<span>When you are sampling from a small finite lot, the hypergeometric distribution applies. The binomial is a poor approximation in this case.
The general equation for the hypergeometric where aCx means the number of combinations of a items selected x at-a-time.:
P(x) =[(aCx)(N-aCn-x)]/NCn
Where
N is the lot size = 20
a is the number of defectives in the lot = 3.
x is the number of defectives in the sample.
n is the sample size = 2.
A. The probability that the first item is defective is
P(x=1) = [(3C1)(17C1)]/(20C2)
= (3)(17)/190 = 0.268
The probability that the second item is defective is
P(x = 1) = [(2C1)(17C1)]/(19C2) = (2)(17)/171 = 0.199.
So the total probability is (0.268)(0.199) = 0.0532
B. The probability that the first item is good is:
P(x = 0) = (3C0)(16C2)]/20C2 = (1)(120)/190 = 0.632
The probability that the second item is defective is
P(x = 0) =[(3C0)(16C2)]/19C2
= (1)(120)/171 = 0.670.
The total probability is 0.632(0.670) = 0.4234</span>
It is option D, "Line h has points on planes R, P, and T".
Look this up on quizlet maybe
Y = -2x + 4
6x + 3y = 6x + 3(-2x + 4) = 6x - 6x + 12 = 12
⇒ y = -2x + 4 == 6x + 3y = 12 ⇒ invinite solutions: 12
⇒ no solutions -12, -4, 0, 4
