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
The answer is 6 and 1/15.
Answer:
Multiply the w x h x l
Step-by-step explanation:
Answer:
1 and -5
Step-by-step explanation:
![~~~~~x^3+3x^2-9x+5=0\\\\\implies x^3-x^2+4x^2-4x-5x+5=0\\\\\implies x^2(x-1)+4x(x-1)-5(x-1)=0\\\\\implies (x-1)(x^2 +4x -5) = 0\\\\\implies (x-1)(x^2+5x -x -5) = 0\\\\\implies (x-1)[x(x+5) -(x+5)] =0\\\\\implies (x-1)(x-1)(x+5)= 0\\\\\implies (x-1)^2(x+5) = 0\\\\\implies x = 1,~~ x = -5](https://tex.z-dn.net/?f=~~~~~x%5E3%2B3x%5E2-9x%2B5%3D0%5C%5C%5C%5C%5Cimplies%20x%5E3-x%5E2%2B4x%5E2-4x-5x%2B5%3D0%5C%5C%5C%5C%5Cimplies%20x%5E2%28x-1%29%2B4x%28x-1%29-5%28x-1%29%3D0%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x%5E2%20%2B4x%20-5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x%5E2%2B5x%20-x%20-5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%5Bx%28x%2B5%29%20-%28x%2B5%29%5D%20%3D0%5C%5C%5C%5C%5Cimplies%20%28x-1%29%28x-1%29%28x%2B5%29%3D%200%5C%5C%5C%5C%5Cimplies%20%28x-1%29%5E2%28x%2B5%29%20%3D%200%5C%5C%5C%5C%5Cimplies%20x%20%3D%201%2C~~%20x%20%3D%20-5)
