<span>assume z = ax for simplicity
z(z) = a(ax) = a^2x
let a^2x = 1/16x and solve for a </span>
The answer to your question would be 150 (c)
Answer:
Bottom left graph
Step-by-step explanation:
We have to use what is called the zero-interval test [test point] in order to figure out which portion of the graph these inequalities share:
−2x + y ≤ 4 >> Original Standard Equation
+ 2x + 2x
_________
y ≤ 2x + 4 >> Slope-Intercept Equation
−2[0] + 0 ≤ 4
0 ≤ 4 ☑ [We shade the part of the graph that CONTAINS THE ORIGIN, which is the right side.]
[We shade the part of the graph that does not contain the origin, which is the left side.]
So, now that we got that all cleared up, we can tell that the graphs share a region in between each other and that they both have POSITIVE <em>RATE OF CHANGES</em> [<em>SLOPES</em>], therefore the bottom left graph matches what we want.
** By the way, you meant
because this inequality in each graph is a <em>dashed</em><em> </em><em>line</em>. It is ALWAYS significant that you be very cautious about which inequalities to choose when graphing. Inequalities can really trip some people up, so once again, please be very careful.
I am joyous to assist you anytime.
By definition of covariance,
![\mathrm{Cov}(X,Y)=\mathbb E[(X-\mathbb E[X])(Y-\mathbb E[Y])]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28X%2CY%29%3D%5Cmathbb%20E%5B%28X-%5Cmathbb%20E%5BX%5D%29%28Y-%5Cmathbb%20E%5BY%5D%29%5D)
![\mathrm{Cov}(X,Y)=\mathbb E[XY-\mathbb E[X]Y-X\mathbb E[Y]+\mathbb E[X]\mathbb E[Y]]=\mathbb E[XY]-\mathbb E[X]\mathbb E[Y]](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28X%2CY%29%3D%5Cmathbb%20E%5BXY-%5Cmathbb%20E%5BX%5DY-X%5Cmathbb%20E%5BY%5D%2B%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D%5D%3D%5Cmathbb%20E%5BXY%5D-%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D)
We have
![\mathbb E[(aX-b)(cY-d)]=\mathbb E[acXY-adX-bcY+bd]](https://tex.z-dn.net/?f=%5Cmathbb%20E%5B%28aX-b%29%28cY-d%29%5D%3D%5Cmathbb%20E%5BacXY-adX-bcY%2Bbd%5D)
![=ac\mathbb E[XY]-ad\mathbb E[X]-bc\mathbb E[Y]+bd](https://tex.z-dn.net/?f=%3Dac%5Cmathbb%20E%5BXY%5D-ad%5Cmathbb%20E%5BX%5D-bc%5Cmathbb%20E%5BY%5D%2Bbd)
![\mathbb E[aX-b]=a\mathbb E[X]-b](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BaX-b%5D%3Da%5Cmathbb%20E%5BX%5D-b)
![\mathbb E[cY-d]=c\mathbb E[Y]-d](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BcY-d%5D%3Dc%5Cmathbb%20E%5BY%5D-d)
![\mathbb E[aX-b]\mathbb E[cY-d]=ac\mathbb E[X]\mathbb E[Y]-ad\mathbb E[X]-bc\mathbb E[Y]+bd](https://tex.z-dn.net/?f=%5Cmathbb%20E%5BaX-b%5D%5Cmathbb%20E%5BcY-d%5D%3Dac%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D-ad%5Cmathbb%20E%5BX%5D-bc%5Cmathbb%20E%5BY%5D%2Bbd)
Putting everything together, we find the covariance reduces to
![\mathrm{Cov}(aX-b,cY-d)=ac(\mathbb E[XY]-\mathbb E[X]\mathbb E[Y])=ac\mathrm{Cov}(X,Y)](https://tex.z-dn.net/?f=%5Cmathrm%7BCov%7D%28aX-b%2CcY-d%29%3Dac%28%5Cmathbb%20E%5BXY%5D-%5Cmathbb%20E%5BX%5D%5Cmathbb%20E%5BY%5D%29%3Dac%5Cmathrm%7BCov%7D%28X%2CY%29)
as desired.