Complentary angle = 90 degrees
90 - 16 = your answer
I would love to help, but what are the options?
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)
.
The graph of the function is a parabola.
The nose comes down as far as y=4 but no farther.
That happens when (x - 2)² = 0 , and THAT happens when x = 2 .
Answer:
y= -3x + 8
Step-by-step explanation:
Put -2x + 6y = -4 into slope-interceltp form
Take the negative reciprocal of the slope.
Use this reciprocal in point-slope form
and you have,
= y= -3x + 8
<em>(EDITED)</em> Steps:
2x+6y = -4
6y=2x - 4
y= 1/3x + 2/3
if two lines are perpendicular,
m1 * m2 + -1
1/3 * m2 = -1
m2 = -3
the line perpendicular to this line is
y = -3x + k ; k ∈ R
the line passes through the point (5,-7)
so,
-7 = -3. (5) +k
-7 = -15 + k
k = 8
∴the equation of a line that passes through the point (5,-7) and is perpendicular to the line -2x+6y=-4 is
y= -3x + 8
Hope this helps, have a nice day/night! :D
