Answer:
Use pythagorean theorem.
Step-by-step explanation:
The length of the shorter sides are 1 & 4
1^2+4^2=c^2
1+16=c^2
17=c^2
So, the length, rounded to the nearest whole inch, would be 4.
To find the perimeter, you need to find the side lengths first.
The short sides are 6 & 7.
6^2+7^2=c^2
36+49=c^2
85=c^2
So the length for that would be 8, in the nearest whole number.
Add all of the side lengths together.
6+7+8= 21
The perimeter is 21.
B - 4.2 < -7.5
subtract b and 4.2
less then sign is <
then <-7.5
Adding all of the areas together:
5 x 6 = 30
4 x 6 = 24
3 x 6 = 18
4^2 + 3^2 = 16 + 9 = 25(square root) = 5
" " = 5
30 + 24 + 18 + 5 + 5 = 82 ft^2
Answer:It would be 3/5 as well.
Step-by-step explanation:
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)
.
