An exponential function displays either a growth or decay behavior going from a low steep to a high steep.
No, because if they were equal the second equation should be 225/(15x3), not (15/3).
Answer:
x= -136
Step-by-step explanation:
Answer:
The answer is 4 minutes
Step-by-step explanation:
I divided 825 by 3 and got 165. If you multiply 165 by all the factors, you get approximately 4 minutes. It's the closest. Try that one.
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
.