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erica [24]
3 years ago
11

(please help!!)

Mathematics
1 answer:
Leviafan [203]3 years ago
4 0

Answer:

x = 300 minutes

Step-by-step explanation:

20 + .15x = 35 + .10x

.05x = 15

 

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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
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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 - <em>α</em>)% 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:

<em>n</em> = 26

<em>s</em> = 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).

3 0
3 years ago
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Answer:

She rode her bike 20 miles... I believe this may be a trick question.

Step-by-step explanation:

Let's break down this problem...

If she is going at a rate of 5 miles per hour while walking than the maximum she could have walked in 5 hours is 25 miles.

If Angela rode her bike for 20 miles, and the got off her bike and walked "<em>the rest of the way</em>" then she should have walked 50 miles. However as we already know her rate is 5 miles per hour while walking, and therefore could only have covered half that distance in 5 hours.

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