Think it of a fraction problem,
5/6y-2=8
+2 +2
Add the reciprocal to cancel out the twos andd 8+2=10
so you have left 5/6y=10 and you'll take the reciprocal of 5 divided by 6 times five and cancels the 5/ out and also multiply 10*5 and you're left with 6y=50 then you just divide by six and there you go you have y=12
hope this helped
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)
.
Answer:
Y = 2
X = 3
Step-by-step explanation:
Substitution:
6(4y-5) - y = 16
24y-30-y=16
23y=46
Y = 2
So then:
6x-2 = 16
6x = 18
X = 3
12/1331
that is the answer. too lazy to explain
