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
55 min/10 km=5.5 min/km
5.5 min/km x 1 km/.621 mile= 5.5/.621=8.86 min/mile
The answer is D.
Answer: 8 batches
explanation: 5 1/3 divided by 2/3 equals 8. To find out how many batches were made, you need to divide the total amount used by the amount needed for one batch.
Answer: It would be B
Step-by-step explanation:
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