Answer:1.01
Step-by-step explanation:
Answer:
A. E(x) = 1/n×n(n+1)/2
B. E(x²) = 1/n
Step-by-step explanation:
The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as
P(x) = {1/n, x = 1,2...n}
Therefore,
Expectation of X
E(x) = summation {xP(×)}
= summation {X×1/n}
= 1/n summation{x}
= 1/n×n(n+1)/2
= n+1/2
Thus, E(x) = 1/n×n(n+1)/2
Value of E(x²)
E(x²) = summation {x²P(×)}
= summation{x²×1/n}
= 1/n
For this case we have the following function:

The first thing you should do is graph the function to see the behavior.
When observing the graph (in the attached image) we observe that the following point belongs to the graph:

To prove it, let's evaluate the point in the function:

The equation is fulfilled and therefore the point belongs to the graph.
Answer:
(6.2, 7.4)
See attached image
Answer:
j = 0 or j = 30
Step-by-step explanation:
j² - 30j = 0
Factorise j out of both terms on the left side.
j (j-30) = 0
Apply zero product rule.
j = 0 or j-30 = 0
Find the value of j.
j = 0 or j = 30