Find the concavity changes of f(x).
Give the x-coordinates.
So, the answer to the multiple choice question is -1 and 5 since those points are where f''(x) changes sign.
Note: You can tell that you're supposed to find the concavity changes of f(x) and not f''(x) because from the graph it's obvious that the concavity of f"(x) doesn't change.
Answer:
Yes
Step-by-step explanation:
A function is a relation for which each value from the set of the first components of the ordered pairs is associated with exactly one value from the set of second components of the ordered pair.
Hope this helps!
Find a common denominator which is 12
1*3=3 4*3=12 so 3 3/12
then 5*2=10 6*2=12 so 2 10/12
2 10/12+3 3/12= 5 13/12
13/12= 1 1/12 plus 5= 6 1/12
so your answer would be 6 1/12
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)
.
