Chapter 5
Lesson 5-1 Classified Ads
Check Your Understanding (Example 1)
Check Your Understanding (Example 2)
Check Your Understanding (Example 3)
Check Your Understanding (Example 4)
Check Your Understanding (Example 5)
Applications
1. Although consumers consider gas mileage,
condition, options, status, and other things when
7. $18,500 × 0.05 = $925
$18,500 – $925 = $17,575
8. 27 – 20 = 7, so there are 7 extra words.
$18 + $0.35(7) = $18 + $2.45 = $20.45
9a. The cost is $46 for 200 characters or less. The
number of characters over 200, use the
expression x – 200.
9b. The graph of the function is shown below.
value of c(x) when x = 200 is 46. The cusp is at
$26 – $2.60 = $23.40
11. $67 × 2.5 = $167.50
$48 + $5(2) = $48 + $10 = $58
12b. $52,900 × 0.08 = $4,232
$52,900 – $4,232 = $48,668
14d. The graph of the function is shown below.
of c(x) when x = 4 is $38. The cusp is at
(4, $38).
.
29 6 75 5 when 5
() .( )
cx xx
=⎨+− >
⎩
17b. $11 × 3 + $5 × (5 – 3) =
$43
each line over 3. To represent the number of
.
11 when 3
33 5 3 when 3
() ()
xx
cx xx
≤
⎧
=⎨+− >
⎩
5 = $12.50 for the first 5 lines and $8 for each
line over 5. To represent the number of lines
Lesson 5-2 Buy or Sell A Car
Check Your Understanding (Example 1)
Check Your Understanding (Example 2)
Check Your Understanding (Example 3)
Check Your Understanding (Example 4)
Check Your Understanding (Example 5)
Check Your Understanding (Example 6)
$8,650 – $6,700 = $1,950
Check Your Understanding (Example 7)
Check Your Understanding (Example 8)
Applications
1. With the tremendous crunching and availability
2a. mean: 7 + 12 + 1 + 7 + 6 + 5 + 11 = 49; 49 ÷ 7
= 7
2b. mean: 85 + 105 + 95 + 90 + 115 = 490; 490 ÷ 5
= 48
2c. mean: 10 + 14 + 16 + 16 + 8 + 9 + 11 + 12 + 3
= 99; 99 ÷ 9 = 11
2d. mean: 10 + 8 + 7 + 5 + 9 + 10 + 7 = 56; 56 ÷ 7
= 8
2e. mean: 45 + 50 + 40 + 35 + 75 = 245; 245 ÷ 5 =
2f. mean: 15 + 11 + 11 + 16 + 16 + 9 = 78; 78 ÷ 6 =
13
+11 15
2 = 26
2 = 13
the data is not skewed. In 2b, the mean is 48
and the median is 95, so the data is skewed. In
the same, so the data is not skewed.
4. $24,600 + $19,000 + $33,000 + $15,000 +
least value, which is $145
5c. Stephanie’s salary is much larger than the rest,
values are below the mean.
7. 90 × 5 = 450
450 – (91 + 82 + 90 + 89) = 450 – 352 = 98
11b. There are 14 values, so the median is the mean
of the seventh-least and eighth-least values. The
seventh-least value is $245. The eighth-least
value is $250.
+245 250
2
$$
= 495
2
$ = $247.50
11e. IQR = Q3 – Q1 = $320 – $210 = $110
boundary. It is $10.
12a. Q2: there are 8 values, so Q2 is the mean of the
fourth-least and fifth-least values. The fourth-
greatest) and seventh-least (second-greatest)
values. The sixth-least value is $400. The
seventh-least value is $450.
+400 450
2
$$
= 850
2
$ = $425
⎜⎟
⎝⎠
8
14. Answers vary. Numbers in the list must have a
16. Answers vary. Numbers in the list must have a
sum of 60 (5 × 12). Third-least number must be
17. Answers vary. Sample: 3, 5, 10, 10, 10, 10, 12,
18. Answers vary. Sample: 1, 18, 20, 21, 22, 109.
Lesson 5-3 Graph Frequency
Distributions
$9,900 1
$10,800 2
$11,000 1
$12,500 2
6k. The graph is shown below.
