Chapter 2, 1 2 continued

continued
45. The price p varies directly with the length * and the
An approximation of the best-fitting line is
y 5 2126.7x 1 6687.
weight w varies directly with the length *,
so p 5 a1* and w 5 a 2*.
y
Solve the second equation for * and substitute the result
into the first equation.
w 5 a2*
p 5 a1*
w
}
a2 5 *
w
p 5 a1 }
a2
1 2
a1
p5}
w
a
2
a1
Because a1 and a2 are both nonzero constants, then }
a
a1
2
is also a nonzero constant. Therefore, p 5 }
w is a direct
a
(1, 6560)
6600
U.S. daily oil production
(thousands of barrels)
Chapter 2,
6400
6200
6000
(7, 5800)
5800
0
0
2
2
variation equation, and you can conclude that the price
of a necklace varies directly with its weight.
Mixed Review for TAKS
46. B;
The sale price is dependent on the original price.
47. G;
0.08x 5 300
x 5 3750
$3750
} 5 $750 in sales per day
5 days
4
6
8
x
Years since 1994
b. When x 5 15:
y 5 2126.7x 1 66875 2126.7(15) 1 6687 ø 4787
You can predict that the daily oil production in 2009 in
the U.S. will be about 4,787,000 barrels.
c. Using the linear regression
feature, the equation
can be rounded to
y 5 2129.8x 1 6702.
Graph the regression
equation with the
scatter plot.
LinReg
y=ax+b
a=-129.8333333
b=6701.555556
r2=.9316903787
r=-.9652410987
Lesson 2.6
Investigating Algebra Activity 2.6 (p. 112)
2.6 Guided Practice (pp. 114–117)
1. a. The scatter plot shows a clear but fairly weak positive
correlation.
b. Because the correlation is fairly weak, the correlation
coefficient is closest to 0.5 (r 5 0.5).
2. a. The scatter plot shows a strong negative correlation.
b. Because the correlation is strong, the correlation
coefficient is closest to 21 (r 5 21).
3. a. The scatter plot shows approximately no correlation.
b. The correlation coefficient is closest to 0 (r 5 0).
4. Sample answer:
a. The line of best fit appears to pass through (1, 6560)
and (7, 5800).
5800 2 6560
2760
m5}
5}
ø 2126.7
721
6
y 2 y1 5 m(x 2 x1)
y 2 6560 5 2126.7(x 2 1)
y 2 6560 5 2126.7x 1 126.7
y ø 2126.7x 1 6687
When x 5 15:
y 5 2129.8x 1 6702 5 2129.8(15) 1 6702 5 4755
You can predict that the daily oil production in 2009 in
the U.S. will be about 4,755,000 barrels.
2.6 Exercises (pp. 117–120)
Skill Practice
1. A line that lies as close as possible to a set of data points
(x, y) is called the best-fitting line for the data points.
2. If y tends to increase as x increases, then the data
have a positive correlation. If y tends to decrease as x
increases, the the data have a negative correlation. If
the points show no obvious pattern, then the data have
approximately no correlation.
3. The data have a negative correlation because as the
x-values increase, the y-values tend to decrease.
4. The data have a positive correlation because as the
x-values increase, the y-values tend to increase.
5. The data have approximately no correlation because the
points show no obvious pattern.
6. By examining the data in a table, you can determine
whether the y-values increase, decrease, or have no
pattern as the x-values increase.
76
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1–5. Answers will vary.
Chapter 2,
continued
7. The correlation coefficient is closest to 0 because the
12. a–b. Sample answer: The line of best fit appears to pass
scatterplot shows approximately no correlation.
through (x1, y1) 5 (25, 75) and (x2, y2) 5 (50, 30).
y2 2 y1
8. The correlation coefficient is closest to 0.5 because the
scatterplot show a fairly weak positive correlation.
2
scatterplot shows a strong negative correlation.
9
through (x1, y1) 5 (1, 10) and (x2, y2) 5 (5, 62).
62 2 10
9
y 2 75 5 2}5 x 1 45
52
m5}
5}
5}
5 13
x 2x
521
4
9
y 5 2}5 x 1 120
1
y 2 y1 5 m(x 2 x1)
An approximation of the best-fitting line is
y 2 10 5 13(x 2 1)
9
y 5 2}5 x 1 120.
y 5 13x 2 13 1 10
y 5 13x 2 3
y
100
An approximation of the best-fitting line is
y 5 13x 2 3.
60
(5, 62)
56
40
42
(50, 30)
20
28
0
14
0
1
2
3
4
28
42
56
x
y 5 2}5 (20) 1 120
y 5 84
y 5 13(20) 2 3
13. a–b. Sample answer: The line of best fit appears to pass
y 5 257
through (x1, y1) 5 (3, 20) and (x2, y2) 5 (15, 102).
