Answer:
1. c
2. a
3. d
4. b
Step-by-step explanation:
For 1, use parchment paper for rotating it around the origin.
For 2, just make the y value positive.
For 3, add 13 to the y value.
For 4, use process of elimination with the answer choices.
(I know right, horrible explanation but you wanted the answers right?)
Answer:
c. and Sagot po PA brainless po pleas
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)
.
Step-by-step explanation: To solve this proportion for n, we can use cross products.
Image provided.
-0.615
-5/8 = 0.625
-0.62
with negatives, the larger the number is, the smaller it is
-0.625 , -0.62 , -0.615....least to greatest
