Check Your Understanding (Example 3)
4
P = D(1 – r)
4
⎛⎞
P
()
1
44
1−
()
r
4
⎛⎞
⎜⎟
P
r +
⎜⎟
1
4
⎛⎞
P
Check Your Understanding (Example 4)
4 years is to the right of the point of intersection.
more than what the car is worth.
Check Your Understanding (Example 6)
26,600 ÷ 2 = 13,300
Applications
1. To some people, the value of a car is more than
2. Use y = a(1 – b)x, where a = 40,000 and
3. Use y = a(1 – b)x, where a = 28,000 and
4. Use y = a(1 – b)x, where a = 19,700 and
5. 100 ÷ 12 ≈ 8.33; 100 months is 8 1
3 years.
b = 0.08625.
y = 21,000(1 – 0.08625)x
6. Use y = a(1 – b)x, where a = D and b = 100
E.
For M years, divide M by 12, y = 12
1100
⎛⎞
−
⎜⎟
⎝⎠
E.
7. 1 – 0.785 = 0.215 = 21.5%;
8. 1 – 0.8625 = 0.1375 = 13.75%;
= 18,547.23(0.8625)6 ≈ $7,635.43;
y = 18,547.23 (1 – 0.1375)w
= 18,547.23(0.8625)w
9a. The minimum on the x-axis is 0, the maximum is
y-axis is 0, the maximum is 20,000, and the
9b. Enter the data from the table into your
9c. 1 – 0.89 = 0.11 = 11%
9d. y = 17,895.97*(0.89)3.5 ≈ 11,902.01
10a. The minimum on the x-axis is 0, the maximum is
10b. Enter the data from the table into your
calculator. Then use the statistics menu to
10c. 1 – 0.89 = 0.11 – 11%
11. 16,000 = A(1 – 0.545)3
12. 6,700 = A(1 – 0.1415)8
6,700 = A(0.8585)8
14. 18,700 = 30,000(1 – r)5
15. 32 ÷ 12 = 2.66667
0.9472256259 = 1 – r
–0.0527743741 = –r
0.053 ≈ r
18. $550 × 12 = $6,600
The yearly expense equation is 6,600x + 10,000.
The exponential depreciation equation is
(1.84, 22,131.10). The value of the car will be
equal to the amount paid after about 1.8 years.
19. $800 × 12 = $9,600
y = 46,600(1 – 0.103)x. Use a graphing
calculator and enter each equation. Use the
(2.46, 35,361.67). The value of the car will be
20a. The minimum on the x-axis is 0, the maximum is
50, and the scale is 1. The minimum on the
20b. Enter the data from the table into your
calculator. Then use the statistics menu to
The equation of the regression line is
y = 2,262.70 × (1.09)x.
20c. An exponential appreciation rate has been used.
Lesson 5-7 Driving Data
Check Your Understanding (Example 1)
Check Your Understanding (Example 2)
0.15517241 × 60 ≈ 9.3
Check Your Understanding (Example 3)
Check Your Understanding (Example 4)
Check Your Understanding (Example 5)
Check Your Understanding (Example 6)
Check Your Understanding (Example 7)
2.675 ÷ 3.8 ≈ 0.70
Applications
1. Although this analyst is probably using a car as
4. 17 ÷ 2 = 8.5
5. 2,100 ÷ 30 = 70
6. Divide miles by miles per gallon: 270
M
7a. 75,421.1 – 74,902.6 = 518.5
10a. x ÷ 42 = 42
x
10b. $2.38 × 42
x
= 238
42
.
x
12b. 12 × $4.08 = $48.96
12c. 17 × $4.15 = $70.55
12d. 26 × $4.30 = $111.80
12e. 15 × $D = $15D
12l. $GP ÷ C = $GP
C
13. $59.76 ÷ $4.15 = 14.4
14.4 × 25 = 360
14. 280 ÷ 1.60934 ≈ 173.984
174 + 42 = 216
216 ÷ 4.3 ≈ 50.23, which is about 50
15c. When hours = 14, miles is about 800.
