The answer is D. Hope it helps!
The sample size needed to obtain a margin of error of 0.05 for the estimation of a population proportion is 384.
Given margin of error of 0.05 and confidence interval of 95%.
We have to find the sample size needed to obtain a margin of error of 0.5 with confidence level of 95%.
Margin of error is the difference between real values and calculated values.
The formula of margin of error is as under:
Margin of error=z*σ/
where
z is the critical value of z for given confidence level
n is sample size
σ is population standard deviation
We have not given population standard deviation so we will use the following formula:
Margin of error=z*/
We have to find z value for 95% confidence level.
z value=1.96
We know that <=1/2
put =1/2
Margin of error=1.96*1/2/
0.05=1.96*0.5/
=0.98/0.05
=19.6
squaring both sides
n=384.16
After rounding off we will get
n=384.
Hence the sample size needed to obtain a margin of error of 0.05 is 384.
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Answer:
15.75$
Step-by-step explanation:
Answer:
625/1000
Step-by-step explanation:
1000 ÷32=31.25
31.25×20=625
625/1000
Answer:
$24
Step-by-step explanation:
30 x .8 equals 24.
Hope this helps!
Plz, Mark Brainliest!