Your answer would be the first one because a reflection is when the object is facing itself
The red corner should face itself
Hope This Helps! :)
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
.
Answer:
The dog drank 5/12 of the bowl
Step-by-step explanation:
Common denominator
3/4 = 9/12
1/3 = 4/12
9/12 - 4/12 = 5/12
Basically, we need to find the lowest common multiple of 6 and 9....and that would be 18.
so 18 minutes from 4 p.m. is 4:18 p.m. <===
-5 4/5, -5.42, 5.34, 5 5/6