In a recent poll of 1200 randomly selected adult office workers, 32% said they had worn a halloween costume to the office at least once. report the 95% confidence interval for the proportion of all adult office workers who have worn a halloween costume to the office at least once. (round final calculations to the nearest tenth of a percent)

Step-by-step explanation:

The confidence interval for population proportion is given by :-

, where = sample proportion.

z* = critical value (Two-tailed)

n= Sample size ;

In the given problem , we are given

n= 1200

Also, we know that the critical value for 95 %confidence interval : z*= 1.96

Now, the 95% confidence interval for the proportion of all adult office workers who have worn a Halloween costume to the office at least once will be :-

Hence, the required confidence interval =

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this