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:
46
Step-by-step explanation:
4 x 2= 8
3 x 5= 15
2(8 +15)= 2(23)=
23 x 2= 46
Answer:
The top option is false.
Step-by-step explanation:
Both segments have a <em>rate</em><em> </em><em>of</em><em> </em><em>change</em><em> </em>[<em>slope</em>] of ⅔. It just that their ratios have unique qualities:
Greatest Common Factor: 2
___ ___
<em>BC</em><em> </em>is at a 4⁄6 slope, and <em>AB</em><em> </em>is at a ⅔ slope. Although their quantities are unique, they have the exact same value.
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