A team of researchers published an article on the study of how vehicles are dispatched based on an airport-based taxi service. The researchers modeled this system with an underlying assumption that travel times of successive trips to and from the terminal are independent exponentially distributed random variables with β = 15 minutes. (a) Find the mean and standard deviation of trip time distribution (b) How likely is it for a particular trip to take more than 25 minutes? (c) If two taxis are dispatched together, what is the probability that both of them will be gone for more than 25 minutes? (d) what is the likelihood of at least of one of the taxis returning within 25?

a. The mean would be 0.067

The standard deviation would be 0.285

b. Would be of 1-e∧-375

c. The probability that both of them will be gone for more than 25 minutes is 1-e∧-187.5

d. The likelihood of at least of one of the taxis returning within 25 is 1-e∧-375

Step-by-step explanation:

a. According to the given data the mean and the standard deviation would be as follows:

mean=1/β=1/15=0.0666=0.067

standard deviation=√1/15=√0.067=0.285

b. To calculate How likely is it for a particular trip to take more than 25 minutes we would calculate the following:

p(x>25)=1-p(x≤25)

since f(x)=p(x≤x)=1-e∧-βx

p(x>25)=1-p(x≤25)=1-e∧-15x25=1-e∧-375

c. p(x>25/2)=1-p(x≤25/2)=1-e∧-15x25/2=1-e∧-187.5

d. p(x≥25)=1-e∧-15x25=1-e∧-375

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3