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Sphinxa [80]
2 years ago
15

len bought a new car for $22,500. after a few years he sold the car at a discount of 25% what are the selling price of the car?

Mathematics
2 answers:
Norma-Jean [14]2 years ago
6 0

Answer:

$16,875

Step-by-step explanation:

what you would have to do is find what 75% of the selling price of the car was before the discount was put on it, so that would look like this: 75% of 22,500 = 16,875

or you could've found 25% of 22,500 and subtracted it from that number like this: 25% of 22,500 = 5,625   22,500 - 5,625 = 16,875

i hope this helps :)

Gala2k [10]2 years ago
4 0

Answer:$16,875

Step-by-step explanation:

25% of 22500 is 5625.

22500-5625=16875.

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Answer:

64.3333333333

Step-by-step explanation:

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The Ohio Department of Agriculture tested 203 fuel samples across the state
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Answer:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

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If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

Step-by-step explanation:

Previous concept

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.  

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The population proportion have the following distribution

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Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

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The proportion estimated would be:

\hat p = \frac{14}{105}= 0.133

And that represent the proportion of failures.

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

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.133 - 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.0475

0.133 + 2.58\sqrt{\frac{0.133(1-0.133)}{105}}=0.2185

The 99% confidence interval would be given by (0.0475;0.2185)

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