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:
Step-by-step explanation:
Cameron, Eliot, Ryan and Ava can ride as their height is more than or equal to 54
Answer:
Converse: if 5=x then x+2=7
Inverse: if x+2 doesnt = 7 then x isnt 5
Contrapositive: if x doesnt equal five then x plus two doesnt equal seven
Step-by-step explanation:
We just did this last week :)
Answer:
Y = 3x^x is a graph that has exponential growth while y = 3^-x has exponential decay.
Y = 3x^x (-∞, 0) and (∞, ∞).
Y = 3x^-x (-∞, ∞) and (∞, 0).
Step-by-step explanation:
The infinity symbols were being used to represent the x and y values of each graph. I will call y = 3^x "graph 1" and y = 3^-x "graph 2".
When graph 1 had positive ∞ for its x value, its y value was reaching towards positive ∞. When its x was reaching for negative ∞, its y was going for 0.
For graph 2, however, when its x was reaching for positive ∞, its x was reaching for 0. When its x was reaching for negative ∞, its y was going for positive ∞.
Here's an image of the graphs: