0.2 + 0.3 + 0.5 = 1
190 x 0.2/1 = 38
190 x 0.3/1 = 57
190 x 0.5/1 = 95
190
0.2: 0.3: 0.5
38: 57: 95
X-15+9-4=x-10 there are ten fewer people after the second stop than the original amount
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
Answer:
The graph of function f of x equals log base 5
So yes , D
