Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week
//0.20113,0.20887[/tex]
Step-by-step explanation:
<u>step 1:-</u>
Given sample size n=200
of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week
Population proportion of employed individuals who work at home at least once per week P = 
Q=1-P= 1-0.205 = 0.705
<u>step 2:-</u>
Now 
=0.0015
<u>step 3:-</u>
<u>Confidence intervals</u>
<u>using formula</u>


=0.20113,0.20887[/tex]
<u>conclusion:</u>-
99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week
//0.20113,0.20887[/tex]
It would be 31+t.
Unless there's more to this question.
Y=6X+9 the A is #1
all you need to do is simp the equation
1/6 of a treat would be your answer
hope this helps