The numerator and denominator must be polynomials so
the answer to your question is - 7x / x^2 ( the last one on the right)
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
.
There are 33/20 or 1.65 pounds left.
Given:
4 3/4 pounds of clay
1 1/10 pounds of clay for a cup
2 pounds of clay for a jar
Convert the mixed fraction into fractions.
4 3/4 = (4*4+3)/4 = 19/4
1 1/10 = (1*10+1)/10 = 11/10
2 = 2/1
19/4 - 11/10 = (19/4 *5/5) - (11/10 * 2/2) = 95/20 - 22/20
= (95-22)/20 = 73/20 or 3.65 pounds after making a cup
73/20 - 2/1 = 73/20 - (2/1 * 20/20) = 73/20 - 40/20 = 73 - 40 / 20 = 33/20 OR 1.65 pounds left after making a jar.
We can also convert the fractions into decimal numbers.
19/4 = 4.75
11/10 = 1.10
2/1 = 2.00
4.75 - 1.1 - 2 = 1.65 pounds left.
Answer:
566.4
Step-by-step explanation:
do multipilication