U. S. consumer are increasingly viewing debit cards as a convenient substitute for cash and checks. The average amount spent annually on a debit card is $7,790. Assume that this average was based on a sample of 100 consumers and that the population standard deviation is $500.

a. At 99% confidence, what is the margin of error?
b. Construct the 99% confidence interval for the population mean amount spent annually on a debit card.

a) 128.75

b) The 99% confidence interval for the population mean amount spent annually on a debit card is between $7,661.25 and $7,918.75

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.

So the answer for a) is 128.75

The lower end of the interval is the sample mean subtracted by M. So it is 7790 - 128.75 = $7,661.25

The upper end of the interval is the sample mean added to M. So it is 7790 + 128.75 = $7,918.75

The 99% confidence interval for the population mean amount spent annually on a debit card is between $7,661.25 and $7,918.75

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

3

step-by-step explanation:

divide 13.5 by 9 to get the scale factor. it equals 1.5

2 * 1.5 = 3