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
The answer is True because just plug in the numbers
C4=15 and C7=14
logical expression
C4>C7
15>14
That reads to 15 is greater than 14, which is true
5.5-3b = 2b-6.25
5.5+6.25 = 2b+3b
11.75 = 5b
11.75/5 = 5b/5
2.35 = b
Answer:
Step-by-step explanation:
b(a + 1) + a = b*a + b + a = ab + b + a
1) b(2a +1 ) = b*2a + b*1 = 2ab + b Not equivalent.
2)a + (a +1)*b = a + ab+ b Equivalent
3) (a +1)(b+ a) = a*(b +a) + 1*(b+a) = ab+ a² +b + a Not equivalent.
4) (a + 1)b + a = ab+ b + a Equivalent
5) a + b(a+1) = a +ab + b Equivalent
6) a + (a +1) + b = a + a + 1 + b = 2a + 1 +b Not equivalent.
7) a(b +1) + b = ab + a + b Equivalent