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:

Interval notation: 
Step-by-step explanation:
<u>First inequality:</u>
<u />
Therefore, this inequality restricts:

<u>Second inequality:</u>

Therefore, this inequality restricts:

Therefore, with both of these restrictions together, we have:
.
Answer:
139
Step-by-step explanation:
First, find all the faces that you must find the areas of. In this case, you need to find the areas of two triangles and three rectangles.
In this triangular prism, the base is one triangle. If the area of the base is 7.75, we can assume that that holds true for both bases, and multiply 7.75 by 2 in order to find the area of both triangles.
Now find the area of each of the three rectangles by multiplying their individual heights and bases by each other. You should get 38, 47.5, and 38 as the areas of the rectangles.
Now add all the individual area's together. (They're bolded for clarity).
Answer:
≈ 58
Step-by-step explanation:
Estimation usually means you want an answer good to one or maybe two significant figures. That usually means you want to round the numbers involved to one or two significant figures.
Doing that here transforms the problem to 180/3 = 60, an answer with one significant figure.
You can improve the estimate a bit by recognizing that it is a little high (the numerator is higher than 175.32, but the denominator is the same). Since the numerator is high by about 5, the estimate is high by about 5/3 or a little less than 2. A closer estimate will be 60 - 2 = 58.
_____
The degree to which you refine the estimate will depend on the error requirements you have. Certainly an estimate of 60 is within 10% of the true value, so is "close enough" for many purposes.