11. a–b. Sample answer: The line of best fit appears to pass
y2 2 y1
42 2 120
102 2 20
82
41
m5}
5}
5}
5}
x 2x
15 2 3
12
6
through (x1, y1) 5 (1, 120) and (x2, y2) 5 (5, 42).
2
278
1
y 2 y1 5 m(x 2 x1)
m5}
5}
5}
5 219.5
x 2x
521
4
2
14
9
5 x
c. When x 5 20:
y2 2 y1
0
c. When x 5 20:
(1, 10)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
(25, 75)
80
y
0
9
y 2 75 5 2}5 (x 2 25)
10. a–b. Sample answer: The line of best fit appears to pass
2
245
1
y 2 y1 5 m(x 2 x1)
9. The correlation coefficient is closest to 21 because the
y2 2 y1
30 2 75
m5}
5}
5}
5 2}5
x 2x
50 2 25
25
1
41
y 2 20 5 }
(x 2 3)
6
y 2 y1 5 m(x 2 x1)
y 2 120 5 219.5(x 2 1)
41
41
41
1
x2}
y 2 20 5 }
6
2
y 5 219.5x 1 19.5 1 120
y 5 219.5x 1 139.5
x 2 }2
y5}
6
An approximation of the best-fitting line is
y 5 219.5x 1 139.5.
An approximation of the best-fitting line is
y
120
y5}
x 2 }2.
6
41
(1, 120)
96
y
120
72
90
1
(15, 102)
48
60
(5, 42)
(3, 20)
24
0
30
0
1
2
3
c. When x 5 20:
y 5 219.5(20) 1 139.5
y 5 2250.5
4
0
5 x
c.
0
4
8
12
16
x
When x 5 20:
41
1
y5}
(20) 2 }2
6
y ø 136
Algebra 2
Worked-Out Solution Key
77
Chapter 2,
continued
14. a–b. Sample answer: The line of best fit appears to pass
16. B; On the scatter plot, a line of best fit might pass
through (x1, y1) 5 (5.6, 120) and (x2, y2) 5 (8.4, 167).
through the points (x1, y1) 5 (10, 21) and
(x2, y2) 5 (30, 11).
y2 2 y1
167 2 120
47
m5}
5}
5}
ø 16.8
x2 2 x1
8.4 2 5.6
2.8
y 2 y1 5 m(x 2 x1)
y2 2 y1
11 2 21
2
1
1
y 2 120 5 16.8(x 2 5.6)
y 2 y1 5 m(x 2 x1)
y 2 120 5 16.8x 2 94.08
y 2 21 5 2}2 (x 2 10)
1
1
y ø 16.8x 1 26
An approximation of the best-fitting line is
y 5 16.8x 1 26
y
y 2 21 5 2}2 x 1 5
1
y 5 2}2 x 1 26
17. The error was made by drawing the line too low. When
(8.4, 167)
160
you sketch a line of best fit, there should be about as
many points above the line as below it.
140
120
100
0
210
5}
5}
5 2}2
m5}
x 2x
30 2 10
20
y
60
(5.6, 120)
40
0
2
4
6
x
8
c.
When x 5 20:
y 5 16.8(20) 1 26
y 5 362
15. a–b. Sample answer: The line of best fit appears to pass
through (x1, y1) 5 (16, 3.9) and (x2, y2) 5 (68, 2.6).
y2 2 y1
2.6 2 3.9
21.3
m5}
5}
5}
ø 20.025
x 2x
68 2 16
52
2
1
y 2 y1 5 m(x 2 x1)
20
0
0
2
4
6
8
x
18. A; The data points would lie closest to a line for the
correlation coefficient that is closest to 1 or 21. The
correlation coefficient r 5 20.96 is the closest to 21.
19. Using the linear regression feature, the equation can
be rounded to y 5 0.051x 1 1.1 Graph the regression
equation with the scatter plot.
y 2 3.9 5 20.025x 1 0.4
y 5 20.025x 1 4.3
An approximation of the best-fitting line is
y 5 0.025x 1 4.3.
LinReg
y=ax+b
a=.0509796028
b=1.138097155
r2=.9891292697
r=.9945497824
y
20. Using the linear regression feature, the equation can be
(16, 3.9)
4
rounded to y 5 2 0.006x 1 98.1. Graph the regression
equation with the scatter plot.
