<–
Solving InequalitiesALGEBRA 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
12
12
313
1-4
>–
<–
<–
>–
>–
<–
Solving InequalitiesALGEBRA 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 20 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
12
1-4
>– <–
>–
<–
Solving InequalitiesALGEBRA 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.
25
1-4
<–
Solving InequalitiesALGEBRA 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
1-4
>–
54. Answers may vary. Sample: 2x + 4 0 or –3x – 3 0
55. a. nob. yes; values of a that are 8 or
greaterc. (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 min3 min per song
90 min5 min per song
<–<–
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.
Solving InequalitiesALGEBRA 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
1-4
Solving InequalitiesALGEBRA 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.
69. 1.61 – 0.1k
910
b + 1212
1-4
1. Solve –2(x – 3) 4. Graph the solution.
2. Solve –5(4 – x) < 5x. Graph the solution.
3. Graph the solution of 3x + 4 1 and –2x + 7 5.
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?
Solving InequalitiesALGEBRA 2 LESSON 1-4
all real numbers
at least 1.98 cm and no more than 2.02 cm
1-4
–<x 1
>–
>– >–
ALGEBRA 2 LESSON 1-5
Solve each equation.
1. 5(x – 6) = 40 2. 5b = 2(3b – 8) 3. 2y + 6y = 15 – 2y + 8
4. 4x + 8 > 20
(For help, go to Lessons 1-3 and 1-4.)
Absolute Value Equations and Inequalities
Solve each inequality.
6. 4(t – 1) < 3t + 5
1-5
5. 3a – 2 a + 6>–
1. 5(x – 6) = 40 2. 5b = 2(3b – 8)
= 5b = 6b – 16
x – 6 = 8 –b = –16x = 14 b = 16
Solutions
ALGEBRA 2 LESSON 1-5Absolute Value Equations and Inequalities
5(x – 6)5
405
5. 3a – 2 a + 6 6. 4(t – 1) < 3t + 53a – a 6 + 2 4t – 4 < 3t + 5
2a 8 4t – 3t < 5 + 4a 4 t < 9
3. 2y + 6y = 15 – 2y + 8 4. 4x + 8 > 202y + 6y + 2y = 15 + 8 4x > 12
10y = 23 x > 3y = 2.3
1-5
>–>–>–>–
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
Solve |15 – 3x| = 6.
|15 – 3x| = 6
15 – 3x = 6 or15 – 3x = –6The value of 15 – 3x can
be 6 or –6 since |6| and |–6| both equal 6.
x = 3 or x = 7Divide each side of both
equations by –3.
–3x = –9 –3x = –21Subtract 15 from each
side of both equations.
Check: |15 – 3x| = 6 |15 – 3x| = 6|15 – 3(3)| 6 |15 – 3(7)| 6
|6| 6 |–6| 66 = 6 6 = 6
1-5
Absolute Value Equations and InequalitiesALGEBRA 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 = –4.5 or x = –13.5Subtract 9 from
each side of both equations. Check: 4 – 2 |x + 9| = –5 4 – 2|x + 9| = –5
4 – 2 |–4.5 + 9| –5 4 – 2 |–13.5 + 9| –5 4 – 2 |4.5| –5 4 – 2 |–4.5| –54 – 2 (4.5) –5 4 – 2 (4.5) –5
–5 = –5 –5 = –5
1-5
|x + 9| = Divide each side by –2.92
x + 9 = or x + 9 = – Rewrite as two equations.92
92
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
Solve |3x – 4| = –4x – 1.
|3x – 4| = –4x – 1
1-5
3x – 4 = –4x – 1or 3x – 4= –(–4x –1)
Rewrite as two equations.
7x – 4 = – 13x – 4
= 4x + 1Solve each equation.
7x = 3–x = 5x = orx = –5
37
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
(continued)
Check: |3x – 4| = –4x – 1 |3x – 4| = –4x – 1
|3( ) – 4| –4( ) – 1 |3(–5) – 4| (–4(–5) –1)37
37
|– | – |–19| 19197
197
– 19 = 19197
197
1-5
=/
The only solution is –5. is an extraneous solution.37
Absolute Value Equations and InequalitiesALGEBRA 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.
x < 1 or x > 4
2x < 2 2x > 8 Solve for x.