15d. When miles = 500, hours is about 9.
16a. Find the number of miles: A2 – A1
16c. Multiply the gallons used (which from part b is
now stored in C2) by the cost per gallon:
=C2*A5
17a. 3.80 × 1.07 ≈ 4.07
17j. 178.50 × 1.07 ≈ 191.00
17k. 178.50 × 0.69 ≈ 123.17
17l. 178.50 × 1.16 ≈ 207.06
17m. 250.00 × 1.07 = 267.50
17q. 5,500.00 × 1.07 = 5,885.00
17r. 5,500.00 × 0.69 = 3,795.00
18a. 85 ÷ 1.07 ≈ 79.44
18b. 1,000 ÷ 0.69 ≈ 1,449.27
18c. 500 ÷ 1.16 ≈ 431.03
18d. 130 ÷ 1 = 130
20b. $2.62 × 3.8 ≈ $9.96
21a. h ÷ 0.69 = 069.
h
22a. 87.42 ÷ x = 87 42.
x
22b. 87 42.
x
× 3.8 = 3 8 87 42.
(
.
)
x
= 332 196.
x
(
)
x
x
x
Lesson 5-8 Driving Safety Data
Check Your Understanding (Example 1)
Check Your Understanding (Example 2)
Road conditions such as rain, snow, and ice affect
Check Your Understanding (Example 4)
2
Extend Your Understanding (Example 4)
Toni’s total stopping distance:
2
72 72
170 5
+=
5 625 75
170 5
+
, ≈ 33.09 + 15 ≈ 48.09
Applications
1. The total braking distance is substantial for a
normal, non-impaired driver. People driving
2d. 5,720 ÷ 60 ≈ 95
3a. 42 × 1 = 42
3b. 42 × 5,280 = 221,760
3f. 5,280 × x = 5,280x
3g. 5,280x ÷ 60 = 5280
,
x
x
x
4a. 80 × 1 = 80
5a. 55 × 1 = 55
5b. 55 × 1,000 = 55,000
6a. x × 1 = x
6b. x × 1,000 = 1,000x
,
x
x
x
x
7. The reaction distance is about 1 foot for each
mile per hour, so a car going 32 mi/h has a
reaction distance of about 32 feet.
8a. 68 × 1 = 68
8b. 68 × 5,280 = 359,040
8c. 359,040 ÷ 60 = 5,984
8d. 5,984 ÷ 60 ≈ 99.73
9b. (0.1 × 52)2 × 5 = (5.2)2 × 5 = 27.04 × 5 = 135.2
9c. 52 + 135.2 = 187.2
10a. The reaction distance is about 1 foot for each
mile per hour, so a car going 40 mi/h has a
reaction distance of about 40 feet.
10b. The reaction distance is about 1 foot for each
reaction distance of about 5 feet.
10f. (0.1 × 40)2 × 5 = (4)2 × 5 = 16 × 5 = 80
10g. (0.1 × 30)2 × 5 = (3)2 × 5 = 9 × 5 = 45
10h. (0.1 × 20)2 × 5 = (2)2 × 5 = 4 × 5 = 20
12.
2
70 70
+ = 4 900 70
+
, ≈ 28.82 + 14 ≈ 42.82;
42.82 < 50, so Martine has enough room ahead
of her to bring the car to a safe stop before the
accident.
13a. Enter the equation as Y1 = X^2/20+X in a
13b. When mi/h = 53, feet is about 185.
13c. When feet = 70, mi/h is about 28.
14a. Enter the equation as Y1 = X^2/170+X/5 in a
graphing calculator. The graph is shown below.
15c. Use the total stopping distance formula and the
kilometers per hour conversion in cell A2.
Lesson 5-9 Accident Investigation
Data
Check Your Understanding (Example 1)
Check Your Understanding (Example 2)
⎝⎠
=
2
MA
⎝⎠
x
Check Your Understanding (Example 3)
Applications
1. Clearly the quote is said in jest, but there is a
2.
55 55 62 62
4
+++ = 234
4 = 58.5
The police were correct since according to the
formula, Rona’s minimum skid speed was
approximately 39.7 miles per hour.
