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3 years ago

What is typically unique in each database table?

Computers and Technology

2 answers:

andrew11 [14]3 years ago

7 0

Answer:

C ID numbers

Explanation:

andriy [413]3 years ago

4 0

ID number APEX answer

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shutvik [7]

IP address i believe, not 100% positive

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3 years ago

What is a spreadsheet program? A spreadsheet program is a computerized version of_________.

BabaBlast [244]

I dont know what exact answer youre looking for because you didnt provide options but..
"A spreadsheet is a sheet of paper that shows accounting or other data in rows and columns; a spreadsheet is also a computer application program that simulates a physical spreadsheet by capturing, displaying, and manipulating data arranged in rows and columns."

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3 years ago

Read 2 more answers

Write a method swaparrayends() that swaps the first and last elements of its array parameter. ex: sortarray = {10, 20, 30, 40} b

Alexus [3.1K]

In the C programming language, you can't determine the array size from the parameter, so you have to pass it in as an extra parameter. The solution could be:

#include <stdio.h>

void swaparrayends(int arr[], int nrElements)
{
int temp = arr[0];
arr[0] = arr[nrElements - 1];
arr[nrElements - 1] = temp;
}

void main()
{
int i;
int myArray[] = { 1,2,3,4,5 };
int nrElements = sizeof(myArray) / sizeof(myArray[0]);
swaparrayends(myArray, nrElements);

for (i = 0; i < nrElements; i++)
{
printf("%d ", myArray[i]);
}

getchar();
}

In higher languages like C# it becomes much simpler:

 static void Main(string[] args)
 {
int[] myArray = {1, 2, 3, 4, 5};
swaparrayends(myArray);
foreach (var el in myArray)
{
Console.Write(el + " ");
}

Console.ReadLine();
}

static void swaparrayends(int[] arr)
{
 int temp = arr[0];
arr[0] = arr.Last();
arr[arr.Length - 1] = temp;
}

3 0

3 years ago

The answer for this question?

Debora [2.8K]

It will flow from 2 to 1

6 0

3 years ago

2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of

allsm [11]

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

n = 1.3333 * 10^9

So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

5 0

3 years ago