7b. If the data has an outlier, then a modified
7c. There are no single points outside the whiskers,
8a. ($22,000 + $19,000 + $18,000 + $16,700 +
8c. There are 5 data values, so the median is the
8d. The 2 values below the median are $16,700 and
9a. Enter the data from the table into your
9b. Use a graphing calculator to find the correlation
9c. Because the correlation coefficient, r, is close to
10a. There are 26 leaves in the table, so 26 students
+ 75 + 75 + 77 + 82 + 82 + 83 + 84) ÷ 26 =
1,615 ÷ 26 ≈ 62.12, or $62.12
10c. There are 26 values, so the median is the mean
2 = $130
2 = $65
10d. The data value 53 occurs 5 times; all other data
10e. $84 – $17 = $67
Q1: there are 13 values in the lower half of the
data, so Q3 is the twentieth-least (seventh-
Q4: maximum value, which is $84
above Q1, so 75% of the students spent $53 or
values are above Q3. 100% – 25% – 25% = 50%
10l. There is 1 data value ($17) below $20.
11c. The data value $226 occurs 3 times, all other
11d. There are 18 data values, so the median is the
12. The sum of the data is xy + 5w + 16 × 4 + 18v =
xy + 5w + 64 + 18v. The number of data values
Lesson 5-4 Automobile Insurance
Check Your Understanding (Example 1)
Check Your Understanding (Example 3)
Check Your Understanding (Example 4)
Applications
will pay the maximum coverage amount,
$25,000.
3c. $410 ÷ 2 = $205
4. $924
× 0.4 = $369.60
$924 × 0.3 = $277.20
9b. The fire hydrant is covered in full; $1,400.
10a. The sign is covered by property damage
insurance (PD).
x – 0.35x.
The monthly premium is −035
.
x
x or 065
.
x
.
14a. With the discount, the annual premium is
quarterly payment = 09 +
xy
.
15b. With the increase, the annual premium is
x
(sixth-greatest) and sixteenth-least (fifth-
greatest) values, 60 and 61, so Q3 is 60.5.
Lesson 5-5 Linear Automobile
Depreciation
Check Your Understanding (Example 1)
Check Your Understanding (Example 2)
Check Your Understanding (Example 3)
Check Your Understanding (Example 4)
Check Your Understanding (Example 5)
In the equation, let y = D and solve for x. The
The expression is 32 000
4000
−
,
Check Your Understanding (Example 6)
to-date earlier in the car’s lifetime.
Applications
1. This somewhat cynical quote compares the
2b. 21
x
2d. The graph is shown below.
3b. 21
21
−
y
−
x
x = 0 11200
70
−
3c. The slope is –1,600 and the y-intercept is
11,200. The equation is y = –1,600x + 11,200.
4a. The vertical scale increases by $7,000 with each
4c. Use (0, 28,000) and (10, 0) to find the slope.
21
The slope is –2,800 and the y-intercept is
28,000. The equation is y = –2,800x + 28,000.
5a. The y-intercept is 85,000, so the maximum value
−
−
6a. The y-intercept is 17,200, so the maximum value
(or its original price) is $17,200.
7a. y = –2,750(5) + 22,000
7b. y = –2,750(8) + 22,000
−
−
−
A
8c. M ÷ 12 = 12
M; M months = 12
Myears
increases by 4 ÷ 2 = 2 with each gridline. The
9c. The vertical scale increases by $16,000 ÷ 5 =
$3,200 with each gridline. The maximum value
(y-intercept is $3,200 × 8 = $25,600. Half the
34,450. The equation is y = –2,650x + 34,450.
10b. 34,560 ÷ 2 = 17,280
2 650
−
,= x
11a. Use (0, A1) and (B1, 0) to find the slope.
The length of time, x, is C1, so the value when
x = C1 is y =
A
1
−
(C1) + A1, and the formula
12a. The depreciation equation is y = –B1x + A1.
12b. The depreciation equation is y = –B1x + A1.
12c. Find the complement of the percent: 100 – E1.
((100 – E1)/100)*A1 = –B1x + A1
The formula is (((100 – E1)/100)*A1 – A1)/–B1.
13a. Amount financed = $54,000 – $8,000 = $46,000.
Use the monthly payment formula:
Substitute p = $46,000, r = 0.04875, t = 4
12(4)
12(4)
0.04875 0.04875
$46,000 1
12 12
0.04875
11
12
⎛⎞⎛ ⎞
+
⎜⎟⎜ ⎟
⎝⎠⎝ ⎠
=
⎛⎞
+−
⎜⎟
⎝⎠
M
The monthly payment is $1,056.74.
13b. The graph is shown below.
13c. Use a graphing calculator to find the point of
intersection are (30.53, 40,261.73). This means
that after a little more than 30.5 months, both the
Lesson 5-6 Historical and
Exponential Depreciation
Check Your Understanding (Example 1)
Check Your Understanding (Example 2)