6a. The formula needs to find average, so use
7. D = 36 45
+= 81
8b. y = 30 50 0.9
×
××
x
= 1, 3 5 0
x
8c. The graph is shown below.
efficiency is 60% (0.6), the speed is about 29
mi/h. When braking efficiency is 80% (0.8), the
×
10. 35 = 30 0.97 0.9×× ×D
11a. 36 = 30 50 0.8×××f
11b. 50 ± 4
2
42 4 8
,.
13. 70 =
26
+
C
14a. r =
2
+
CM
2
31 3
+
+ ≈ 41.5
14b. S = 15fr = 15 1.02 41.5×× ≈ 25.2
16a. AE = AB – EB = 2r – M
16c. 2r – M(M) =
2
()
CD
()
2r – M =
4
()
CD
M
()
r =
()CD M
17a. =(B2^2/(8*B3)+B3/2)
17b. =sqrt(15*B1*B4)
8
x =
8
x=
2
x
18b. S = 15fr =
15 1.05 2
××
x
Assessment
Really? Really! Revisited
1. Answers vary. Sample graph is shown
below.
4. Answers vary.
Applications
character over 100. To represent the number of
34 0 09 100 when 100
⎩
= $49 + $19 + $30
3. 6 – 3 = 3, so there are 3 extra lines.
$d + $a(3) = d + 3a
4a. The y-intercept is 36,000, so the maximum value
(or its original price) is $36,000.
4b. The amount of value it loses each year is the
4c. 0 = –2,400x + 36,000
5a. Use (0, 43,500) and (12, 0) to find the slope.
5b. $43,500 ÷ 4 = $10,875
–32,625 = –3,625x
6a. $350 + $350 + $390 + $400 + $500 + $500 +
6b. There are 14 values, so the median is the mean
6c. The data value $500 occurs 3 times; all other
6d. Q2: from part b, the median is $550.
6e. IQR = Q3 – Q1 = $800 – $400 = $400
6g. $800 + 1.5($400) = $800 + $600 = $1,400;
There is 1 data value ($1,700) above $1,400.
7a. Amount financed = $37,800 – $7,000 = $30,800.
7b. The graph is shown below.
7c. Use a graphing calculator to find the point of
intersection of the lines. Rounded to the nearest
hundredth, the coordinates of the point of
car is less than what was invested in it.
8b. (11 + 18 + 19 + 19 + 20 + 20 + 23 + 34 + 36 +
36 + 37 + 37 + 38 + 40 + 41 + 42 + 55 + 55 + 56
8c. There are 25 values, so the median is the
thirteenth-least value, which is 38.
seventh-least value. The sixth-least value is 20.
The seventh-least value is 23.
20 23
2
+ = 43
2 = 21.5
11d. The medical bills are covered by PIP insurance.
8g. 38 is Q2. By definition, 50% of the values are
8h. IQR = Q3 – Q1 = 57 – 21.5 = 35.5 12,239.74 ≈ A, which rounds to $12,200
8i. 38 is Q2. 57 is Q3. By definition, 25% of the
13a. 1 – 0.815 = 0.185 = 18.5%
y = 28,158.50(1 – 0.185)3.25
= 28,158.50 (0.815)3.25 ≈ $14,500
9a. The fence is covered by property damage.
9b. The fence is covered in full; $7,000.
13d. y = 28,158.50(1 – 0.185)y
= 28,158.50 (0.815)y
9c. The bus is covered by property damage and is
covered in full, $20,000.
14. 980 ÷ 25 = 39.2
9d. The car is covered by collision insurance.
175 × $1.49 = $260.75 10a. The minimum on the x-axis is 0, the maximum is
10, and the scale is 1. The minimum on the
16. The reaction distance is about 1 foot for each
37.85 > 30, so Marlena does not have enough
The equation of the regression line is
y = 31,985.36*(0.91)x.
19. D = 69 70 74
3
3 = 71
107.3 ≈ D
11b. Because the hand was injured, the bodily injury
2
2