A 95 percent confidence interval for the difference between the proportion of high school seniors who play a varsity sport in the school district and high school juniors who play a varsity sport in the school district is to be calculated. What is the standard error of the difference

Hello your question is incomplete below is the complete question

In a large school district, 16 of 85 randomly selected high school seniors play a varsity sport. in the same district, 19 of 67 randomly selected high school juniors play a varsity sport. a 95 percent confidence interval for the difference between the proportion of high school seniors who play a varsity sport in the school district and high school juniors who play a varsity sport in the school district is to be calculated. What is the standard error of the difference

answer : 0.0695

Step-by-step explanation:

First step : Determine the sample proportions

For P1 = 16 / 85 = 0.188

For P2 = 19 / 67 = 0.284

Next determine the Standard error of the difference using the relation below

Standard Error =  ------- ( 1 )

where : P1 = 0.188

             P2 = 0.284

             n1 = 85 high school seniors

             n2 = 67  high school juniors

Input values into equation 1

Standard error of the difference = 0.0695

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3

i believe the answer is the polygon, heptagon.