Answer:
First one is wrong. x should be 16.
Second one is not completely visible, so cannot say.
Step-by-step explanation:
In the last step, the divison by 4 yields:
x/16 = 1, so x = 16
Answer:
Step-by-step explanation:
Simply substitute the x and y values into the equation, Infinity is a number that gets closer & closer to zero and the y-intercept is always 2.
Infinity is a number that gets closer & closer to zero and the y-intercept is always 2.
y=-2(2)(-infity)+2
y=2
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)
.
Q1 of company A = 2.5
Q3 of company A = 8
Interquatile range = (Q3 - Q1)/2 = (8 - 2.5)/2 = 5.5/2 = 2.75
Q1 of company B = 2
Q3 of company B = 5.5
Interquatile range = (5.5 - 2)/2 = 3.5/2 = 1.75
Therefore, t<span>he interquartile range for Company A employees is 2 more than the interquartile range for Company B employees.</span>
