Answer:
The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).
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 z-score that has a p-value of 
.
Out of 100 people sampled, 42 had kids.
This means that 
99% confidence level
So 
, z is the value of Z that has a p-value of 
, so 
.
The lower limit of this interval is:
The upper limit of this interval is:
The 99% confidence interval for the true population proportion of people with kids is (0.293, 0.547).
Think your answer would be C or D.
it's your choice to decide between those two now.
Answer:
use the distributive property to distribute the 8
-16+16n+3n=40+5n
-16+19n=40+5n
-16=40+5n-19n
-16=40-14n
-14n+40= -16
-14n= -16-40
-14n= -56
n= -56/-14
n= 4
Ok,
f(0.35)= 7f/20
f(-5.2)=-26f/5
f(10)= 10f
f(-0.5)= -f/2
as for the last question I am not quite sure, sorry....hope I helped a little :)
Answer:
You may just add 2 in the spot. Moreover, you get no solution.
Step-by-step explanation:
Hope this helped!
