Answer:
72.4%
Step-by-step explanation:
The essay is 25% of your grade, and the rest is 75% of your grade.
25 (3.3/4) + 75 (0.69) ≈ 72.4
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
.
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
.
Answer:
C. 0 6 10
Step-by-Step-Explanation:
the number on the left of the indicated operation is the original set
the number '2' next to R shows that you need to multiply certain numbers by 2
the subscript on the R's shows which row the multiplication needs to be done on...
so in this case, leave the second row alone: 1 3 4
but multiply the first row by 2: 0 6 10
137 paquetes de vasos, aunque el último nomas contiene 7 vasos se necesita el paquete para que no queden sin empacar
You can just use a calculator but when you add the numbers all together you get 323