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Burka [1]
3 years ago
6

PLEASE HELPS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
blsea [12.9K]3 years ago
4 0

Answer: If I had to guess it would be the fourth one down

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In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo
larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

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Thus, 1-α=0.99

α=0.01

Now we need to determine z_{\frac{\alpha}{2}}=z_{0.005}

Now by using z score table we find that  z_{\frac{\alpha}{2}}=2.58

The boundaries of the confidence interval are:

\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64

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3 years ago
N the triangle below, what is the value<br> of x?
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Step-by-step explanation:

can u help me

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