Answer:
Solution given:
f(x)=x²
g(x)=x+5
h(x)=4x-6
now
23:
(fog)(x)=f(g(x))=f(x+5)=(x+5)²=x²+10x+25
24:
(gof)(x)=g(f(x))=g(x²)=x²+5
25:
(foh)(x)=f(h(x))=f(4x-6)=(4x-6)²=16x²-48x+36
26:
(hof)(x)=h(f(x))=h(x²)=4x²-6
27;
(goh)(x)=g(h(x))=g(4x-6)=4x-6+5=4x-1
28:
(hog)(x)=h(g(x))=h(x+5)=4(x+5)-6=4x+20-6=4x-14
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)
.
Answer:
1950
Step-by-step explanation:
if we do 325 gallons of water times 6 buckets of water we get 1,950 gallons of water in a fraction 325/1950
hope this helps
Answer:
x = the number
Step-by-step explanation:
15.) 2x + 9 = 25
16.) 3x - 6 = 21
17.) 7x = 12 + 3x
18.) 4x = 18 + x
19.) 2x + 24 = 8x
20.) 30 - x = 4x
