For every x value, there should be only one y value. Or, for every input there is only one output.
A - Incorrect - there are two 2's for the x values and each of them has a different y value
B - Correct - all of the x values are different and all of the y values are different
C - Incorrect - there are two 11's for the x values and each of them has a different y value
D - Incorrect - there are two 3's for the x values and each of them has a different y value
Hope this helps! :)
Well there is no following but I can tell you that the probability of getting tails is 48/100 which they probably want simplified which jusst means divide both numerator and denominator(top and bottom numbers) by Z (whatever number works) untill it can't be a whole number anymore(a number without a decimal number), simply put, divide top and bottom numbers by the GCM(Greatest Common Multiple) which in this case is 4. 48/4=12, 100/4=25 48/100=12/25 you can't simplify that any more so the answer is 12/25
The Answer Is 12/25 or 48%
Hope This Helps.
Answer:
200
Step-by-step explanation:
6,000 divided by 30 is 200
Let

be the random variable indicating whether the elevator does not stop at floor

, with

Let

be the random variable representing the number of floors at which the elevator does not stop. Then

We want to find

. By definition,
![\mathrm{Var}(Y)=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%28Y%29%3D%5Cmathbb%20E%5B%28Y-%5Cmathbb%20E%5BY%5D%29%5E2%5D%3D%5Cmathbb%20E%5BY%5E2%5D-%5Cmathbb%20E%5BY%5D%5E2)
As stated in the question, there is a

probability that any one person will get off at floor

(here,

refers to any of the

total floors, not just the top floor). Then the probability that a person will not get off at floor

is

. There are

people in the elevator, so the probability that not a single one gets off at floor

is

.
So,

which means
![\mathbb E[Y]=\mathbb E\left[\displaystyle\sum_{i=1}^nX_i\right]=\displaystyle\sum_{i=1}^n\mathbb E[X_i]=\sum_{i=1}^n\left(1\cdot\left(1-\dfrac1n\right)^m+0\cdot\left(1-\left(1-\dfrac1n\right)^m\right)](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BY%5D%3D%5Cmathbb%20E%5Cleft%5B%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5EnX_i%5Cright%5D%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En%5Cmathbb%20E%5BX_i%5D%3D%5Csum_%7Bi%3D1%7D%5En%5Cleft%281%5Ccdot%5Cleft%281-%5Cdfrac1n%5Cright%29%5Em%2B0%5Ccdot%5Cleft%281-%5Cleft%281-%5Cdfrac1n%5Cright%29%5Em%5Cright%29)
![\implies\mathbb E[Y]=n\left(1-\dfrac1n\right)^m](https://tex.z-dn.net/?f=%5Cimplies%5Cmathbb%20E%5BY%5D%3Dn%5Cleft%281-%5Cdfrac1n%5Cright%29%5Em)
and
![\mathbb E[Y^2]=\mathbb E\left[\left(\displaystyle\sum_{i=1}^n{X_i}\right)^2\right]=\mathbb E\left[\displaystyle\sum_{i=1}^n{X_i}^2+2\sum_{1\le i](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BY%5E2%5D%3D%5Cmathbb%20E%5Cleft%5B%5Cleft%28%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En%7BX_i%7D%5Cright%29%5E2%5Cright%5D%3D%5Cmathbb%20E%5Cleft%5B%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En%7BX_i%7D%5E2%2B2%5Csum_%7B1%5Cle%20i%3Cj%7DX_iX_j%5Cright%5D%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En%5Cmathbb%20E%5B%7BX_i%7D%5E2%5D%2B2%5Csum_%7B1%5Cle%20i%3Cj%7D%5Cmathbb%20E%5BX_iX_j%5D)
Computing
![\mathbb E[{X_i}^2]](https://tex.z-dn.net/?f=%5Cmathbb%20E%5B%7BX_i%7D%5E2%5D)
is trivial since it's the same as
![\mathbb E[X_i]](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BX_i%5D)
. (Do you see why?)
Next, we want to find the expected value of the following random variable, when

:

If

, we don't care; when we compute
![\mathbb E[X_iX_j]](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BX_iX_j%5D)
, the contributing terms will vanish. We only want to see what happens when both floors are not visited.

![\implies\mathbb E[X_iX_j]=\left(1-\dfrac2n\right)^m](https://tex.z-dn.net/?f=%5Cimplies%5Cmathbb%20E%5BX_iX_j%5D%3D%5Cleft%281-%5Cdfrac2n%5Cright%29%5Em)

where we multiply by

because that's how many ways there are of choosing indices

for

such that

.
So,