Answer:
The solution set is (4, 3)
Step-by-step explanation:
To solve by substitution, start by solving the second equation for y.
2x - y = 5
-y = -2x + 5
y = 2x - 5
Now that we have this, put that expression in for y in the first equation and solve for x.
5x - 2(2x - 5) = 14
5x - 4x + 10 = 14
x + 10 = 14
x = 4
Now that we have the value of x, input it into either equation and solve for y.
2x - y = 5
2(4) - y = 5
8 - y = 5
-y = -3
y = 3
Answer:
2 × 7 =14
Step-by-step explanation:
step by step that is the answer
On a graph, go up 7 units and to the right 2 units from the origin
Answer:
Step-by-step explanation:
so what you need to do is subtracked 26 with 46 then you get the ansewer 20 but after that you have to do 20 divided by 4 and you will get your answer.
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)
.
