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
If angle a is 24ft and angle c is 30ft, then angle b is 126ft.
Answer:
16/15
Step-by-step explanation:
Simplify the following:
8/3 - 8/5
Put 8/3 - 8/5 over the common denominator 15. 8/3 - 8/5 = (5×8)/15 + (3 (-8))/15:
(5×8)/15 + (3 (-8))/15
5×8 = 40:
40/15 + (3 (-8))/15
3 (-8) = -24:
40/15 + (-24)/15
40/15 - 24/15 = (40 - 24)/15:
(40 - 24)/15
| 3 | 10
| 4 | 0
- | 2 | 4
| 1 | 6:
Answer: 16/15
