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torisob [31]
3 years ago
10

A random sampling of a company's monthly operating expenses for n=36 months produced a sample mean of $5474 and a stan-dard devi

ation of S 764. Find a 90% upper confidence bound for the company's mean monthly expenses.
Mathematics
1 answer:
lilavasa [31]3 years ago
4 0

Answer with explanation:

Mean of the sample(m) = $ 5474

Standard deviation of the sample (S)=764

Number of observation(n)=36

Z_{90 \text{Percent}}=Z_{0.09}=0.5359

z_{score}=\frac{\Bar x-\mu}{\frac{S}{\sqrt{n}}}\\\\0.5359=\frac{5474- \mu}{\frac{764}{\sqrt{36}}}\\\\0.5359=\frac{5474- \mu}{\frac{764}{6}}\\\\764 \times 0.5359=6 \times (5474- \mu)\\\\409.4276=32844-6 \mu\\\\6 \mu=32844 -409.4276\\\\ 6 \mu=32434.5724\\\\ \mu=\frac{32434.5724}{6}\\\\ \mu=5405.76

So, Mean Monthly Expenses of Population =$ 5405.76, which is 90% upper confidence bound for the company's mean monthly expenses.

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<h2>2</h2>

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I’m using elimination so I’m trying to get rid of y
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Answer:

Step-by-step explanation:

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If we are looking first for the temp after 20 minutes, we have to solve for the k value.  That's what we will do first, given the info that we have:

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Filling in to solve for k:

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On your calculator, raise e to that power and multiply that number by 70:

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