8.8 more than the quotient of five and X
1 answer:
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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.
Answer:
(-4, -1/2)
Step-by-step explanation:
to calculate the midpoints bt of the line use the ormula:
Where (x1, x2) is one coordinate point and (y1, y2) is anither coordinate point. Any two points will work, but I chose A (-5,-4) and B (-3, 3).
(-4, -1/2)