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Tcecarenko [31]
1 year ago
14

A car dealership pays $8,350 for a car. They Mark up the price by 17.4% to get the retail price. What is the retail price of the

car at this dealership? Round the answer to tge nearest (hundredth) secon decimal place.
Mathematics
1 answer:
olga55 [171]1 year ago
6 0

A car dealership pays the sum of $8,350 for a car

The dealer mark up the car price by 17.4%

The retail price = Mark up percentage x original price of the car

Mark up = 17.4%

= 17.4 + 10

= 117.4%

Convert 117.4% to decimal

= 117.4/100

= 1.174

The retail price = $8, 350 x 1.174

= $9,802.90

The retail price of the car is $9,802.90

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Answer:

a) The critical value is z = 2.575.

b) The 99% confidence interval for the mean repair cost for the TVs is ($66.31, $110.61).

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.99}{2} = 0.005

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha. This is our critical value

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a. Find the critical value that should be used in constructing the confidence interval.

The critical value is z = 2.575.

b. Construct the 99% confidence interval. Round your answer to two decimal places.

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample. So, in this problem

M = 2.575*\frac{17.20}{\sqrt{4}} = 22.15

The lower end of the interval is the mean subtracted by M. So it is 88.46 - 22.15 = $66.31

The upper end of the interval is the mean added to M. So it is 88.46 - 22.15 = $110.61

The 99% confidence interval for the mean repair cost for the TVs is ($66.31, $110.61).

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