The RF is 9. You could go to Easycalculations.com/statics/cumulative/relativefrequency to double check :)
5x+6y-16
Hope this helps...Mark brainliest please
Answer:
nj +k = p
Step-by-step explanation:
n = (p-k)/j
Multiply each side by j
nj = (p-k)/j *j
nj = p-k
Add k to each side
nj +k p-k+k
nj +k = p
Answer:
An event with a probability of 0 is impossible.
An event with a probability of 1 is certain.
Step-by-step explanation:
Probability is typically expressed in terms of a fraction between 0 and 1 where the denominator is the total number of outcomes and the numerator is the number of desired outcomes. Since probability is expressed as a fraction, if the probability is 0, that means it is impossible, or there is no chance that the event can happen. However, if the probability is 1, that means that the event is certain to happen and the odds are completely in your favor that the event will happen.
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 ![E[L]=1.375n](https://tex.z-dn.net/?f=E%5BL%5D%3D1.375n)
.
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 ![\mathrm{Var}[L]=1.859n](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5BL%5D%3D1.859n)
.
