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
The distance of it away from zero cannot be a negative distance, distance is always positive therefor the absolute value will always be positive.
Short answer: no, absolute value is always positive
I hope this helps :)
Answer:
The first one is equivalent [the 6x+48 = 2(3x+24)] and the second one is <u>NOT</u> equivalent [the 7x+21 ≠ 2(5x+3)]
Step-by-step explanation:
Just follow distributive property to solve these. You can ignore the first expression in both until you have to compare the answers.
1. 6x+48 and 2(3x+24)
2(3x+24) ---> 2(3x) + 2(24) ---> <u>6x + 48</u>
Bring in the first expression ~ <u>6x+48 and 6x+48 </u>
They are the same, so they are equivalent
2. 7x+21 and 2(5x+3)
2(5x+3) ---> 2(5x) + 2(3) ----> 10x + 6
Bring in the first expression ~ <u>7x+21 and 10x + 6</u>
They are NOT the same, so they are NOT equivalent
