1/2(5 sqrt 2)(4 sqrt 2)= 20
For large sample confidence intervals about the mean you have:
xBar ± z * sx / sqrt(n)
where xBar is the sample mean z is the zscore for having α% of the data in the tails, i.e., P( |Z| > z) = α sx is the sample standard deviation n is the sample size
We need only to concern ourselves with the error term of the CI, In order to find the sample size needed for a confidence interval of a given size.
z * sx / sqrt(n) = width.
so the z-score for the confidence interval of .98 is the value of z such that 0.01 is in each tail of the distribution. z = 2.326348
The equation we need to solve is:
z * sx / sqrt(n) = width
n = (z * sx / width) ^ 2.
n = ( 2.326348 * 6 / 3 ) ^ 2
n = 21.64758
Since n must be integer valued we need to take the ceiling of this solution.
n = 22
Answer:
I Believe that your answer is D. Sorry if i'm wrong.
Step-by-step explanation:
Answer:
About 2105.98
Step-by-step explanation:
31% of 3354 is 1039.74
3354-1039.74=2314.26
9% of 2314.26 is 208.2834
2314.26-2018.2834=2105.9766
There isn't a rectangular prism. Guess 12.
