5.
f(K) = D^3 => f(25) = 125 => 25 * t = 125 ( because K is directly proportional with D^3 )=> t = 125 / 25 => t = 5 => f(25) = 25 * 5 => K * 5 = D^3 ;
6.
f(L) = F^3 => f(2) =3^3 =>f(2) = 27 => 2 / t =27 => t = 2 / 27 => t = 0.074 => f(2) = 2 / 0.074 => K / 0.074 = F^3 ;
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
.
I’m not sure but is it 7/10 since out of 10 times she hit an ace 7 times already so there is a 7/10 chance she would hit another ace
answer = ) 100.80
Step-by-step explanation:
I subtracted the people who payed on the 1st of the month which was 4 from the total 12 memebers. I got 8, then that 8 payed on the 2nd of the month so I multiplied 8 by 12.60
Answer:
What do u mean by legs please can you explain it further