Answer:
117m
Step-by-step explanation:
I attached a file.
Since the rotation is about the y-axis, I'll integrate by dy.
![\displaystyle y=x^3\\x=\sqrt[3]y\\\\V=\pi \int \limits_1^8(2^2-(\sqrt[3]y)^2)\, dy\\V=\pi \Big[4x-\dfrac{3}{5}x^{\tfrac{5}{3}}\Big]_1^8\\V=\pi \left(4\cdot8-\dfrac{3}{5}\cdot8^{\tfrac{5}{3}\right-\left(4\cdot1-\dfrac{3}{5}\cdot1^{\tfrac{5}{3}\right)\right)\\V=\pi \left(32-\dfrac{96}{5}-\left(4-\dfrac{3}{5}\right)\right)\\V=\pi \left(\dfrac{64}{5}-\dfrac{17}{5}\right)\\V=\dfrac{47\pi}{5}](https://tex.z-dn.net/?f=%20%5Cdisplaystyle%20y%3Dx%5E3%5C%5Cx%3D%5Csqrt%5B3%5Dy%5C%5C%5C%5CV%3D%5Cpi%20%5Cint%20%5Climits_1%5E8%282%5E2-%28%5Csqrt%5B3%5Dy%29%5E2%29%5C%2C%20dy%5C%5CV%3D%5Cpi%20%5CBig%5B4x-%5Cdfrac%7B3%7D%7B5%7Dx%5E%7B%5Ctfrac%7B5%7D%7B3%7D%7D%5CBig%5D_1%5E8%5C%5CV%3D%5Cpi%20%5Cleft%284%5Ccdot8-%5Cdfrac%7B3%7D%7B5%7D%5Ccdot8%5E%7B%5Ctfrac%7B5%7D%7B3%7D%5Cright-%5Cleft%284%5Ccdot1-%5Cdfrac%7B3%7D%7B5%7D%5Ccdot1%5E%7B%5Ctfrac%7B5%7D%7B3%7D%5Cright%29%5Cright%29%5C%5CV%3D%5Cpi%20%5Cleft%2832-%5Cdfrac%7B96%7D%7B5%7D-%5Cleft%284-%5Cdfrac%7B3%7D%7B5%7D%5Cright%29%5Cright%29%5C%5CV%3D%5Cpi%20%5Cleft%28%5Cdfrac%7B64%7D%7B5%7D-%5Cdfrac%7B17%7D%7B5%7D%5Cright%29%5C%5CV%3D%5Cdfrac%7B47%5Cpi%7D%7B5%7D%20)
At 13% significance level, there isn't enough evidence to prove the administrators to claim that the mean score for the state's eighth graders on this exam is more than 280.
<h3>How to state hypothesis conclusion?</h3>
We are given;
Sample size; n = 78
population standard deviation σ = 37
Sample Mean; x' = 280
Population mean; μ = 287
The school administrator declares that mean score is more (bigger than) 280. Thus, the hypotheses is stated as;
Null hypothesis; H₀: μ > 280
Alternative hypothesis; Hₐ: μ < 280
This is a one tail test with significance level of α = 0.13
From online tables, the critical value at α = 0.13 is z(c) = -1.13
b) Formula for the test statistic is;
z = (x- μ)/(σ/√n)
z = ((280 - 287) *√78 )/37
z = -1.67
c) From online p-value from z-score calculator, we have;
P[ z > 280 ] = 0.048
d) The value for z = -1.67 is smaller than the critical value mentioned in problem statement z(c) = - 1.13 , the z(s) is in the rejection zone. Therefore we reject H₀
e) We conclude that at 13% significance level, there isn't enough evidence to prove the administrators to claim that the mean score for the state's eighth graders on this exam is more than 280.
Read more about Hypothesis Conclusion at; brainly.com/question/15980493
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Answer:
95% Confidence interval: (0.0429,0.0791)
Step-by-step explanation:
We are given the following in the question:
Sample size, n = 679
Number of anonymous websites, x = 42
95% Confidence interval:
Putting the values, we get:
is the required confidence interval for proportion of all new websites that were anonymous.
Answer:
There are a total of 2 hours and 50 min.
Step-by-step explanation:
1 hour = 60 min
60 x 2 = 120 min
60 x 3 = 180 min - too much
so take away 10 from 180 min to get 170
