All categories

3 years ago

Distinguish between application and system software.

Computers and Technology

1 answer:

Bingel [31]3 years ago

7 0

Answer:

kuch

Explanation:

⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄⇆⇄

You might be interested in

Emotions can have Impact on your driving?

xeze [42]

A negative impact because it makes u not focus on the road

5 0

4 years ago

After 4th collision in ethernet, find the maximum time that the adapter waits until sensing the channel again for a 100 mbps bro

VARVARA [1.3K]

Waiting time for a 100 Mbps broadcast channel is 0.512 ms

Recall that with the CSMA/CD protocol, the adapter waits K * 512 bit times after a collision,

where K is drawn randomly. For K = 100,

The one bit time for 10 Mbps is 1/10000000 s = 0.0001 ms

The one bit time for 100 Mbps is 1/100000000 s = 0.00001 ms

Waiting time for a 10 Mbps broadcast channel is K*512*0.0001= 5.12 ms

Waiting time for a 100 Mbps broadcast channel is K*512*0.00001= 0.512 ms

<h3>What is a collision in Ethernet?</h3>

A collision happens on a half-duplex Ethernet network when two devices on the same network attempt to communicate data at the exact same time.
The two transmitted packets are "collapsed" by the network, which results in the network discarding both of them.
On Ethernets, collisions are unavoidable.

A collision is a momentary interaction between two bodies or more than two bodies at once that modifies the motion of the bodies involved as a result of the internal forces at work.
Collisions entail the application of force (there is a change in velocity).

Learn more about collision in Ethernet: brainly.com/question/14123270

#SPJ4

3 0

2 years ago

You are the network administrator for Slimjim, a peripheral device company. The network uses Linux, and you need information on

oksano4ka [1.4K]

Step 1 of 4

What command did you use to accomplish this task?

First of all, let us ask ourselves what manual pages or man pages are. Manual pages in Unix/Linux are sometimes referred to as an online form of software documentation. They are a set of pages that try to explain and expound every command available on your system. Further, these pages explain to you the specifics of what each command does and how to run them. And in this case, we can use two commands to search for all the commands that have the word "copy" in their name or description. The commands are “man –k copy” or “apropos copy”

Step 2 of 4

Are there any commands in this list that only a root user can execute? How are they indicated?

Yes. There are several commands in this list that only a root user can execute. If you would want to find out what these commands are, the best way to do that is to look at those manual pages that are indicated by the manual page section number. If it has a number 8, then a root user can execute this command

Let’s compare the cp command and the cpio command

If you do an apropos or a man page on the commands above, you will spot a few differences. The cp command is used to copy files and directories from source to target location while the cpio command copies files to and from archives. The available options are –a –b –d –f –i –l –L –p and several others. On the other hand, the available options for cpio are –0 –a –A –b –B –C –c –d –E and more.

Access and read the manual pages on three other commands that interest you either by using the command name or searching for them by related keyword (try using apropos)

affcopy command

NAME

affcopy - reorder and recompress AFF files

SYNOPSIS

affcopy [options] file1 file

affcopy [options] file1 file2 file3 ... dir

affcopy [options] file1 file2 file3 ... dir1 dir2...

OPTIONS

General options:

-d Print debugging information as well.

-h Help, print this message.

-L Recompress pages (preen) with LZMA (smaller but slower).

-m Just copy the missing segments.

bcopy command

NAME

bcopy - copy byte sequence

SYNOPSIS

#include <strings.h>

void bcopy(const void *src, void *dest, size_t n);

DESCRIPTION

The bcopy() function copies n bytes from src to dest. The result is correct, even when both areas overlap.

RETURN VALUE

None.

cifssd command

NAME

cifsdd - convert and copy a file over SMB

SYNOPSIS

cifsdd [OPERAND]...

cifsdd OPTION

DESCRIPTION

This tool is part of the samba(7) suite.

Copy a file, converting and formatting according to the operands.

bs=BYTES

read and write up to BYTES bytes at a time (default: 4096)

ibs=BYTES

read up to BYTES bytes at a time (default: 4096)

obs=BYTES

write BYTES bytes at a time (default: 4096)

6 0

3 years ago

Compute change

Illusion [34]

Answer:

Explanation:

The following code is written in Java and modified to do as requested. It asks the user to enter a number that is saved to the amountToChange and then uses the % operator to calculate the number of 5 dollar bills and the number of 1 dollar bills to give back. A test case has been provided and the output can be seen in the attached image below.

import java.util.Scanner;

class ComputingChange {

public static void main(String[] args) {

Scanner scnr = new Scanner(System.in);

int amountToChange;

int numFives;

int numOnes;

amountToChange = scnr.nextInt();

numFives = amountToChange / 5;

/* Your solution goes here */

numOnes = amountToChange % 5;

System.out.print("numFives: ");

System.out.println(numFives);

System.out.print("numOnes: ");

System.out.println(numOnes);

}

5 0

3 years ago

Scientists measure an object’s mass in kilograms and its weight in newtons. If you knowthe amount of mass of an object in kilogr

Phoenix [80]

Answer:

public class WeightCalculator {

private static DecimalFormat df2 = new DecimalFormat("#.##");

public static void main(String args[]){

//create scanner to receive user input

Scanner in = new Scanner(System.in);

//alert user to input mass

System.out. println("Enter the Mass in Kg");

//collect the input as a string

String s = in. nextLine();

//convert string to double and multiply by 9.8

Double weight = Double.valueOf(s)* 9.8;

//Print out answer formatted to 2 decimal place

System.out. println("The Weight is " + df2.format(weight) + "N");

//conditional statement to check if weight is less than 100

if (weight < 100){

System.out. println("Oops, it is too light");

}

//conditional statement to check if weight is more than 500

if (weight > 500){

System.out. println("Oops, it is too heavy");

}

7 0

4 years ago