Afast food restaurant executive wishes to know how many fast food meals teenagers eat each week. they want to construct a 85% confidence interval for the mean and are assuming that the population standard deviation for the number of fast food meals consumed each week is 1.4 . the study found that for a sample of 502 teenagers the mean number of fast food meals consumed per week is 3.1 . construct the desired confidence interval. round your answers to one decimal place. lower end point and upper end point

The 85% confidence interval for the mean number of fast food meals each week is (3.01 meals, 3.19 meals).

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 M as such

In which is the standard deviation of the population and n is the length of the sample. So

The lower end of the interval is the mean subtracted by M. So it is 3.1 - 0.09 = 3.01 meals.

The upper end of the interval is the mean added to M. So it is 3.1 + 0.09 = 3.19 meals.

The 85% confidence interval for the mean number of fast food meals each week is (3.01 meals, 3.19 meals).

a)(x-2)^2+(y-1)^2=9

step-by-step explanation:

(picture explains)

0.0076, if you meant 4.125 divided by 0.03125 it would be 132

step-by-step explanation:

first you have to move the decimal in 0.03125 over until its behind the 5 which is 5 places over. now the number is 3125

then you have to move the decimal in 4.125, 5 places over. now the number is 4,112,500.

now you divide like you would normally would and get 0.0076