Answer:
If you mean 10/9 then the answer should be 1.1 repeating itself
Step-by-step explanation:
Answer:
-2
Assumption:
Find the value of x such that
.
Step-by-step explanation:



Combine like terms:

This is not too bad too factor on the left hand side since 2(2)=4 and 2+2=4.


So we need to solve:

Subtract 2 on both sides:

Let's check:






0 was the desired output of
.
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
.
h = -4.9t^2 + vt
In our problem,
v = 12
t = 2
Let's plug our numbers into the equation.
h = -4.9(2)^2 + (12)(2)
h = -19.6 + 24
h = 4.4 m