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:
subtract 3 from both sides
Step-by-step explanation:
Step-by-step explanation:
5 rows of 4 tomato plants= 5×4= 20 tomato plants
4 rows of 3 cabbage plants= 4×3= 12 cabbage plants
Well , in 3 hours , the amount of the tank that will be filled with pipe A = 8,000 x 3 = 24,000 L
Pipe could empty that 24,000 L in 3 hours.
So the capacity is around 6 - 8 L . .. . we need additional information to find out for sure
hope this helps
Answer:
-4
Step-by-step explanation:
5 x(-4) = -20
50 - 20 = 35