1-5
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
Solve –2|x + 1| + 5 –3. Graph the solution.
1-5
>–
|x + 1| 4 Divide each side by –2 and reverse the inequality.
<–
–2|x + 1| + 5 –3>–
–2|x + 1| –8Isolate the absolute value expression. Subtract 5 from each side.
>–
–4 x + 1 4Rewrite as a compound inequality.
<– <––5 x 3Solve for x.
<– <–
Absolute Value Equations and Inequalities
The area A in square inches of a square photo is required to satisfy 8.5 A 8.9. Write this requirement as an absolute value inequality.
ALGEBRA 2 LESSON 1-5
1-5
<– <–
Write an inequality.–0.2 A – 8.7 0.2<– <–
Rewrite as an absolute value inequality.|A – 8.7| 0.2<–
Find the tolerance.8.9 – 8.52 = 0.4
2 = 0.2
Find the average of the maximum andminimum values.
8.9 + 8.52 = 17.4
2 = 8.7
Absolute Value Equations and Inequalities
Pages 36–38 Exercises
ALGEBRA 2 LESSON 1-5
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
53
32
2332
32
16. x < –12 or x > 6
17. x –3 or x 13
18. y –9 or y 15
19. All real numbers are solutions.
1-5
>–<–
>–<–
Absolute Value Equations and InequalitiesALGEBRA 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. –3 w
27. – t
28. |h – 1.4| 0.1
29. |k – 50.5| 0.5
30. |C – 27.5| 0.25
23
13
12
12
1115
1715
1-5
>–<–
<– <–
<– <–
<–
<–
<–
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
31. |b – 52.2| 2.5
32. |m – 1250| 5033. |d – 0.11885|
0.00015
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
32143
163
13
52118
7136
23
46. All real numbers are solutions.
47. t < – or t > 2
48. All real numbers are solutions.
49. x < – or x >
32
32
12
1-5
>–
<–
<–
<–
<– <–
<–
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
50. x –8.4 or x 9.6
51. –3.5 x 7.5
52. –5 < x < 11
53. x < –32 or x > 22
54. Region 10 s 0.1, |s – 0.05| 0.05; Region 20.1 s 1, |s – 0.55| 0.45; Region 31 s 3, |s – 2| 1;Region 43 s 6, |s – 4.5| 1.5;
55. |C – 28.75| 0.25; 28.5 C 29.0
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
>–<–
<– <–
<– <– <–
<– <– <–
<– <– <–
<– <– <–
<– <– <–
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
57. Answers may vary. Sample: |x – 1| 0; | x | < –5
58. |x – 36.8| 0.05, 36.75 x 36.85
59. |x – 9.55| 0.02, 9.53 x 9.57
60. x is in inches: |x – 3600| 4,3596 x 3604.
61. – ,
62. ,
63. ,
64. (–6 x –5) or (5 x 6)
65. (x –6) or (–5 < x < 5) or (x 6)
66. x
b + ca
b + ca
ab + dc
–ab + dc
ac + dab
ac – dab
52
67. D
68. F
69. C
70. G
71. [2] |x – 3| 5
–5 x – 3 5
–2 x 8
–1, 0, 1, 2, 3, 4, 5, 6, 7, 8
[1] only includes –2 x 8
and does not show work1-5
>–
<–<– <–
<–<– <–
<–<– <–
<– <–<– <–
<–>–
>–
<–<–<–<–
<–
<– <–
Absolute Value Equations and InequalitiesALGEBRA 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 12
75
1-5
>–
<–
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
77. x
78. t 7
79. 36
80. –
81. Inverse Prop. of Add.
82. Closure Prop. Of Mult.
83. Comm. Prop. of Mult.
1114
415
1-5
>–
<–
Absolute Value Equations and InequalitiesALGEBRA 2 LESSON 1-5
1. Solve |3x + 1| = 4.
2. Solve |–2x + 3| + 7 > 9. Graph the solution.
3. Solve |4x – 12| 8. Graph the solution.
4. A machinist is to drill a hole with diameter D inches, that satisfies 0.16 D 0.19. Express the specification using an absolute value inequality.
1-5
<–
<– <–
1 x 5–< –<
|D – 0.175| 0.015–<
– , 153
x < or x >12
52
1-6
ALGEBRA 2 LESSON 1-6
Write each number as a percent.
