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Elis [28]
2 years ago
6

What would be an ideal scenario for using edge computing solutions?

Computers and Technology
1 answer:
Dima020 [189]2 years ago
7 0

An ideal scenario for using edge computing solutions is: B. a school computer lab with workstations connected to a local network.

<h3>What is edge computing?</h3>

Edge computing can be defined as a distributed computing system that involves the deployment of computing and storage resources closer to the sources of data.

An advantage of edge computing is that helps to significantly improve or enhance response time and save bandwidth.

In this context, an ideal scenario for using edge computing solutions is a school computer lab with workstations connected to a local network.

Read more on edge computing here: brainly.com/question/23858023

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What is an enterprise system
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Answer:

Explanation:

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5 0
3 years ago
Read 2 more answers
C++
nikitadnepr [17]

Answer:

Check the explanation

Explanation:

<u>The Code</u>

#include <fstream>

#include <iostream>

#include <cmath>

#include <cstring>

#include <cstdlib>

#include <ctime>

using namespace std;

//Function Declarations

void fillArrayWithRandNos(int nos[],int size);

void displayArray(int nos[],int size);

double mean(int nos[],int size);

double variance(int nos[],int size);

double median(int nos[],int size);

int mode(int nos[],int size);

void histogram(int nos[],int size);

int main() {

  //Declaring variables

const int size=100;

srand(time(NULL));

 

// Creating array dynamically

int* nos = new int[size];

 

//Calling the functions

fillArrayWithRandNos(nos,size);

displayArray(nos,size);

cout<<"Mean :"<<mean(nos,size)<<endl;

cout<<"variance :"<<variance(nos,size)<<endl;

cout<<"Median :"<<median(nos,size)<<endl;

cout<<"Mode :"<<mode(nos,size)<<endl;

histogram(nos,size);

 

  return 0;

}

void fillArrayWithRandNos(int nos[],int size)

{

  for(int i=0;i<size;i++)

  {

      nos[i]=rand()%(45) + 55;

  }

}

void displayArray(int nos[],int size)

{

  cout<<"Displaying the array elements :"<<endl;

  for(int i=0;i<size;i++)

  {

      cout<<nos[i]<<" ";

      if((i+1)%10==0)

      cout<<endl;

  }

}

double mean(int nos[],int size)

{

  double sum=0;

  for(int i=0;i<size;i++)

  {

      sum+=nos[i];

  }

  return sum/size;

}

double variance(int nos[],int size)

{

  double avg=mean(nos,size);

 

double variance,sum=0;

for(int i=0;i<size;i++)

{

sum+=pow(nos[i]-avg,2);

}

//calculating the standard deviation of nos[] array

variance=(double)sum/(size);

return variance;

}

double median(int nos[],int size)

{

      //This Logic will Sort the Array of elements in Ascending order

  int temp;

  for (int i = 0; i < size; i++)

{

for (int j = i + 1; j < size; j++)

{

if (nos[i] > nos[j])

{

temp = nos[i];

nos[i] = nos[j];

nos[j] = temp;

}

}

}

 

int middle;

float med;

middle = (size / 2.0);

if (size % 2 == 0)

med = ((nos[middle - 1]) + (nos[middle])) / 2.0;

else

med = (nos[middle]);

return med;

}

int mode(int nos[],int size)

{

  int counter1 = 0, counter2, modevalue;

for (int i = 0; i < size; i++) {

counter2 = 0;

for (int j = i; j < size; j++) {

if (nos[i] == nos[j]) {

counter2++;

}

if (counter2 > counter1) {

counter1 = counter2;

modevalue = nos[i];

}

}

}

if (counter1 > 1)

return modevalue;

else

return 0;

}

void histogram(int nos[],int size)

{

  int hist[9]={0};

  for(int i=0;i<size;i++)

  {

      if(nos[i]>=55 && nos[i]<=59)

      {

      hist[0]++;

      }

      else if(nos[i]>=60 && nos[i]<=64)

      {

      hist[1]++;

      }

      else if(nos[i]>=65 && nos[i]<=69)

      {

      hist[2]++;

      }

      else if(nos[i]>=70 && nos[i]<=74)

      {

      hist[3]++;

      }

      else if(nos[i]>=75 && nos[i]<=79)

      {

      hist[4]++;

      }

      else if(nos[i]>=80 && nos[i]<=84)

      {

      hist[5]++;

      }

      else if(nos[i]>=85 && nos[i]<=89)

      {

      hist[6]++;

      }

      else if(nos[i]>=90 && nos[i]<=94)

      {

      hist[7]++;

      }

      else if(nos[i]>=95 && nos[i]<=99)

      {

      hist[8]++;

      }

  }

     

  cout<<"Displaying the count of numbers in each interval:"<<endl;

 

      int cnt=0;

  for(int i=55;i<=99;i+=5)

  {

  cout<<i<<"-"<<i+4<<"|"<<hist[cnt]<<endl;

 

      cnt++;

  }

 

  cout<<"Displaying the histogram :"<<endl;

  cnt=0;

  for(int i=55;i<=99;i+=5)

  {

  cout<<i<<"-"<<i+4<<"|";

  for(int j=0;j<hist[cnt];j++)

  {

      cout<<"*";

  }  

      cout<<endl;

      cnt++;

  }          

     

 

 

}

#########

___________________________

The output can be seen in the attached image below.

8 0
3 years ago
You should process the tokens by taking the first letter of every fifth word,starting with the first word in the file. Convert t
zhenek [66]

Answer:

See explaination

Explanation:

import java.io.File;

import java.io.IOException;

import java.util.Scanner;

import java.util.StringTokenizer;

public class SecretMessage {

public static void main(String[] args)throws IOException

{

File file = new File("secret.txt");

StringBuilder stringBuilder = new StringBuilder();

String str; char ch; int numberOfTokens = 1; // Changed the count to 1 as we already consider first workd as 1

if(file.exists())

{

Scanner inFile = new Scanner(file);

StringTokenizer line = new StringTokenizer(inFile.nextLine()); // Since the secret.txt file has only one line we dont need to loop through the file

ch = line.nextToken().toUpperCase().charAt(0); // Storing the first character of first word to string builder as mentioned in problem

stringBuilder = stringBuilder.append(ch);

while(line.hasMoreTokens()) { // Looping through each token of line read using Scanner.

str= line.nextToken();

numberOfTokens += 1; // Incrementing the numberOfTokens by one.

if(numberOfTokens == 5) { // Checking if it is the fifth word

ch = str.toUpperCase().charAt(0);

stringBuilder = stringBuilder.append(ch);

numberOfTokens =0;

}

}

System.out.println("----Secret Message----"+ stringBuilder);

}

}

}

5 0
3 years ago
Which of the following is not a property of a WAN:
docker41 [41]

Answer:

Option d is the correct answer for the above question.

Explanation:

  • WAN is a type of network that facilities network connection all over the world. It can connect every computer which is in anywhere in the world while the other technology like LAN and MAN is used for small area locations.
  • The above question asked that about that statement which is not the property of WAN and that is option d because it states that WAN is used to connect only small areas of computers, but it can connect all the computers of the world, while the other property is valid for WAN.
6 0
3 years ago
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