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evablogger [386]
3 years ago
8

A construction-based client would like to develop an application that can analyze an image of machinery and overlay information

that will assist in the repair. Which form of Extended Reality can best assist the client in this situation?1-Mixed Reality2-Haptic Reality3-Augmented Reality4-Virtual Reality
Computers and Technology
1 answer:
Elden [556K]3 years ago
4 0

A form of Extended Reality which can best assist the client with information on the repair is: 3. Augmented Reality.

<h3>What is Extended Reality?</h3>

Extended reality refers to an umbrella terminology that is used to describe all real and virtual physical environments (realities) and human-machine interactions through the use of computer technologies and wearables.

<h3>The forms of Extended Reality.</h3>

In Computer technology, there are four (4) main types of Extended Reality and these include:

  • Mixed Reality
  • Haptic Reality
  • Virtual Reality
  • Augmented Reality

Augmented Reality is mainly applied in maintenance and repair by analyzing an image of machineries and provide information that will assist in the repair process.

Read more on Augmented Reality here: brainly.com/question/9054673

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What's the code?
lys-0071 [83]

Answer:

Explanation:

#include <iostream>

using namespace std;

// Recipe of single portion salad

int main()

{

   float Qing[3]={0.0,0.0,0.0};

   string ItemName[3]={" "," "," "};

   int qty=0;

   cout<<"Please enter 3 Ingredients required for Salad and Quantity required for a single serve"<<endl;

   for (int i=0;i<3;i++)

   {

       cout<<"Enter the ingredient number "<<(i+1)<<" :";

       cin>>ItemName[i];

       cout<<"Qty required for single serve (in Oz) :";

       cin>>Qing[i];

   }

   cout<<"Number of servings required :";

   cin>>qty;

   cout<<endl<<"Total Quantities required for "<<qty<<" servings"<<endl;

   for (int i=0;i<3;i++)

   {

       cout<<ItemName[i]<<" Qty for "<<qty<<" servings :"<<(Qing[i]*qty)<<" Oz."<<endl;

   }

   return 0;

}

// You can run this after compiling without any problem.

3 0
3 years ago
Which principle suggests that specific single responsibility interfaces are better than one general purpose interface?
ElenaW [278]

Answer:Interface segregation principle

Explanation: Interface-segregation principle (ISP) is amongst the major principles of the object-oriented design which describes that none of the users/clients can be forced for indulging and depending on the unknown methods or methods that they don't have knowledge about.

It functions by making the interfaces visible to the user that specifically fascinates them and keeping other smaller interfaces.Interfaces are made by splitting process and making the small interfaces from them.

8 0
3 years ago
Step 1: Configure the initial settings on R1. Note: If you have difficulty remembering the commands, refer to the content for th
Andru [333]

Answer:

The configuration of the R1 is as follows

Explanation:

Router>enable

Router#show running-config

Router#show startup-config

Router#configure terminal

Router(config)#hostname R1

R1(config)#line console 0

R1(config-line)#password letmein

R1(config-line)#login

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R1(config)#enable secret itsasecret

R1(config)#service password-encryption

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3 0
3 years ago
Create an array of numbers filled by the random number generator. (value = (int)(Math.random() * 100 + 1);) Print the array and
Y_Kistochka [10]

Answer:

Explanation:

the following is the code to run this (JAVA)

MeanStandardDev.java

import java.util.Random;

import java.util.Scanner;

public class MeanStandardDev {

public static void main(String[] args) {

// Declaring variables

int N;

double lower, upper, min, max, mean, stdDev;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

// Getting the input entered by the user

System.out.print(" How many Random Numbers you want to generate :");

N = sc.nextInt();

System.out.print("Enter the Lower Limit in the Range :");

lower = sc.nextDouble();

System.out.print("Enter the Upper Limit in the Range :");

upper = sc.nextDouble();

// Creating Random class object

Random rand = new Random();

double nos[] = new double[N];

// this loop generates and populates 10 random numbers into an array

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

nos[i] = lower + (upper - lower) * rand.nextDouble();

}

//calling the methods

min = findMinimum(nos);

max = findMaximum(nos);

mean = calMean(nos);

stdDev = calStandardDev(nos, mean);

//Displaying the output

System.out.printf("The Minimum Number is :%.1f\n",min);

System.out.printf("The Maximum Number is :%.1f\n",max);

System.out.printf("The Mean is :%.2f\n",mean);

System.out.printf("The Standard Deviation is :%.2f\n",stdDev);

}

//This method will calculate the standard deviation

private static double calStandardDev(double[] nos, double mean) {

//Declaring local variables

double standard_deviation=0.0,variance=0.0,sum_of_squares=0.0;

/* This loop Calculating the sum of

* square of eeach element in the array

*/

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

{

/* Calculating the sum of square of

* each element in the array    

*/

sum_of_squares+=Math.pow((nos[i]-mean),2);

}

//calculating the variance of an array

variance=((double)sum_of_squares/(nos.length-1));

//calculating the standard deviation of an array

standard_deviation=Math.sqrt(variance);

return standard_deviation;

}

//This method will calculate the mean

private static double calMean(double[] nos) {

double mean = 0.0, tot = 0.0;

// This for loop will find the minimum and maximum of an array

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

// Calculating the sum of all the elements in the array

tot += nos[i];

}

mean = tot / nos.length;

return mean;

}

//This method will find the Minimum element in the array

private static double findMinimum(double[] nos) {

double min = nos[0];

// This for loop will find the minimum and maximum of an array

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

// Finding minimum element

if (nos[i] < min)

min = nos[i];

}

return min;

}

//This method will find the Maximum element in the array

private static double findMaximum(double[] nos) {

double max = nos[0];

// This for loop will find the minimum and maximum of an array

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

// Finding minimum element

if (nos[i] > max)

max = nos[i];

}

return max;

}

}

the OUTPUT should give;

How many Random Numbers you want to generate :10

Enter the Lower Limit in the Range :1.0

Enter the Upper Limit in the Range :10.0

The Minimum Number is :1.1

The Maximum Number is :9.9

The Mean is :6.30

The Standard Deviation is :2.98

cheers i hope this helps!!!

4 0
3 years ago
Need help with these
kolbaska11 [484]
The first one is d the second one is true the third one is false
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3 years ago
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