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3 years ago

Please help 25 points the problem is below

Mathematics

1 answer:

andrew11 [14]3 years ago

3 0

The formula of an area of a rectangle:

$A=wl$

We have l = 2w + 40

Substitute:

$A=w(2w+40)=2w^2+40w$

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Mr. Good Wrench advertises that a customer will have to wait no more than 30 minutes for an oil change. A sample of 26 oil chang

Andru [333]

Answer:

The 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

Step-by-step explanation:

The (1 - α)% confidence interval for the population standard deviation is:

$CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}$

The information provided is:

n = 26

s = 4.8 minutes

Confidence level = 90%

Compute the critical values of Chi-square as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.10/2, (26-1)}=\chi^{2}_{0.05, 25}=37.652$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{1-0.10/2, (26-1)}=\chi^{2}_{0.95, 25}=14.611$

*Use a Chi-square table.

Compute the 90% confidence interval for the population standard deviation waiting time for an oil change as follows:

$CI=\sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2, (n-1)}}}\leq \sigma\leq \sqrt{\frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2, (n-1)}}}$

$=\sqrt{\frac{(26-1)\times 4.8^{2}}{37.652}}\leq \sigma\leq \sqrt{\frac{(26-1)\times 4.8^{2}}{14.611}}\\\\=3.9113\leq \sigma\leq 6.2787\\\\\approx 3.9 \leq \sigma\leq6.3$

Thus, the 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

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3 years ago

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Answer:

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Step-by-step explanation:

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4 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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3 years ago

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