1. 2. 1
3. 0.0043 4.
5. 1.04 6. 3
(For help, go to Skills Handbook page 842.)
Probability
38
56
1 400
1. = 3 ÷ 8 = 0.375 = 0.375(100%) = 37.5%
2. 1 = = 11 ÷ 6 = 1.83 = 1.83(100%) = 183 %
3. 0.0043 = 0.0043(100%) = 0.43%
4. = 1 ÷ 400 = 0.0025 = 0.0025(100%) = 0.25%
5. 1.04 = 1.04(100%) = 104%
6. 3 = 3(100%) = 300%
Solutions
ALGEBRA 2 LESSON 1-6Probability
1 400
56
116
13
38
1-6
ProbabilityALGEBRA 2 LESSON 1-6
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
ProbabilityALGEBRA 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
ProbabilityALGEBRA 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.
1-6
26 6 equally likely outcomes are in
the sample space.
2 outcomes result in a multiple of 3.
13=
ProbabilityALGEBRA 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 bB BB Bbb Bb bb
Gene fromFather
Gene fromMother
Let B represent the dominant gene for brown eyes. Let b represent the recessive gene for blue eyes.
1-6
The sample space contains four equally likely outcomes {BB, Bb, Bb, bb}.
14
The outcome bb is the only one for which a child will have blue eyes. So, P(blue eyes) = .
14The theoretical probability that the child will have blue eyes is , or 25%.
ProbabilityALGEBRA 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)area of circle with radius 4r
=16 r 2 – 9 r 2
16 r 2
1-6
=16 r 2
7 r 2
=(4r)2 – (3r)2
(4r2)
= 7 16
The theoretical probability of hitting the outer ring is , or about 44%. 7 16
Probability
Pages 42–45 Exercises
1. 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%
ALGEBRA 2 LESSON 1-6
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%.
161340
179340
21134116745
268116745
26823
134
1-6
ProbabilityALGEBRA 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
ProbabilityALGEBRA 2 LESSON 1-6
6. , or 30%
7. , or 50%
8. , or 80%
9. , or 80%
10. , or 38.4%11. , or 15.2%
12. , or 82.4%13. , or 56%
14. , or 61.6%
31012
454548
12519
125103125142577
125
12
15. {Gg, Gg, gg, gg}; , or 50%
16. {Gg, Gg, Gg, Gg}; 1, or 100%
17. , or 6.25%18. , or 37.5%
19. , or 25%
20. , or 75%
21. a. 1
b. 0
116
381434
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 .520
14
1-6
ProbabilityALGEBRA 2 LESSON 1-6
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 .
24. , or 78.9%25. , or 35.4%26. , or 29.3%
12
1161475214743147
27. , or 21.1%28.
29.
30.
31. 0
32.
33. 1
34.
35.
311471612
23
16
13
49
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.
49
49
13616
1-6
ProbabilityALGEBRA 2 LESSON 1-6
39. 6.4%
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
42. or 6.7%
14
34
115
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
aa + b
ab – a
12
12
13
1323
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
1-6
ProbabilityALGEBRA 2 LESSON 1-6
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
19
no. of favorable outcomestotal no. of outcomes = .5
9
49
59
49
1-6
ProbabilityALGEBRA 2 LESSON 1-6
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 263. –15 x –3
53
125
245
13
1-6
>–
<– <–
<–
<– >–
<– <–
ProbabilityALGEBRA 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?
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
49
16
17
Tools of AlgebraALGEBRA 2 CHAPTER 1
1-A
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
163
83
15. 4
16. x = , a b17. x = 3c2, c 0
18. x = a(b – 1) + 5, a 0
19. x = , a b
20. r =
21. $138
22. 9
2aa – b
c – 2a – b
s2 h
=/
=/
=/
=/
Tools of AlgebraALGEBRA 2 CHAPTER 1
23. x –4
24. x > 7
25. t >
26. x < –16 or x >
12
25
27. 1 < d < 2
28. –1,
29. –
30. 3 p 7
32
53
31. w < – or w > 0
32.
33. 0
34.
35. 1
36. about 40%
43
25
35
1-A
>–
<– <–
Recommended