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
Answer:
D. is the answer
Step-by-step explanation:
C. x³-4x²-16x+24.
In order to solve this problem we have to use the product of the polynomials where each monomial of the first polynomial is multiplied by all the monomials that form the second polynomial. Afterwards, the similar monomials are added or subtracted.
Multiply the polynomials (x-6)(x²+2x-4)
Multiply eac monomial of the first polynomial by all the monimials of the second polynomial:
(x)(x²)+x(2x)-(x)(4) - (6)(x²) - (6)(2x) - (6)(-4)
x³+2x²-4x -6x²-12x+24
Ordering the similar monomials:
x³+(2x²-6x²)+(-4x - 12x)+24
Getting as result:
x³-4x²-16x+24
Answer:
it's after 10pm, so will I still get brainliest?
