Shota built a time travel machine, but he can't control the duration of his trip. Each time he uses the machine he has a 0.80.80, point, 8 probability of staying in the alternative time for more than an hour. During the first year of testing, Shota uses his machine 202020 times. Assuming that each trip is equally likely to last for more than an hour, what is the probability that at least one trip will last less than an hour? Round your answer to the nearest hundredth.

0.99 probability that at least one trip will last less than an hour

Step-by-step explanation:

For each time that the machine is used, there are only two possible outcomes. Either it lasts more than an hour, or it does not. The probability of a trip lasting more than an hour is independent of other trips. So the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

In which is the number of different combinations of x objects from a set of n elements, given by the following formula.

And p is the probability of X happening.

0.8 probability of lasting more than one hour.

So

20 trips

So

Assuming that each trip is equally likely to last for more than an hour, what is the probability that at least one trip will last less than an hour?

Either all the trips last for more than an hour, or at least one does not. The sum of the probabilities of these outcomes is decimal 1. So

We want P(X < 20). So

In which

0.9885 probability that at least one trip will last less than an hour

Rouding to the nearest hundreth.

0.99 probability that at least one trip will last less than an hour

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5

answer:     true

step-by-step explanation: