Answer:
5 is the leading coefficient
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:
$10 dollars
Step-by-step explanation:
10x9=90
Given:
The two functions are:


To find:
The type of transformation from f(x) to g(x) in the problem above and including its distance moved.
Solution:
The transformation is defined as
.... (i)
Where, a is horizontal shift and b is vertical shift.
- If a>0, then the graph shifts a units left.
- If a<0, then the graph shifts a units right.
- If b>0, then the graph shifts b units up.
- If b<0, then the graph shifts b units down.
We have,


The function g(x) can be written as
...(ii)
On comparing (i) and (ii), we get

Therefore, the type of transformation is translation and the graph of f(x) shifts 2 units up to get the graph of g(x).
Answer:
2
Step-by-step explanation: