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Lena [83]
3 years ago
8

Planes frequently push back from the gate on time but then wait 2 feet from the gate until it is time to queue up for takeoff. T

his increases fuel consumption and increases the time that passengers must sit in a cramped plane awaiting takeoff.
Which of the following performance metrics would, if emphasized in evaluations, incentivize airlines to engage in such practices?

a. A performance metric that measures customer satisfaction, based on customer comfort while on the plane
b. A performance metric that measures timeliness of the flight, where a flight is considered "on time" as long as the flight is boarded and away from the gate by the scheduled departure time
c. A performance metric that measures timeliness of the flight, where a flight is considered "on time" as long as the plane takes off by the scheduled departure time.
Business
1 answer:
ira [324]3 years ago
4 0

Answer:

b. A performance metric that measures timeliness of the flight, where a flight is considered "on time" as long as the flight is boarded and away from the gate by the scheduled departure time

Explanation:

We are told that airplanes make a mock depart by exiting the boarding gates, but they stay on the runway for long periods of time. This is due to the fact that airlines measure which planes are on time based on the moment that they left the boarding gate, not when they actually lift into the air. it happened to me once and it was extremely unpleasant to just sit without moving for more than one hour. I doubt any passenger likes these type of situations.

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3 years ago
Two plumbers received a job. at first, one of the plumbers worked alone for 1 hour, and then they worked together for the next 4
nika2105 [10]

Answer:

The first worker complete the job in 25 days himself

The second worker complete the job in 20 days himself

Explanation:

r1= rate of work done by first worker

r2=rate of work done by second worker

W= total work done

t days= time taken by the 1st worker to complete the job

r1(t)=W

r1=W/t (1)

Then the time taken by the 2nd worker to complete the job is t-5 days.

r2(t−5)=W

r2=W/(t−5) (2)

If 1st worker do the job for 1 hour and then both the worker do the job for 4 hours and 40% work is done, So

r1(1)+(r1+r2)(4)=4W/10

r1+4r1+4r2=4W/10

5r1+4r2=4W/10 (3)

Substitute equation 1 and 2 into (3)

5W/t+4{W/(t−5)}=4W/10

Multiply through by 10(t-5)

5/t+4/t−5=4/10

5(10)(t−5)+4(10)(t)=4(t−5)(t)

50t−250+40t=4t^2−20t

4t^2−110t+250=0

Solving the above quadratic equation using factorization method

4t^2−100t−10t+250=0

4t(t−25)−10(t−25)=0

(t−25)(4t−10)=0

t=25 or t=2.5

2.5 can't be the answer because if we take 2.5 days in which 1st worker completes the work, then the second worker will complete the work in -2.5days which is wrong.

The first worker completes the job in 25 days by himself and the second worker completes the job in 20 days by himself.

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3 years ago
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