Answer:
P(Y=1|X=3)=0.125
Step-by-step explanation:
Given :
p(1,1)=0
p(2,1)=0.1
p(3,1)=0.05
p(1,2)=0.05
p(2,2)=0.3
p(3,2)=0.1
p(1,3)=0.05
p(2,3)=0.1
p(3,3)=0.25
Now we are supposed to find the conditional mass function of Y given X=3 : P(Y=1|X=3)
P(X=3) = P(X=3,Y=1)+P(X=3,Y=2) +P(X=3,Y=3)
P(X=3)=p(3,1) +p(3,2) +p(3,3)
P(X=3)=0.05+0.1+0.25=0.4
Hence P(Y=1|X=3)=0.125
Answer:
2.
Step-by-step explanation:
The common difference is 5 so the first number is 7 - 5 = 2.
30 since that’s the lowest number both 3 and 10 can fit in
Answer: See below
Step-by-step explanation:
27. -(a-3)
28. (b-1)(b+3)
29. (c+4)(c+5)
30. d(d+5)
31. -(3/4)(2e-5)
Sorry - I don't have time to enter the details. Look for areas where the expressions can be factored in a manner that forms as many equivalent expressions in both the numerator and denominator.
For example: In problem 30:
(5d-20)/(d^2+d-20) * [??]/20d = 1/4
Factor:
<u>(5(d-4))</u> <u>d(d+5)</u> = 1/4
(d-4)(d+5<u>)</u> 20d
The (d-4), d+5, and d terms cancel, leaving
5/20 = 1/4
