Answer:
Step-by-step explanation:
You can't ever let this go negative. At least not at the grade you are in.
It can be 0.
So the domain must start at x = 7
sqrt(5*7 - 35) = sqrt(0) = 0
x can have any value (including 7) between 7 and infinity. If you choose a number less than 7 (like 6) the square root will go negative and that's not to be done.
So the interval is
7 ≤ x < ∞
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
.
B. I hope this help please let me know if it's correct.
Simply multiply the baby's original weight, which is 7.25 lbs, by 2.5
7.25lbs x 2.5 = 18.125 lbs
The baby weighted 18.125 pounds at the end of 8 months.
Hope this helps and May the Force Be With You!
-Jabba
Answer:
B
Step-by-step explanation:
Because you divide b to e
1.3