Answer:
E(X) = 6
Var(X) = 3.394
Step-by-step explanation:
Let X represent the number of carp caught out of the 20 fishes caught. Now, if we are to assume that each
of the (100, 20) ways to catch the 20 fishes will be equally likely.
Thus, we can say that X fulfills a hypergeometric
distribution with parameters as follows;
n = 20, N = 100, k = 30
Formula for expected mean value in hypergeometric distribution is;
E(X) = nk/N
E(X) = (20 × 30)/100
E(X) = 6
Formula for variance is;
Var(X) = (nk/N) × [((n - 1)(k - 1)/(N-1))) + (1 - nk/N)]
Var(X) = ((20 × 30)/100) × [((20 - 1)(30 - 1)/(100 - 1)) + (1 - (20 × 30/100)]
Var(X) = 6 × 0.5657
Var(X) = 3.394
We have to start by creating an equation.
Cost of System = x
Cost of games = 3x
Thus x + 3x = $500
Now we find x.
4x = 500,
x = 500/4 = 125
So the system is 125,
The games, 125*3 = 375
46.06 would be the answer
C < 5
there will be an open circle on 5....it is open because there is no equal sign in the problem.....the shading will be to the left....because less then is shaded to the left
