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Answer:
We need a sample size of at least 719
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of
.
So it is z with a pvalue of
, so 
Now, find the margin of error M as such

In which
is the standard deviation of the population and n is the size of the sample.
How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?
This is at least n, in which n is found when
. So






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We need a sample size of at least 719
Answer:
M=p-5n
Step-by-step explanation:
Here we are given that M+5n=p
We are asked to solve this equation for M. In order to do that we will follow these steps.
Subtracting 5n from both sides we get
M+5n-5n=p-5n
M=p-5n
Hence , now we have our M as dependent variable, whose value depends on the values of p and n. One or both of them can be independent variable/.
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