A sample size of n = 64 is drawn from a population whose standard deviation is o = 5.6.
Find the margin of error for a 99% confidence interval for u.

The margin of error for a 99% confidence interval for the population mean is 1.8025.

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.

In this problem:

So

The margin of error for a 99% confidence interval for the population mean is 1.8025.

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.

i think it is c, not positive though

step-by-step explanation: