A survey of an urban university showed that 750 or 1,100 students sampled attended a home football game during the season. Using the 90% level of confidence, what is the confidence interval for the proportion of students attending a football game?

The 90% confidence interval for the proportion of students attending a football game is (0.6587, 0.7049).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

For this problem, we have that:

A survey of an urban university showed that 750 or 1,100 students sampled attended a home football game during the season. This means that

90% confidence interval

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for the proportion of students attending a football game is (0.6587, 0.7049).

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.

answer: the correct answer is -32

step-by-step explanation: