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Sliva [168]
3 years ago
5

Plz help it’s due tonight!!!

Mathematics
1 answer:
Nataliya [291]3 years ago
5 0
LKM≈TEJ since it makes the triangles congruent, although the angles in each triangle are not corresponding equal angles
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this is the stuff i meed help on

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3 years ago
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 - <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).

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Step-by-step explanation:

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The shirt was $7.45 he pays with a $20 how much he getting back
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$12.55

Step-by-step explanation:

20-7.45=12.55

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4 years ago
Through the point (5,1) with a slope of 4/5​
Nataly_w [17]

Answer:

y-1=4/5(x-5)

Step-by-step explanation:

y-1=4/5(x-5)

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