First find g(2).
g(2) = (2+4)2 = 12
Because g(2) is 12, h(g(2)) = h(12).
h(12) = 2(12)-7 = 17
answer: 17
12 students/1 teacher = 276 students/x teachers
Students are on top of the ratio and teachers are on the bottom of the ratio. It must remain the same on both sides of the equation.
12 x = 276
divide by 12
x =23
Answer:
C
Step-by-step explanation:
Hey There!
So the answer is C because essentially answer choice c is saying that you can predict what the amount of problems that the next kid would complete in 30 minutes because all of the samples have a mean of 15.6-15.9 or that all of the samples are reliable because all of the means are near each other
And if you didnt know the mean is the average of all of the numbers in the sample
There C is your answer
Answer:
9??
Step-by-step explanation:
Sorry this could be wrong
Answer:
Integration of I= 
=![[\frac{logx}{n+1} x^{(n+1)}]-[\frac{1}{(n+1)^{2}}x^{(n+1)}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7D%20x%5E%7B%28n%2B1%29%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7Dx%5E%7B%28n%2B1%29%7D%5D)
Step-by-step explanation:
Given integral is I=
Take logx=t
I= 
I= 
Using integration by part,
![I= (t)\int [e^{(n+1)t}]\, dt-\int[\frac{d}{dt}{t}\times\int (e^{(n+1)t})]\\\\I= (t) [\frac{1}{n+1}e^{(n+1)t}]-\int[1\times\frac{1}{n+1}e^{(n+1)t}]\,dt\\\\I=[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=I%3D%20%28t%29%5Cint%20%5Be%5E%7B%28n%2B1%29t%7D%5D%5C%2C%20dt-%5Cint%5B%5Cfrac%7Bd%7D%7Bdt%7D%7Bt%7D%5Ctimes%5Cint%20%28e%5E%7B%28n%2B1%29t%7D%29%5D%5C%5C%5C%5CI%3D%20%28t%29%20%5B%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5Cint%5B1%5Ctimes%5Cfrac%7B1%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D%5C%2Cdt%5C%5C%5C%5CI%3D%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
Writing in terms of x
I=![[\frac{t}{n+1}e^{(n+1)t}]-[\frac{1}{(n+1)^{2}}e^{(n+1)t}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bt%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29t%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29t%7D%5D)
I=![[\frac{logx}{n+1}e^{(n+1)logx}]-[\frac{1}{(n+1)^{2}}e^{(n+1)logx}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7B%28n%2B1%29logx%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7B%28n%2B1%29logx%7D%5D)
I=![[\frac{logx}{n+1}e^{logx^{(n+1)}}]-[\frac{1}{(n+1)^{2}}e^{logx^{(n+1)}}]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Blogx%7D%7Bn%2B1%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D-%5B%5Cfrac%7B1%7D%7B%28n%2B1%29%5E%7B2%7D%7De%5E%7Blogx%5E%7B%28n%2B1%29%7D%7D%5D)
I=
Thus,
Integration of I= =
