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:
The potato salad costs 4.63
Step-by-step explanation:
We have 1.75lb
1=2.75
then 75% of 2.75 is 2.0625
the answer is 4.63
Answer:
I think the answer is B. f(x) = -1/3x - 4
Step-by-step explanation:
Use the given functions to set up and simplify
4−16.
XF(x)=X
Fx
1 − 7 = −6
2 − 10 = −8
3 − 13 = −10
4 − 16 = −12
This is a guess just so you know...
I think the answer is (4,-6) because that is the point at which the two lines intersect sooooo....
I'm going with option 3.
Answer:
y2-y1/x2-x1 or rise/run
Step-by-step explanation:
I learned this in algebra