<span>To find the confidence interval, add and subtract the margin of error from the mean.
With mean 18.7 and margin of error 5.9, you have 95% confidence the answer is between 12.8 and 24.6.</span>
X=1,
X=-3
Use Photomath it helps
Answer: E(Y) = 1.6 and Var(Y)=1.12
Step-by-step explanation:
Since we have given that
X 0 1 2
P(X) 0.4 0.4 0.2
Here, number of games = 2
So, 
Since
are independent variables.
so, ![E[Y]=2E[X]\\\\Var[Y]=2Var[X]](https://tex.z-dn.net/?f=E%5BY%5D%3D2E%5BX%5D%5C%5C%5C%5CVar%5BY%5D%3D2Var%5BX%5D)
So, we get that
![E(X)=0.4\times 0+0.4\times 1+0.2\times 2=0.8\\\\and Var[x]=E[x^2]-(E[x])^2\\\\E[x^2]=0\times 0.4+1\times 0.4+4\times 0.2=1.2\\\\So, Var[x]=1.2-(0.8)^2\\\\Var[x]=1.2-0.64=0.56](https://tex.z-dn.net/?f=E%28X%29%3D0.4%5Ctimes%200%2B0.4%5Ctimes%201%2B0.2%5Ctimes%202%3D0.8%5C%5C%5C%5Cand%20Var%5Bx%5D%3DE%5Bx%5E2%5D-%28E%5Bx%5D%29%5E2%5C%5C%5C%5CE%5Bx%5E2%5D%3D0%5Ctimes%200.4%2B1%5Ctimes%200.4%2B4%5Ctimes%200.2%3D1.2%5C%5C%5C%5CSo%2C%20Var%5Bx%5D%3D1.2-%280.8%29%5E2%5C%5C%5C%5CVar%5Bx%5D%3D1.2-0.64%3D0.56)
So, E[y]=2×0.8=1.6
and Var[y]=2×0.56=1.12
Hence, E(Y) = 1.6 and Var(Y)=1.12
Perimeter (P)= 2Length (L) +2Width (W)
P=2L + 2W
P=(2y)+(3y-5)
P=5y-5
A. 5y-5
Make is a proportion
6/18=28/x
Cross multiply
18•28=6•x
504=6x
x=84