Answer:
3a²b
Step-by-step explanation:
The product of two algebraic
terms is 6a3b2. If one of the terms is
2ab, find the other term.
Let us represent
First term = a
Other term = b
a × b = 6a³b²
a = 2ab
b = ?
b = 6a³b²/a
b = 6a³b²/2ab
b = 6a³b²/2a¹b¹
b = (6 ÷ 2) × a^3 - 1 × b ^2 - 1
b = 3a²b
Therefore, the other term = 3a²b
Find out what the lowest number is that each of the numbers on the bottom of the fractions can divide into evenly.
The material costs 5.37 for 1 yard.
Multiply the cost of 1 yard by the number of yards bought:
5.37 per yard x 4.75 yards = $25.51 Total cost.
The answer would be D.
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)
.
20 x=5 I hope this helps
22.x=15
