round you volume measure to a whole number. like if it was 80.4 it would be 80
The answer is true. Hope this helps!!!
Answer:
50 degrees
Step-by-step explanation:
Okay! So first off, because this is NOT a right triangle, we can't use soh cah toa. That means we can either use the law of sines or the law of cosines.
Because we only have sides here and no angles, we are forced to use the law of cosines.
c^2 = a^2 + b^2 − 2ab cos(C)
c = 6
b = 7.5
a = 6.5
36 = 6.5^2 + 7.5^2 - 2(6.5)(7.5)cos(C)
36 = 98.5 - 97.5cos(C)
-62.5 = -97.5cos(C)
0.641 = cos(C)
angle C = cos^-1(0.641)
angle C = 50.13 which is around 50 degrees
Remember! A dumb thing i always used to do in geometry was use radians instead of degrees, be sure to use degrees here because you are looking for degrees. Radians are for things involving not degrees, but PI.
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
.
Asymptotes are lines that continues to approach a given curve but never actually meets it! For the second figure it is the Y axis since the curves slowly appear to approach the Y axis! Not sure about the first figure(probably the X axis)!