We wish to estimate what percent of adult residents in a certain county are parents. Out of 100 adult residents sampled, 8 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places.

0.0101 < p < 0.1499

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 zscore that has a pvalue of .

Out of 100 adult residents sampled, 8 had kids. Based on this, construct a 99%.

This means that

99% confidence level

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:

Express your answer in tri-inequality form.

0.0101 < p < 0.1499

i believe the answer to be 48

step-by-step explanation:

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2