Answer:
3
Step-by-step explanation:
Hundreds Tens Ones
5 3. 8
ok so here we go:
1) 
or 
2 )
so the answer is (0, -4) or -4
3) y=
4) y=[tex]\frac{3}{2}x+1[tex] so the answer is (0, 1) or 1
<span>Income earned over one year is principal * int rate so we have
0.12*A + 0.13*B +0.14*C = 400
and A + B + C = 3000
and A+B = C</span>
Answer: The surface area of the planter will be 120.5 square feet.
In our shape, we will have 5 sides. The base and the four sides going up from each edge of the base.
The base will be 7 x 5 = 35 square feet.
The front and back side will each be 7 x 0.75 = 5.25 square feet.
The left and right side will each be 5 x 0.75 = 3.75 square feet.
If we add up the 5 faces we get:
35 + 5.25 + 5.25 + 3.75 + 3.75 = 120.5 square feet
The expected length of code for one encoded symbol is
where 
is the probability of picking the letter 
, and 
is the length of code needed to encode . is given to us, and we have
so that we expect a contribution of
bits to the code per encoded letter. For a string of length 
, we would then expect ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n)
.
By definition of variance, we have
![\mathrm{Var}[L]=E\left[(L-E[L])^2\right]=E[L^2]-E[L]^2](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3DE%5Cleft%5B%28L-E%5BL%5D%29%5E2%5Cright%5D%3DE%5BL%5E2%5D-E%5BL%5D%5E2)
For a string consisting of one letter, we have
so that the variance for the length such a string is
"squared" bits per encoded letter. For a string of length , we would get ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n)
.
