Answer:
variable(s) determine the amount of potential energy an object has?
Explanation:
Answer:
Following code will store the largest value in array parkingTickets in the variable mostTickets
mostTickets = parkingTickets[0];
for(int k = 0; k<parkingTickets.length; k++)
{
if(parkingTickets[i]>mostTickets)
{
mostTickets = parkingTickets[i];
}
}
Explanation:
In the above code segment, initially the number of tickets at first index is assumed as largest value of tickets in array.
Then using a for loop each value in the array parkingTickets is compared with the current mostTickets value.
If the compared value in parkingTickets array is larger than the current mostTickets value. Then that value is assigned to mostTickets.
This process is repeated for all elements in array.
Thus after looping through each element of array the largest value in array will get stored in mostTickets variable.
Answer:
b) The elevator prioritizes the requests that are on the way where it’s going, but also based on a first come first served principle.
Explanation:
Elevator routing often looks complex, as the elevator has to decide whether to go to the person who has waited the longest or the one who is closest? There are also issues of how long should the rider spend.
A simple algorithm can be designed where the elevator prioritizes the requests that are on the way where it’s going, but also based on a first come first served principle. Once it exhausts the request in its current direction, it then switches directions if there are requests in that direction. This way the elevator is able to tackle the complex issue based on the simple algorithm.
