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
.
<span>For "The probability a business major is female" - you're looking for the probability of being female. That the person is a business major is already given. So, P(A|B)
</span>For "The probability a female student is majoring in business" - you're looking for the probability of being majoring in business. That the person is a female is already given. So, P(B|A)
Each sandwich is toasted for 2 minutes and it takes 3 seconds to flip it.
The first two sandwiches will be toasted in 2 minutes (toasting) and 6 seconds (flipping).
The two will then be switched which will take 10 seconds.
The next two will again take 2 minutes and 6 seconds.
The final sandwich will take 5 seconds to be placed.
The final sandwich will also take 2 minutes to toast and 3 seconds to flip.
Total time:
2 + 2 +2 = 6 minutes
6 + 10 + 6 + 3 = 25 seconds
Total time is 6 minutes and 25 seconds.
Answer:

Step-by-step explanation:
We have the expression:

The first thing we want to do, is to have the same denominator in both equations, then we need to multiply the first term by (2/2), so the denominator becomes 4*x
We will get:

Now we can directly add the terms to get:

We can't simplify this anymore
Answer:
D. $845.00
Step-by-step explanation:
$13 per hour x 65 hrs total = $845 before taxes and deduction.