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3 years ago

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Khaliq produces and sells toy cars. The material for each car costs him $4. He has been selling them for $8 each and averaging s

ales of 40 per week. He has been told by Kiarash that he could charge more but has found that for each $0.50 increase in price, he would lose 4 sales each week. How much should Khaliq charge per car in order to have the greatest profit? What is the greatest
profit?

1 answer:

mylen [45]3 years ago

7 0

Answer:

He should charge 8.50 for each car for a profit of 162 dollars.

Step-by-step explanation:

Plot y = (x - 4)(40 - 8(x - 8)) on a graphing calculator such as desmos

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A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

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Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

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$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

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$\sigma_x = \frac{\sigma}{\sqrt n}$

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Calculate margin of error (E)

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$Lower = \bar x - E$

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Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

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<em>Notice that the margin of error decreases when the sample size increases.</em>

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