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Novay_Z [31]
3 years ago
6

One car model costs $12,000 and costs an average of $0.10 per mile to maintain. Another car model costs $14,000 and costs $0.08

per mile to maintain. If one of each model is driven the same number of miles, after how many miles would the total cost of one model be the same as the other?
Mathematics
1 answer:
BlackZzzverrR [31]3 years ago
3 0
Answer:
See a solution process below:
Explanation:
Let's call the number of miles driven we are looking for
m
.
The the total cost of ownership for the first car model is:
12000
+
0.1
m
The the total cost of ownership for the second car model is:
14000
+
0.08
m
We can equate these two expressions and solve for
m
to find after how many miles the total cost of ownership is the same:
12000
+
0.1
m
=
14000
+
0.08
m
Next, we can subtract
12000
and
0.08
m
from each side of the equation to isolate the
m
term while keeping the equation balanced:
−
12000
+
12000
+
0.1
m
−
0.08
m
=
−
12000
+
14000
+
0.08
m
−
0.08
m
0
+
(
0.1
−
0.08
)
m
=
2000
+
0
0.02
m
=
2000
Now, we can divide each side of the equation by
0.02
to solve for
m
while keeping the equation balanced:
0.02
m
0.02
=
2000
0.02
0.02
m
0.02
=
100000
After 100,000 miles the total cost of ownership of the two cars would be the same.
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Answer:

The solution is x=1,y=2,z=3

Step-by-step explanation:

The given system of equations is ;\

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3x+4y−5z=−4...(2)

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x=\frac{8+3y-4z}{2}...(4)

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3(\frac{8+3y-4z}{2})+4y-5z=-4

Multiply through by;

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Expand;

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Simplify;

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Equation (4) in (3)

4(\frac{8+3y-4z}{2})-5y+6z=12

2(8+3y-4z)-5y+6z=12

16+6y-8z-5y+6z=12

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Put equation (6) into equation (5)

17(2z-4)-22z=-32

34z-68-22z=-32

34z-22z=-32+68

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z=3

Put z=3 into equation (6)

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Step-by-step explanation:

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The following data on average daily hotel room rate and amount spent on entertainment (The Wall Street Journal, August 18, 2011)
Zielflug [23.3K]

Answer:

a. Predicted Amount = $109.46

b. Confidence Interval = (94.84,124.08)

c. Interval = (110.6883,188.8517)

Step-by-step explanation:

Given

ŷ = 17.49 + 1.0334x.

SSE = 1541.4

a.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46 --- Approximated

Predicted Amount = $109.46

b.

ŷ = 17.49 + 1.0334(89)

ŷ = 109.4626

ŷ = 109.46

First we calculate the standard deviation

variance = SSE/(n-2)

v = 1541.4/(9-2)

v = 1541.4/7

v = 220.2

s = √v

s = √220.2

s = 14.839

Then we calculate mean(x) and ∑(x - (mean(x))²

X --- Y -- Mean(x) --- ∑(x - (mean(x))²

148 -- 161 -- 43-- 1849

96 || 105|| -9 || 81

91 ||101 || -14 || 196

110 || 142 || 5 || 25

90 || 100 || -15 || 225

102 || ||120 ||-3|| 9

136 || 167 ||31 ||961

90 || 140 ||-15 ||225

82 || 98 ||-23 || 529

Sum 945 || 1134|| 0 ||4100

Mean (x) = 945/9 = 105

∑(x - (mean(x))² = 4100

α = 1 - 95% = 5%

α/2 = 2.5% = 0.025

tα,df = n − 2 = t0.025,7 =2.365

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Confidence Interval = (109.46 ± 14.62)

Confidence Interval = (94.84,124.08)

c.

ŷ = 17.49 + 1.0334(128)

ŷ = 149.7652

ŷ = 149.77

Interval = 149.77 ± 2.365 * 14.839 √((1/9)+ (128-105)²/4100

Interval = 149.77 ± 39.0817

Interval = (110.6883,188.8517)

3 0
3 years ago
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