Historical evidence indicates that times between fatal accidents on scheduled American domestic passenger flights have an approximately exponential distribution. Assume that the mean time between accidents is 52 days. (a) If one of the accidents occurred at 12:00 am on October 1 of a randomly selected year in the study period, what is the probability that another accident occurred that same month

The probability that another accident will occur that same month = 0.44907

Step-by-step explanation:

Let consider X to be the random variable that follows an exponential distribution with a mean of 52.

The mean of the exponential distribution

Thus, the probability distribution of X is:

Suppose X represents the time interval between the accident that happened in October 1  and the following accident.

Then, the probability that another accident will occur in the same month is the same as the event that the time between the two accidents are less than 31 (given that the month of October have 31 days)

Thus, we can calculate the required probability as follows:

Thus, the probability that another accident will occur that same month = 0.44907

$25

step-by-step explanation:

25 - 4.50 = 20.5

20.5 / 2 = 10.25

10.25 + 10 = 20.25

20.25 - 12.50 = 7.75

therefore the answer is $25 as shown as the first number.

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph