now, let's recall the rational root test, check your textbook on it.
so p = 18 and q = 1
so all possible roots will come from the factors of ±p/q
now, to make it a bit short, the factors are loosely, ±3, ±2, ±9, ±1, ±6.
recall that, a root will give us a remainder of 0.
let us use +3.
well, that one worked... now, using the rational root test, our p = 6, q = 1.
so the factors from ±p/q are ±3, ±2, ±1
let's use 3 again
and of course, we can factor x²-x-2 to (x-2)(x+1).
(x-3)(x-3)(x-2)(x+1).
Answer:
0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.
The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.
Step-by-step explanation:
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested.
First six not defective, each with 0.98 probability.
7th defective, with 0.02 probability. So
0.0177 = 1.77% probability that the first defect is caused by the seventh component tested.
Find the expected number and variance of the number of components tested before a defective component is found.
Inverse binomial distribution, with
Expected number before 1 defective(n = 1). So
Variance is:
The expected number of components tested before a defective component is found is 50, with a variance of 0.0208.
