Answer
Part A: The right triangles are smiliar because If the lengths of the hypotenuse and a leg of a right triangle are proportional to the corresponding parts of another right triangle, then the triangles are similar.
Part B: 2z:2x or 2z/2x. You can write z/x as z:x. Then just multiply both sides by the same number to get an equivalent ratio. Let's multiply by 2. then the answer will be: 2z:2x or 2z/2x
Step-by-step explanation:
Mid segment of triangle = 1/2 base
base = x here, so
1/2x = 15
multiply both sides by 2,
x=30
answer: x=30
Answer:
exact form: 13/5
Decimal form: 2.6
x= 2 3/5
Step-by-step explanation:
3x+4x-7=8-3x+11
7x-7=8-3+11
7x-19-3x
7x-7 + 7 =-3x+19+7
7x=-3+26
10x=26
x= 13/5
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)
.
C 4and 12 are the factors of 48
