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
Keep in mind x >0 since x is a measurement of length.
Answer:
p^(m+n)
Step-by-step explanation:
p^m * p^n
We know that when they are multiplied when the bases are the same, we add the exponents
p^(m+n)
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