Answer:
y = - 
x + 6
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c
Given
14x + 6y = 36 ( subtract 14x from both sides )
6y = - 14x + 36 ( divide all terms by 6 )
y = - 
x + 6, that is
y = - x + 6 ← in slope- intercept form
Answer:
n = -5, 3.
Step-by-step explanation:
n^2 + 2n - 15 = 0
Let's check if we can factor this quadratic:
We need 2 numbers whose product is -15 and whose sum is +2.
There are 2 such numbers and they are +5 and -3, so the factors are:
(n + 5)(n - 3) = 0.
n + 5 = 0 or n - 3 = 0
therefore the zeroes are -5 and 3.
Answer:
a constant is a data item whose value cannot change during the program execution just as its name implies that the value is constant a variable is a data item whose value can change during the program's execution . Thus as its name implies the the value can vary
Answer:
Yes it is, because...
Step-by-step explanation:
This is an inequality, so you can treat it as an algebraic expression.
The first step is to multiple both sides by 2, to get rid of the 2 on the left side.
Your equation will now look like this:
y >= 2y - 22
The next step is to get all the y variables to one side, now that its a lot more simplified. Subtract 2y from both sides to get:
-y >= -22
Finally, cancel out the negative on both sides of the equation to get the y as a positive y, all by itself. This will get you:
y <= 22
REMINDER: when you divide by a negative number, such as in this case dividing by -1 on both sides, the inequality sign will flip!
y = 18 works because it is less than 22. (:
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)
.
