Suppose a caravan (5 cars) travels 100 km, beginning in front of one tollbooth, passing through a second tollbooth, and nishing just after a third tollbooth. The distance between two tollbooths is 50 km. Each car takes 12 sec to serve. The tollbooth starts to serve the caravan when the entire caravan is lined up. The caravan can only dispatch a tollbooth after all cars in the caravan are served. What is the end-to-end delay (from when the caravan is lined up before 1st tollbooth till the caravan is served by the 3rd tollbooth)

1 hr 2 min 30 secs

Step-by-step explanation:

Assume Propagation speed = 100 km/hr

number of cars = 5

distance between booths = 50 km

Total distance = 100 km

take taken to serve each car = 12 secs

Determine the end to end delay

first calculate the propagation delay = Total distance / propagation speed

                                                             = 100 / 100 = 1  hr

next calculate : Time taken to serve 5 cars in each Tollbooth

= 5 * 12 secs = 60 secs

Next calculate : transmission delay ( time taken by three tollbooth to reach 5 cars )

= 60 secs * 3 = 180 secs = 2 min 30 secs

therefore end-end delay

= propagation delay + transmission delay

= 1 hr 2 min 30 secs

the equation for this situation is y=122+45. to solve the equation you would add 45 and 122 to get a total of 167lbs. in conclusion, grace's weight eight weeks ago was 167lbs.  

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8