A survey found that 31 of 60 randomly selected women and 38 of 73 randomly selected men follow a regular exercise program. Find a 95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program.

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

(-0.1740 , 0.1672)

Step-by-step explanation:

Step(i):-

Given A survey found that 31 of 60 randomly selected women

First sample proportion

Second sample proportion

Level of significance = 0.05

Z₀.₀₅ = 1.96

Step(ii):-

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

where

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

((0.5166-0.520-1.96(0.08706) , 0.5166-0.520+1.96(0.08706))

(-0.1740 , 0.1672)

Final -

95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program

(-0.1740 , 0.1672)

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!