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Artist 52 [7]
1 year ago
10

Enabling encryption of all data on a desktop or laptop computer is generally considered:______.

Computers and Technology
1 answer:
Nezavi [6.7K]1 year ago
4 0

Enabling encryption of all data that are stored on a desktop or laptop computer is generally considered: essential for any computer.

<h3>What is information security?</h3>

Information security can be defined as a preventive practice which is typically used to protect an information system (IS) that use, store or transmit information, from potential theft, attack, damage, or unauthorized access, especially through the use of a body of technologies, encryption, frameworks, processes and network engineers.

<h3>The types of security control.</h3>

In Cybersecurity, there are three (3) different types of security control and these include the following:

  • Management security.
  • Physical security.
  • Operational security.

In this context, we can infer and logically deduce that the use of encryption, endpoint security, firewalls, intrusion detection systems (IDS), and biometric fingerprint readers is generally considered essential for any computer that stores data, unless they aren't sensitive information.

Read more on information security here: brainly.com/question/14286078

#SPJ4

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A technician is configuring the Windows computers on a network to print to a printer that is directly connected to the network v
finlep [7]

Answer:

Network Printer

Explanation:

A network printer is a type of printer that is accessible by network connections, making it available for use by other computers connected to that network. The network can either come from the printer or from a local computer provided to create network for the printer. Most of them contains a network interface card that allow them connect directly to a local network from which computers can then access them. This is different from a local printer that is directly connect to the computer via USB cable.

Network printer is very advantageous in that multiple network computers can access the printer at thesame time.

7 0
3 years ago
Write a C console application that will be used to determine if rectangular packages can fit inside one of a set of spheres. You
MatroZZZ [7]

Answer:

#include <cmath>

#include <iostream>

using namespace std;

int getSphereSize(double length, double breadth, double height) {

   double diagonal = sqrt(length * length + breadth * breadth + height * height);

   if (diagonal <= 4)

       return 4;

   if (diagonal <= 6)

       return 6;

   if (diagonal <= 8)

       return 8;

   if (diagonal <= 10)

       return 10;

   if (diagonal <= 12)

       return 12;

   return 0;

}

int main() {

   double length, breadth, height;

   int sphereCounts[5] = {0};

   int sphereSize;

   while (true) {

       // Get dimensions of the box

       cout << "Enter the dimensions of the box:\n";

       cout << "Length: ";

       cin >> length;

       cout << "Breadth: ";

       cin >> breadth;

       cout << "Height: ";

       cin >> height;

       if (length <= 0 || breadth <= 0 || height <= 0)

           break;

       sphereSize = getSphereSize(length, breadth, height);

       if (sphereSize == 0)

           cout << "The box cannot fit in any of the spheres";

       else

           cout << "The box can fit in the " << sphereSize << "-inch sphere";

       // Increment the counter

       if (sphereSize == 4)

           sphereCounts[0]++;

       else if (sphereSize == 6)

           sphereCounts[1]++;

       else if (sphereSize == 8)

           sphereCounts[2]++;

       else if (sphereSize == 10)

           sphereCounts[3]++;

       else if (sphereSize == 12)

           sphereCounts[4]++;

       cout << "\n\n";

   }

   cout << "\nNumber of 4-inch spheres: " << sphereCounts[0];

   cout << "\nNumber of 6-inch spheres: " << sphereCounts[1];

   cout << "\nNumber of 8-inch spheres: " << sphereCounts[2];

   cout << "\nNumber of 10-inch spheres: " << sphereCounts[3];

   cout << "\nNumber of 12-inch spheres: " << sphereCounts[4];

   cout << endl;

   return 0;

}

Explanation:

The "cmath" library is included in the c++ program. The getSphereSize function is used to return the sphere size the rectangle dimension can fit into. It program continuously prompts the user for the length, breadth, and height of the rectangle and passes the values to the getSphereSize function in the while but breaks if any or all of the variable value is zero.

The sizes of the sphere objects in inches are collected in an array of five integer values of zeros and are incremented by one for every match with a rectangle.

7 0
2 years ago
A technician is buying a PC that will host three VMs running at the same time with the current configuration. The technician bel
raketka [301]

Answer:

B. SSD

Explanation:

VMs or virtual machines are virtually created environment for multiple operating system on a host operating system. The host operating system has an extension called Hyper-V. The hyper-V manager helps to manage the virtual machine's activities.

The VMs can hold applications in their respective containers, which requires a partition of the storage. So when more VMs are configured, more storage memory is required to store their individual data. The VMs in a computer system can share a network interface card.

7 0
3 years ago
The birthday paradox says that the probability that two people in a room will have the same birthday is more than half, provided
poizon [28]

Answer:

The Java code is given below with appropriate comments for explanation

Explanation:

// java code to contradict birth day paradox

import java.util.Random;

public class BirthDayParadox

{

public static void main(String[] args)

{

   Random randNum = new Random();

   int people = 5;

   int[] birth_Day = new int[365+1];

   // setting up birthsdays

   for (int i = 0; i < birth_Day.length; i++)

       birth_Day[i] = i + 1;

 

   int iteration;

   // varying number n

   while (people <= 100)

   {

       System.out.println("Number of people: " + people);

       // creating new birth day array

       int[] newbirth_Day = new int[people];

       int count = 0;

       iteration = 100000;

       while(iteration != 0)

       {

           count = 0;

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               // generating random birth day

               int day = randNum.nextInt(365);

               newbirth_Day[i] = birth_Day[day];

           }

           // check for same birthdays

           for (int i = 0; i < newbirth_Day.length; i++)

           {

               int bday = newbirth_Day[i];

               for (int j = i+1; j < newbirth_Day.length; j++)

               {

                   if (bday == newbirth_Day[j])

                   {

                       count++;

                       break;

                   }

               }

           }

           iteration = iteration - 1;

       }

       System.out.println("Probability: " + count + "/" + 100000);

       System.out.println();

       people += 5;

   }

}

}

4 0
3 years ago
Please help guys I'm so lost ​
PilotLPTM [1.2K]

Answer:

The correct answer is C ( W * 5 )

4 0
2 years ago
Read 2 more answers
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