If two dices are rolled once, you would have a total number of 6*6=36 possibilities.
To get a sum greater than 10:
A) one of your dices could be showing a 5, the other a 6. That’s two possibilities. Dice A being the 5 or dice B being the 5.
B) both of your dices could be showing 6s.
That’s one possibility.
So your overall possibility to get a sum greater than 10 is (1+2)/36 3/36=1/12
One twelfth.
There are 6 reading groups
24 students ÷ 4 students each group= 6 reading groups
Answer:
30% of 90
30% x 90
30/100 =3/10
3/10 x 90 = 270/10
270 = 27
90 - 27 = 63
Keisha has to read 63 more minutes to reach her goal.
Answer:
Step-by-step explanation:
I don't say u must have to mark my ans as brainliest but if it has really helped u plz don't forget to thank me my frnd..
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)
.
