B as it goes through the correct and average amount of points on the graph, hope this helps
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
.
Factor the expression
Use the properties of radicals
Simplify the roots
Calculate
And your solution will be 20 overall 5
Message me if you want me to write it all down!
Answer:
2 and 4
Step-by-step explanation:
Explanation
A function is defined as a relation between a set of inputs having one output each,in this case our independent variable is the number of days( d) and the dependent variable would be the number of pizzas
so
Step 1
let the function

therefore, to know the number of pizzas are in the stock after 5 days we need to evaluate the equation for d=3
so

so, after five day there willbe 28 pizzas in stock
I hope this helpsyou