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]
Answer:
192 markers
Step-by-step explanation:
if 8 = 4% then 2=1%
so we can assume there are 200markers in the package.
8-200= 192 markers
Answer:
Ikr >:(
Step-by-step explanation:
Solution :
The data is normally distributed.
The standard deviation is 18 days
Here the data is normally distributed and 54 days is 3 days of standard deviation.
Therefore, the percentage of the births that would be 
within the 54 days of the mean 
length is given by :
= P( -3 < Z < 3)
= 0.9544
= 95 %
Therefore, about 95% of the births would be within 54 days of the men length.
Answer: x=-4.8
Step-by-step explanation:
Given
-10x+4=52
Subtract 4 on both sides
-10x+4-4=52-4
-10x=48
Divide -10 on both sides
-10x/-10=48/-10
x=-4.8
Hope this helps!! :)
Please let me know if you have any questions
