X-y=50 and 2x=3y
x=y+50 so 2(y+50) = 3y
2y+100=3y
y=100
x-100=50
x=150
so our numbers are 150 and 100
150-100=50 and 150 is 3*50 and 100 is 2*50 so fits all requirements!
hope this helps you! thank you!
Answer:
3/4
Step-by-step explanation:
1/2 : 2/3
X : 3/3
3/4 : 3/3
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
Since 2+8 =10 you would just have 3 left to add so it would be 10+3=13