The Bernoulli distribution is a distribution whose random variable can only take 0 or 1
- The value of E(x2) is p
- The value of V(x) is p(1 - p)
- The value of E(x79) is p
<h3>How to compute E(x2)</h3>
The distribution is given as:
p(0) = 1 - p
p(1) = p
The expected value of x2, E(x2) is calculated as:

So, we have:

Evaluate the exponents

Multiply

Add

Hence, the value of E(x2) is p
<h3>How to compute V(x)</h3>
This is calculated as:

Start by calculating E(x) using:

So, we have:


Recall that:

So, we have:

Factor out p

Hence, the value of V(x) is p(1 - p)
<h3>How to compute E(x79)</h3>
The expected value of x79, E(x79) is calculated as:

So, we have:

Evaluate the exponents

Multiply

Add

Hence, the value of E(x79) is p
Read more about probability distribution at:
brainly.com/question/15246027
Answer:
Bella saves more per chore than sweet t
Step-by-step explanation:
Becuase i just did it
Answer:
3 1/2
Step-by-step explanation:
We are told that Emerson struck out 112 times in 350 at-bats. We are asked to find the percent of strike outs per at-bat.
Let us find out 112 is what percent of 350.



Therefore, Emerson struck out 32% of the at-bats.
Answer:
$5.82
Step-by-step explanation: