A random sample found that 30% of 150 Americans were satisfied with the gun control laws in 2018. Compute a 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 Fill in the blanks appropriately. A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is ( , ) (round to 3 decimal places)

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).

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 .

For this problem, we have that:

98% 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:

A 98% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2018 is (0.213, 0.387).

answer: 1inch=7days  

12inch=12*7=84days

step-by-step explanation:

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)