Team B because mode is how often a number appears in a list and 60 is greater than 48 or 30, which are the modes of Team A and Team C
Hope this helps!
<span>Let be A= 6x6 − x3y4 − 5xy5 and B= 4x5y + 2x3y4 + 5xy5
and when we do their difference, it is A - B =
6x6 − x3y4 − 5xy5 -( 4x5y + 2x3y4 + 5xy5)=6x6 − x3y4 − 5xy5 - 4x5y - 2x3y4 - 5xy5 = 6x6 - x3y4 - 2x3y4 - 5xy5 -4xy5 -5xy5=6x6 - 3x3y4 -14xy5, so the final solution is A - B =6x6 - 3x3y4 -14xy5, the degree of this is equal to the degree of - 3x3y4, and it is 3+4=7, the answer is
The difference has 3 terms and a degree of 7.</span>
Step-by-step explanation:
Given that,
a)
X ~ Bernoulli
and Y ~ Bernoulli 
X + Y = Z
The possible value for Z are Z = 0 when X = 0 and Y = 0
and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0
If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,
Therefore Z is a Bernoulli random variable
b)
If X and Y can not be both equal to 1
then,
or 
and 

c)
If both X = 1 and Y = 1 then Z = 2
The possible values of the random variable Z are 0, 1 and 2.
since a Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution
Answer:

Step-by-step explanation:
Distribute (2h − 3k)(h + 5k).
2h^2 + 10hk - 3hk - 15k^2
2h^2 + 7hk 15k^2