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slega [8]
3 years ago
10

An application with a 150 GB relational database runs on an EC2 Instance. While the application is used infrequently with small

peaks in the morning and evening, which storage type would be the most cost-effective option for the above requirement?
Computers and Technology
1 answer:
kupik [55]3 years ago
4 0

Answer:

the bandman is hered that mybaddba

You might be interested in
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.

3 0
3 years ago
Gaming related
Blababa [14]

Answer:

yes affects your K/D

Explanation:

winner winner chicken dinner

please mark me please brainliest or mark thanks

5 0
2 years ago
The code size of 2-address instruction is________________.? 5 bytes? 7 bytes? 3 bytes? 2 bytes
Finger [1]

Answer:

7 bytes

Explanation:

<u>2 Address Instruction</u>

The 2 address instruction consist 3 components in the format.

One is opcode,other two are addresses of destination and source.

<u>Example-</u>

load b,c | Opcode   destination address,source address

add a,d  | Opcode   destination address,source address

sub c,f    | Opcode   destination address,source address

Opcode consists of 1 bytes whereas destination address and source address consist of 3 bytes each.

(1+3+3) bytes=7 bytes

5 0
3 years ago
HELP!!!<br> THIS HAPPENS EVERY TIME
Lelu [443]
Idek but hope u figure it out!
4 0
3 years ago
Read 2 more answers
Which printout will result from the snippet of code?
Anna35 [415]

Answer:

These are the supplies in the list:  

[‘pencil’, ‘notebook’, ‘backpack’, ‘pen’, ‘calculator’]

Explanation:

The line return (\n) character will be in the output (so there will be a change of line), but it will NOT be visible as it would have been interpreted as a special character.

So the output will be on 2 different lines, with no \n visible.

If the command would have been: print('These are the supplies in the list:\n', supplies), with single quotes (') instead of double quotes (") then then \n would have been printed but not interpreted as a special character.  At least in most computer language.  Since we don't know of which language the question refers to, we can't be sure at 100%.

7 0
3 years ago
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