To do these, start by looking at the "b" value -6.
divide it by 2
-6/2 = -3
now square this number
(-3)^2 = 9
this is what you need for the "c" value
there is only a 5 for the c value so add 4 to both sides of the equation. ( +4 = +4)
y +4 = x^2 -6x +5 +4
y +4 = x^2 -6x +9
y +4 = (x -3)^2
y = (x -3)^2 - 4
vertex ( 3, -4) upwards facing like a bowl, because the "a" value is positive. So the vertex is the minimum, lowest point on the graph.
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
.
I believe the answer is 14
9514 1404 393
Answer:
b = 71 m
A = 83°
C = 29°
Step-by-step explanation:
Many calculators can solve triangles. Apps are available for phone and tablet, or on the internet, like the one used here. In general, it takes less time to use one of these than to type your question into Brainly.
Given two sides and the angle between them, the Law of Cosines is the appropriate relation to use for finding the third side.
b = √(a² +c² -2ac·cos(B))
b = √(76² +37² -2·76·37·cos(67.75°)) ≈ √5015.48
b ≈ 70.82005 ≈ 71 . . . meters
__
One a side and its opposite angle are known, the remaining angles are found using the Law of Sines.
sin(A)/a = sin(B)/b
A = arcsin(a·sin(B)/b) = arcsin(76·sin(67.75°)/70.82005) ≈ 83.33°
A ≈ 83°
C = arcsin(37·sin(67.75°)/70.82005) ≈ 28.92°
C ≈ 29°
Or, you can find the remaining angle from 180° -68° -83° = 29°.
Answer:
4
Step-by-step explanation:
- If you take the lowest half of the data (from the lower extreme to the median) and find the median of these numbers, that is called the lower quartile or quartile 1.
- If you take the upper half of the data (from the upper extreme to the median) and find the median of these numbers, that is called the upper quartile or quartile 3.
- The interquartile range is the difference between quartile 1 and quartile 3.
- 4, 5, 6, 8, 9, 10, 11, 13
- Median=8.5
- Q1=6.5
- Q3=10.5
- 10.5-6.5=4
I also added my PowerPoint if you didn't get what I was saying.