Answer:
32 m
Step-by-step explanation:
The distance across the river is give as x and we are interested in finding the value of x.
Both the triangles are similar by AA Postulate.
Therefore,
20/x = 10/16 (by c.s.s.t.)
2/x = 1/16
x = 16*2
x = 32 m
Take 5 × 3 to equal 15
add the 1 +4.5+2.9+1
to equal 12.4
2.6
Hi there! The third option shows us the best setup.
We can find the area of a triangle with the standard formula area = 1/2 * base * height.
To be able to fill in this formula, we need to have a base and a height. We can't easily find a triangle with given base and height, so we must look for another option.
We can also take the area of a square (length × width) and then divide the answer by 2. This is possible when we take the third setup. Hence the third answer is correct.
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)
.
