Plsss im becoming a expert ok peepeepoopoo
You basically have to count how many days received the amount of mail listed in the left column. Then out of the total amount of mail received, you would compute the fraction of the days during which a given amount of mail was received.
For instance, on every day, at least 5 pieces of mail were counted, so there are 0 days on which 0-4 pieces of mail were received.
On the other hand, on days 5-7 (15, 18, 19), the amount of mail received fell within the 15-19 range. This means out of 7 days, 15-19 pieces of mail were received on 3 of those days. So the frequency with which this occurred was
![\dfrac37\approx0.43](https://tex.z-dn.net/?f=%5Cdfrac37%5Capprox0.43)
.
So what you should get in the right column is
Answer: a) 683 b) 1067
Step-by-step explanation:
The confidence interval for population proportion is given by :-
![p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}](https://tex.z-dn.net/?f=p%5Cpm%20z_%7B%5Calpha%2F2%7D%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
a) Given : Significance level :![\alpha=1-0.95=0.05](https://tex.z-dn.net/?f=%5Calpha%3D1-0.95%3D0.05)
Critical value : ![z_{\alpha/2}}=\pm1.96](https://tex.z-dn.net/?f=z_%7B%5Calpha%2F2%7D%7D%3D%5Cpm1.96)
Margin of error : ![E=0.03](https://tex.z-dn.net/?f=E%3D0.03)
Formula to calculate the sample size needed for interval estimate of population proportion :-
![n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.2(0.8)(\dfrac{1.96}{0.03})^2=682.951111111\approx683](https://tex.z-dn.net/?f=n%3Dp%281-p%29%28%5Cdfrac%7Bz_%7B%5Calpha%2F2%7D%7D%7BE%7D%29%5E2%5C%5C%5C%5C%3D0.2%280.8%29%28%5Cdfrac%7B1.96%7D%7B0.03%7D%29%5E2%3D682.951111111%5Capprox683)
Hence, the required sample size would be 683 .
b) If no estimate of the sample proportion is available then the formula to calculate sample size will be :-
![n=0.25(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.25(\dfrac{1.96}{0.03})^2=1067.11111111\approx1067](https://tex.z-dn.net/?f=n%3D0.25%28%5Cdfrac%7Bz_%7B%5Calpha%2F2%7D%7D%7BE%7D%29%5E2%5C%5C%5C%5C%3D0.25%28%5Cdfrac%7B1.96%7D%7B0.03%7D%29%5E2%3D1067.11111111%5Capprox1067)
Hence, the required sample size would be 1067 .
AC = 18
BC = 6x
AB = 2x + 4
Write and solve:
18 + 6x + 2x + 4
18 + 8x + 4 = 180
You can solve further:
18 + 8x = 176
8x = 158
x = 19 & 3/4
Not 100% sure what the question is asking, but hopefully this helps.
Answer:
c
Step-by-step explanation:
There is 6 students in the 8hr zone.
Gosh, do I have to explain? A is a, c is c!
4+6+6+6.5+6.5+6.5+6.5+7+7+7+7+7.5+7.5+7.5+8+8+8+8+8+8.5=141.
141÷20=7.05. So, no, it is not b.