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:
1.5
Step-by-step explanation:
6 divided by 4
The answer is 4 because if u divide both equations by eachother you get 4
Short side length = x
Medium side = x + 7
Long side = 5x
Total 49, so
x + x + 7 + 5x = 49
If you combine like terms:
7x + 7 = 49
Subtract 7 from both sides
7x = 42
Divide both sides by 7
x = 6 inches.
