"A simple random sample of 49 8th graders at a large suburban middle school indicated that 84% of them are involved with some type of after school activity. Find the margin of error associated with a 98% confidence interval that estimates the proportion of them that are involved in an after school activity."'

The margin of error associated with a 98% confidence interval that estimates the proportion of them that are involved in an after school activity is of 0.1219.

Step-by-step explanation:

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

In which

z is the z-score that has a p-value of .

The margin of error is of:

A simple random sample of 49 8th graders at a large suburban middle school indicated that 84% of them are involved with some type of after school activity.

This means that

98% confidence level

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

Margin of error:

The margin of error associated with a 98% confidence interval that estimates the proportion of them that are involved in an after school activity is of 0.1219.

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.

b. would be the answer : )

step-by-step explanation:

step 1: add -16x to both sides.

16x−4y+−16x=64+−16x  

−4y=−16x+64

step 2: divide both sides by -4.

−4y /-4 = -16x + 64/-4

answer:   y=4x−16

hope i'm right