3
(68, 2.6)
2
LinReg
y=ax+b
a=-.0059565349
b=98.13247812
r2=.9977613019
r=-.9988800238
1
0
0
20
40
60
80
x
c. When x 5 20:
y 5 20.025(20) 1 4.3
y 5 3.8
21. a. Sample answer: Measuring the depth of water at
different times while filling a swimming pool; the
number of gallons of milk you buy and the total cost.
b. The age of a car and its current value; the number of
miles you have driven since you last put gas in the tank
and the amount of gas left in the tank.
c. The height of a person and the month they were
born; the age of a person and the number of vehicles
they own.
78
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
y 2 3.9 5 20.025(x 2 16)
continued
22. It is not logical to use the best fitting line to make
predictions from the data because the positive correlation
is very weak. The closer the value or r is to 1 or 21, the
stranger the correlation. Because the data set has a weak
correlation, the data are not strongly related.
23. x and z would have a negative correlation because as x
increases, y tends to increase, and as y increases, z tends
to decrease. So, as x increases, z tends to decrease.
An approximation of the best-fitting line is
y 5 20.00186x 1 212.1.
y
215
Boiling point (8F)
Chapter 2,
195
0
24. The line of best fit appears to pass through
When x 5 14,000:
2.8
m5}
5}
5}
5 0.4
7
x 2x
720
y 5 20.00186(14,000) 1 212.1
1
y 5 0.4x 1 19.7
An approximation of the bestfitting line is y 5 0.4x 1 19.7.
Population (millions)
y 2 y1 5 m(x 2 x1)
y 2 19.7 5 0.4(x 2 0)
y ø 186.1
y
24
22
27. a. (0, 37), (4, 49), (8, 57), (12, 64), (14, 67), (18, 72),
20
(22, 77)
(0, 19.7)
18
0
b.
0
2
4
6
Years since 1997
25. The line of best fit appears to pass through
(x1, y1) 5 (0, 2240) and (x2, y2) 5 (6, 2850).
y2 2 y1
2850 2 2240
So, the boiling point of water at an elevation of 14,000
feet is about 186.18F.
(7, 22.5)
x
Number of countries
2
610
5}
5}
ø 101.7
m5}
x 2x
620
6
2
y 2 y1 5 m(x 2 x1)
(0, 41)
20
0
6
12
18
24
x
(x1, y1) 5 (0, 41) and (x2, y2) 5 (16, 70).
y2 2 y1
An approximation of the best-fitting line is
y 5 101.7x 1 2240.
70 2 41
29
5}
5}
ø 1.8
m5}
x 2x
16 2 0
16
2
1
y 2 y1 5 m(x 2 x1)
(6, 2850)
y 2 41 5 1.8(x 2 0)
2800
y 5 1.8x 1 41
2600
An approximation of the best-fitting line is
y 5 1.8x 1 41.
When x 5 34:
2400
2200 (0, 2240)
0
2
y 5 1.8(34) 1 41
4
6
y 5 102.2
x
Years since 1997
26. The line of best fit appears to pass through
(x1, y1) 5 (2000, 208.4) and (x2, y2) 5 (12,000, 189.8).
y2 2 y1
189.8 2 208.4
218.6
29.3
5 }}
5}
5}
m5}
x 2x
12,000 2 2000
10,000
5000
1
5 20.00186
y 2 y1 5 m(x 2 x1)
y 2 208.4 5 20.00186(x 2 2000)
y 5 20.00186x 1 3.72 1 208.4
y 5 20.00186x 1 212.12
So, you can predict that about 102 countries will
participate in 2014.
28. a. There is a negative correlation between the year and
the number of cassettes shipped.
Cassette shipments (millions)
Average annual public
college tuition (dollars)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
40
c. The line of best fit appears to pass through
y 5 101.7x 1 2240
2
(16, 70)
60
Years since 1980
y 2 2240 5 101.7(x 2 0)
0
y
80
0
1
y
4000 8000 12,000 x
0
Elevation (feet)
(x1, y1) 5 (0, 19.7) and (x2, y2) 5 (7, 22.5).
22.5 2 19.7
(12,000, 189.8)
185
Problem Solving
y2 2 y1
(2000, 208.4)
205
y
400
300
200
100
0
0
4
8
12
16
x
Year since 1988
Algebra 2
Worked-Out Solution Key
79
Chapter 2,
continued
b. There is a positive correlation between the year and the
number of CDs shipped.
5. Let (x1, y1) 5 (0, 7) and (x2, y2) 5 (23, 22).
y2 2 y1
1
y 2 y1 5 m(x 2 x1)
600
y 2 7 5 3(x 2 0)
y 5 3x 1 7
400
6. Let (x1, y1) 5 (29, 9) and (x2, y2) 5 (29, 0).
200
y2 2 y1
0
4
8
12
16
029
29
5}
5}
Undefined
m5}
x 2x
0
29 2 (29)
x
2
Years since 1988
cassette shipments. As CD shipments increased,
cassette shipments decreased. Over time it has become
cheaper and easier to make CDs. In addition, they last
longer and sound better than cassettes.
7. y 5 ax
1
25a
8.
y 5 ax
4
8 5 a(22)
a result of technological advancement, not a cause.
Giving people more personal computers will not
necessarily impact all of the other factors needed for
higher life expectancies.
24 5 a
y 5 ax
y
4
216
}5a
5
Area 5 * 3 w 5 55 3 20 5 1100 ft2
The area of the garden is 1100 square feet.
31. H;
16
x.
So, y 5 2}
5
10. y 5 ax
y
1
}5a
3
Quiz 2.4 –2.6 (p.120)
1. y 5 mx 1 b
y 5 25x 1 3
3. y 2 y1 5 m(x 2 x1)
y 2 6 5 4(x 2 (23))
y 5 4x 1 12 1 6
y 5 4x 1 18
y 2 y1 5 m(x 2 x1)
y 2 (24) 5 27(x 2 1)
y 5 27x 1 7 2 4
y 5 27x 1 3
Algebra 2
Worked-Out Solution Key
y 5 2x 1 12
1
23
1
2. y 5 mx 1 b
x
21
4 5 a(12)
The y-intercept is 2.
x
21
216 5 a(5)
1
width 5 }2 (150 2 110) 5 20 feet
y
So, y 5 24x.
9.
30. B;
x
21
So, y 5 2x.
b. No. The number of personal computers per capita is
Mixed Review for TAKS
y
2 5 a(1)
29. a. The number of personal computers per capita in a
country probably indicates the level of technological
advancement in that country. As technology advances,
so do improvements in medicine and communication,
which can result in higher life expectancies.
1
Because the slope is undefined, the line is vertical. A
vertical line that passes through (29, 9) has the equation
x 5 29.
c. There is a negative correlation between CD and
80
29
So, y 5 }3 x.
11. When x 5 4 and y 5 12:
y 5 ax
12 5 a(4)
35a
An equation is y 5 3x.
When x 5 8:
y 5 3(8)
y 5 24
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
CD shipments (millions)
2
800
0
4.
22 2 7
5}
5}
53
m5}
x 2x
23 2 0
23
y
Chapter 2,
continued
12. When x 5 23 and y 5 28:
When x 5 11:
y 5 3.5(11) 1 40
y 5 ax
y 5 78.5
28 5 a(23)
So, you can predict that the average price of a ticket in
2010 will be about $78.50.
8
}5a
3
8
An equation is y 5 }3 x.
Lesson 2.7
When x 5 8:
Investigating Algebra Activity 2.7
8
y 5 }3 (8)
(pp. 121–122)
1. The graphs of the form y 5 a{x{where a < 0 are
64
1
, or 21 }3
y5}
3
reflections of y 5{x{in the x-axis, but are either
13. When x 5 40 and y 5 25:
wider or narrower depending on whether
y 5 ax
21 < a < 0 or a < 21. When a 5 21, the graph
25 5 a(40)
is a simple reflection of y 5{x{in the x-axis.
1
2}8 5 a
1
An equation is y 5 2}8 x.
When x 5 8:
1
y 5 2}8 (8)
y 5 21
2. y 5 {x{ 1 6 has the same shape as the graph of
14. When x 5 12 and y 5 2:
y 5 {x{, but is shifted 6 units vertically.
y 5 ax
2 5 a(12)
1
6
}5a
When x 5 8:
1
3. y 5 {x{ 2 4 has the same shape as the graph of
y 5 }6 (8)
4
y 5 {x{, but is shifted 24 units vertically.
1
y 5 }3 , or 1 }3
15. The line of best fit appears to pass through
(x1, y1) 5 (0, 40) and (x2, y2) 5 (5, 57.5).
y2 2 y1
57.5 2 40
17.5
5}
5}
5 3.5
m5}
5
x 2x
520
2
1
y 2 y1 5 m(x 2 x1)
y 2 40 5 3.5(x 2 0)
y 5 3.5x 1 40
4. y 5 {x 2 3{has the same shape as the graph of
y 5 {x{, but is shifted 3 units horizontally.
An approximation of the best-fitting line is
y 5 3.5x 1 40.
Ticket price (dollars)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
An equation is y 5 }6 x.
y
65
(5, 57.5)
55
45
(0, 40)
35
0
0
2
4
x
Years since 1999
Algebra 2
Worked-Out Solution Key
81