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 is 18 unless u wont it in letters
$13 because Isabella will be awarded $104 making the ratio 1:8
Answer:
Therefore,
Step-by-step explanation:
Given:
In ΔABC, BC=4 cm,
angle b=angle c, and
angle a=20°
To Find:;
AC = ?
Solution:
Triangle sum property:
In a Triangle sum of the measures of all the angles of a triangle is 180°.
We know in a Triangle Sine Rule Says that,
In Δ ABC,
substituting the given values we get
Therefore,
Answer:
3
Step-by-step explanation:
The dilation factor is 3
