Answer:
61
Step-by-step explanation:
It should be A!! Hope this helps!!
Answer:
e. 1.28
Step-by-step explanation:
We have that to find our
level, that is the subtraction of 1 by the confidence interval divided by 2. So:
![\alpha = \frac{1-0.8}{2} = 0.1](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B1-0.8%7D%7B2%7D%20%3D%200.1)
Now, we have to find z in the Ztable as such z has a pvalue of
. This is our critical value.
So it is z with a pvalue of
, so ![z = 1.28](https://tex.z-dn.net/?f=z%20%3D%201.28)
The correct answer is:
e. 1.28
![Prove\ that\ the\ assumption \is \true for\ n=1\\1^3=\frac{1^2(1+1)^2}{4}\\ 1=\frac{4}{4}=1\\](https://tex.z-dn.net/?f=Prove%5C%20that%5C%20the%5C%20assumption%20%5Cis%20%5Ctrue%20for%5C%20n%3D1%5C%5C1%5E3%3D%5Cfrac%7B1%5E2%281%2B1%29%5E2%7D%7B4%7D%5C%5C%201%3D%5Cfrac%7B4%7D%7B4%7D%3D1%5C%5C)
Formula works when n=1
Assume the formula also works, when n=k.
Prove that the formula works, when n=k+1
![1^3+2^3+3^3...+k^3+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k+1)^2}{4}+(k+1)^3=\frac{(k+1)^2(k+2)^2}{4} \\\frac{k^2(k^2+2k+1)}{4}+(k+1)^3=\frac{(k^2+2k+1)(k^2+4k+4)}{4} \\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+4k^3+4k^2+2k^3+8k^2+8k+k^2+4k+4}{4}\\\\\frac{k^4+2k^3+k^2}{4}+k^3+3k^2+3k+1=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+2k^3+k^2}{4}+\frac{4k^3+12k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\\frac{k^4+6k^3+13k^2+12k+4}{4}=\frac{k^4+6k^3+13k^2+12k+4}{4}\\](https://tex.z-dn.net/?f=1%5E3%2B2%5E3%2B3%5E3...%2Bk%5E3%2B%28k%2B1%29%5E3%3D%5Cfrac%7B%28k%2B1%29%5E2%28k%2B2%29%5E2%7D%7B4%7D%20%5C%5C%5Cfrac%7Bk%5E2%28k%2B1%29%5E2%7D%7B4%7D%2B%28k%2B1%29%5E3%3D%5Cfrac%7B%28k%2B1%29%5E2%28k%2B2%29%5E2%7D%7B4%7D%20%5C%5C%5Cfrac%7Bk%5E2%28k%5E2%2B2k%2B1%29%7D%7B4%7D%2B%28k%2B1%29%5E3%3D%5Cfrac%7B%28k%5E2%2B2k%2B1%29%28k%5E2%2B4k%2B4%29%7D%7B4%7D%20%5C%5C%5Cfrac%7Bk%5E4%2B2k%5E3%2Bk%5E2%7D%7B4%7D%2Bk%5E3%2B3k%5E2%2B3k%2B1%3D%5Cfrac%7Bk%5E4%2B4k%5E3%2B4k%5E2%2B2k%5E3%2B8k%5E2%2B8k%2Bk%5E2%2B4k%2B4%7D%7B4%7D%5C%5C%5C%5C%5Cfrac%7Bk%5E4%2B2k%5E3%2Bk%5E2%7D%7B4%7D%2Bk%5E3%2B3k%5E2%2B3k%2B1%3D%5Cfrac%7Bk%5E4%2B6k%5E3%2B13k%5E2%2B12k%2B4%7D%7B4%7D%5C%5C%5Cfrac%7Bk%5E4%2B2k%5E3%2Bk%5E2%7D%7B4%7D%2B%5Cfrac%7B4k%5E3%2B12k%5E2%2B12k%2B4%7D%7B4%7D%3D%5Cfrac%7Bk%5E4%2B6k%5E3%2B13k%5E2%2B12k%2B4%7D%7B4%7D%5C%5C%5Cfrac%7Bk%5E4%2B6k%5E3%2B13k%5E2%2B12k%2B4%7D%7B4%7D%3D%5Cfrac%7Bk%5E4%2B6k%5E3%2B13k%5E2%2B12k%2B4%7D%7B4%7D%5C%5C)
Since the formula has been proven with n=1 and n=k+1, it is true. ![\square](https://tex.z-dn.net/?f=%5Csquare)
Answer:
My name is Melanie
Step-by-step explanation: