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:
x^3-6x^2-10x+100
Step-by-step explanation:
Step 1: Distribute x to (x²-6x+20).
x(x²-6x+20)=x^3-6x^2+20x
Step 2: Distribute 5 to (x²-6x+20).
5(x² − 6x + 20)=5x²-30x+100
Step 3: Combine x^3-6x^2+20x and 5x²-30x+100.
x^3-6x^2-10x+100
hope it helps you <3
Answer:
this might be it : 12xy+228y
Step-by-step explanation:
you multiply 2 by 3 and get 6 and then multiply by 2 again and add they y
and then multiply 52 by 2 and gte 104 and them multiply by 2 and get 228 and add they y
33% is ur answer
sourse: Yahoo m8
