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Natasha_Volkova [10]
3 years ago
8

If a pilot is converting standard time to UTC time and is given the time 1730 UTC, what would EST be?

Computers and Technology
2 answers:
bulgar [2K]3 years ago
6 0

The answer is 12:30 hours

Coordinated Universal Time which is abbreviated as (UTC) is a successor to the Greenwich Meridian Time. Offset, UTC is zero hours ahead of GMT and is universally used. It regulates clocks and time. EST (Eastern Standard Time) is minus 5 hours from UTC. Thus, 17:30 in EST will be 12:30 hours






MAVERICK [17]3 years ago
6 0

Answer:

4:30PM

Explanation:

I believe this because I think pilots use military time if I'm not to be mistaken...

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Contrast the following terms (provide examples): Partial dependency; transitive dependency
kiruha [24]

Answer:

Explanation:

Transitive dependency

In this case, we have three fields, where field 2 depends on field 1, and field three depends on field 2.

For example:

Date of birth --> age --> vote

Partial dependency

It is a partial functional dependency if the removal of any attribute Y from X, and the dependency always is valid

For example:

Course and student these tables have a partial dependency, but if we have the field registration date, this date will depend on the course and student completely, we must create another table with the field registration date to remove this complete dependency.

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5 0
2 years ago
write a 2d array c program that can capture marks of 15 students and display the maximum mark, the sum and average​
bekas [8.4K]

Answer:

#include <stdio.h>  

int MaxMark(int* arr, int size) {

   int maxMark = 0;

   if (size > 0) {

       maxMark = arr[0];

   }

   for (int i = 0; i < size; i++) {

       if (arr[i] > maxMark) {

           maxMark = arr[i];

       }

   }

   return maxMark;

}

int SumMarks(int* arr, int size) {

   int sum = 0;

   for (int i = 0; i < size; i++) {

       sum += arr[i];

   }

   return sum;

}

float AvgMark(int* arr, int size) {

   int sum = SumMarks(arr, size);

   return (float)sum / size;

}

int main()

{

   int student0[] = { 7, 5, 6, 9 };

   int student1[] = { 3, 7, 7 };

   int student2[] = { 2, 8, 6, 1, 6 };

   int* marks[] = { student0, student1, student2 };

   int nrMarks[] = { 4, 3, 5 };

   int nrStudents = sizeof(marks) / sizeof(marks[0]);

   for (int student = 0; student < nrStudents; student++) {              

       printf("Student %d: max=%d, sum=%d, avg=%.1f\n",  

           student,

           MaxMark(marks[student], nrMarks[student]),

           SumMarks(marks[student], nrMarks[student]),

           AvgMark(marks[student], nrMarks[student]));

   }

   return 0;

}

Explanation:

Here is an example using a jagged array. Extend it to 15 students yourself. One weak spot is counting the number of marks, you have to keep it in sync with the array size. This is always a problem in C and would better be solved with a more dynamic data structure.

If you need the marks to be float, you can change the types.

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2 years ago
Which image file format is the most widely used?
Ostrovityanka [42]

Answer: jpeg because its a lot of people use it for quaity

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