Answer:
Part 1) 
Part 2) 
Part 3) 
Step-by-step explanation:
step 1
Find the measure of length side FG
In the right triangle EFG
we know that
----> by SOH (opposite side divided by the hypotenuse)
substitute the given values
step 2
Find the measure of length side EF
In the right triangle EFG
we know that
----> by CAH (adjacent side divided by the hypotenuse)
substitute the given values
step 3
Find the measure of angle G
we know that
---> by complementary angles in a right triangle
3 to the negative second power is written like this:
.
When simplifying an expression with a negative exponent, you take a positive version of the exponent, in this case that would be 2, and apply that to the base.
After doing that, we have 9.
The next step is to put one over that number.
In this case, now after doing that the answer is
.
Hope this helps!
Answer: These ARE congruent to each other.
Step-by-step explanation: Do you see the ''tick marks?'' They both have 1 set of 2, and 1 set of 1. If that makes sense. They also have a "angle" in one of their acute angles.
Adding 2 to each value of the random variable 
makes a new random variable 
. Its mean would be
![E[X+2]=E[X]+E[2]=E[X]+2](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3DE%5BX%5D%2BE%5B2%5D%3DE%5BX%5D%2B2)
since expectation is linear, and the expected value of a constant is that constant. ![E[X]](https://tex.z-dn.net/?f=E%5BX%5D)
is the mean of , so the new mean would be
![E[X+2]=10+2=12](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3D10%2B2%3D12)
The variance of a random variable is
![V[X]=E[X^2]-E[X]^2](https://tex.z-dn.net/?f=V%5BX%5D%3DE%5BX%5E2%5D-E%5BX%5D%5E2)
so the variance of would be
![V[X+2]=E[(X+2)^2]-E[X+2]^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5B%28X%2B2%29%5E2%5D-E%5BX%2B2%5D%5E2)
We already know ![E[X+2]=12](https://tex.z-dn.net/?f=E%5BX%2B2%5D%3D12)
, so simplifying above, we get
![V[X+2]=E[X^2+4X+4]-12^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5BX%5E2%2B4X%2B4%5D-12%5E2)
![V[X+2]=E[X^2]+4E[X]+4-12^2](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DE%5BX%5E2%5D%2B4E%5BX%5D%2B4-12%5E2)
![V[X+2]=(V[X]+E[X]^2)+4E[X]-140](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3D%28V%5BX%5D%2BE%5BX%5D%5E2%29%2B4E%5BX%5D-140)
Standard deviation is the square root of variance, so ![V[X]=3^2=9](https://tex.z-dn.net/?f=V%5BX%5D%3D3%5E2%3D9)
.
![\implies V[X+2]=(9+10^2)+4(10)-140=9](https://tex.z-dn.net/?f=%5Cimplies%20V%5BX%2B2%5D%3D%289%2B10%5E2%29%2B4%2810%29-140%3D9)
so the standard deviation remains unchanged at 3.
NB: More generally, the variance of 
for 
is
![V[aX+b]=a^2V[X]+b^2V[1]](https://tex.z-dn.net/?f=V%5BaX%2Bb%5D%3Da%5E2V%5BX%5D%2Bb%5E2V%5B1%5D)
but the variance of a constant is 0. In this case, 
, so we're left with ![V[X+2]=V[X]](https://tex.z-dn.net/?f=V%5BX%2B2%5D%3DV%5BX%5D)
, as expected.
You Need 7 Gallons Of Fuel For The Truck
