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yanalaym [24]
3 years ago
11

A new car is purchased for 19000 dollars. The value of the car depreciates at 11.25% per year. What will the value of the car be

, to the nearest cent, after 14 years.
Mathematics
1 answer:
Ainat [17]3 years ago
7 0

Answer:

$3573. 64

Step-by-step explanation:

Convert 11.25% to the decimal fraction 0.1125.  Then subtract this result from 1.0000:  0.8875.

After 14 years, the formerly $19000 car will have the value

$19000(0.8875)^14 = $3573. 64

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A random sample of ten 2011 sports cars is taken and their city mileage is recorded. The results are as follows: 20 21 25 21 21
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Answer:

ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=24.8

The sample deviation calculated s=4.367

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

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Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that t_{\alpha/2}=1.833

And the margin of error would be given by:

ME = 1.833 * \frac{4.367}{\sqrt{10}}= 2.531

5 0
4 years ago
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