Prove that if m + n and n + p are even integers, where m, n, and p are integers, then m + p is even.
m=2k-n, p=2l-n
Let m+n and n+p be even integers, thus m+n=2k and n+p=2l by definition of even
m+p= 2k-n + 2l-n substitution
= 2k+2l-2n
=2 (k+l-n)
=2x, where x=k+l-n ∈Z (integers)
Hence, m+p is even by direct proof.
Answer:
E’ is (11,-1)
Step-by-step explanation:
Here, we want to get the new coordinates of E
The translation is 4 units right ( add 4 to x-value) and 3 units down ( subtract 3 from y-value)
So for E, we have
(7 + 4, 2-3)
= (11, -1)
Answer:
0
Step-by-step explanation:
The formula gives you a relation between S and n. For every n, you can perform the calculation and obtain an S.
So, for n=0 we get 0/2 (0+1) which equals 0. So (0,0) is an (n,S) pair.
Answer:
a) 
And replacing we got:
b) 
And then the expected value would be:
Step-by-step explanation:
We assume the following distribution given:
Y 0 1 2 3
P(Y) 0.60 0.25 0.10 0.05
Part a
We can find the expected value with this formula:
And replacing we got:
Part b
If we want to find the expected value of 
we need to find the expected value of Y^2 and we have:
And replacing we got:
And then the expected value would be:
In three-dimensional geometry, skew lines are two lines that do not intersect and are not parallel. A simple example of a pair of skew lines is the pair of lines through opposite edges of a regular tetrahedron. Two lines that both lie in the same plane must either cross each other or be parallel, so skew lines can exist only in three or more dimensions. Two lines are skew if and only if they are not coplanar. Hope this helps!! :)
