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labwork [276]
1 year ago
11

The term used to describe whereby old and new media are available via the integration of personal computers and high speed satel

lite based phone or cable links is:________
Computers and Technology
1 answer:
garri49 [273]1 year ago
6 0

The term used to describe whereby old and new media are available via the integration of personal computers and high speed satellite based phone or cable links is: media convergence.

<h3>What's a good illustration of media convergence? </h3>
  • Smartphones, laptops, and ipads are the finest instances of media convergence since they combine several forms of digital media, including radio, cameras, TVs, music, and more, into a single, straightforward gadget.
  • The blending of formerly separate media platforms and technologies through digitization and computer networking is referred to as media convergence. Another name for this is technical convergence.
  • Media ownership concentration, sometimes referred to as media consolidation or media convergence, is the process through which a smaller number of people or organisations come to control a larger portion of the mainstream media.
  • According to recent study, there is a rising amount of consolidation in the media sectors, which are already highly concentrated and controlled by a very limited number of companies.

To learn more about media convergence, refer to the following link:

brainly.com/question/25784756

#SPJ4

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For this project you will write a Java program that will run a simple math quiz. Your program will generate two random integers
ElenaW [278]

Answer:

Here is the JAVA program:

import java.util.Scanner;   //to accept input from user

public class FunWithBranching {   //class name

   public static void main(String[] args) {  //start of main function

   Scanner input = new Scanner(System.in);  //creates Scanner object

   System.out.print("Enter your name: "); //prompts user to enter name

   String userName = input.nextLine();  //stores the name

   System.out.print("Welcome " + userName + "! Please answer the following questions:\n");   //prompts user to answer the questions

   int randomOperand1 =  (int)(20 * Math.random()) + 1;  //generates random number for operand 1 (1-20)

   int randomOperand2 =  (int)(20 * Math.random()) + 1;  //generates random number for operand 2 (1-20)

   int randomAdd = randomOperand1 + randomOperand2;  //performs addition of two random numbers and stores result in randomAdd

   int randomMul = randomOperand1 * randomOperand2;  //performs multiplication of two random numbers

   int randomDiv = randomOperand1 / randomOperand2;  //performs division of two random numbers

   int randomMod = randomOperand1 % randomOperand2;  //performs modulo of two random numbers

   int correctAns = 0;  //stores number of correct answers

   System.out.print(randomOperand1 + " + " + randomOperand2 + " = ");  //displays random number + random number

   int AdditionGuess = input.nextInt();   //reads the answer of addition from user and stores it in AdditionGuess

   if (AdditionGuess == randomOperand1 + randomOperand2) {  //if the user answer is correct

   System.out.println("Correct!");  //displays correct

   correctAns++;  //adds 1 to the count of correctAns every time the user gives correct answer of addition

 } else {  //if user answer is incorrect

   System.out.println("Wrong!");  //displays wrong

   System.out.println("The correct answer is " + randomAdd);   }  //displays the correct answer of addition

   System.out.print(randomOperand1 + " * " + randomOperand2 + " = ");  //displays random number * random number  

   int MultiplicationGuess = input.nextInt();   //reads the answer of multiplication from user and stores it in MultiplicationGuess

 if (MultiplicationGuess == randomOperand1 * randomOperand2) {  //if the user answer is correct

     System.out.println("Correct!");  //displays correct

     correctAns++;  //adds 1 to the count of correctAns every time the user gives correct answer of multiplication

 }else{  //if user answer is incorrect

       System.out.println("Wrong!");  //displays wrong

       System.out.println("The correct answer is " + randomMul);   }  //displays the correct answer of multiplication

   System.out.print(randomOperand1 + " / " + randomOperand2 + " = ");  //displays random number / random number  

   int DivisionGuess = input.nextInt();   //reads the answer of division from user and stores it in DivisionGuess

   if (DivisionGuess == randomOperand1 / randomOperand2) {  //if the user answer is correct

       System.out.println("Correct!");  //displays correct

       correctAns++;   //adds 1 to the count of correctAns every time the user gives correct answer of division

       }else{  //if user answer is incorrect

           System.out.println("Wrong!");  //displays wrong

           System.out.println("The correct answer is " + randomDiv);     }  //displays the correct answer of division

      System.out.print(randomOperand1 + " % " + randomOperand2 + " = ");  //displays random number % random number  

       int ModGuess = input.nextInt();  //reads the answer of modulo from user and stores it in ModGuess

           if (ModGuess == randomOperand1 % randomOperand2) {  //if the user answer is correct

           System.out.println("Correct!");  //displays correct

           correctAns++;   //adds 1 to the count of correctAns every time the user gives correct answer of modulo

           }else{  //if user answer is incorrect

               System.out.println("Wrong!");  //displays wrong

               System.out.println("The correct answer is " + randomMod);    }  //displays the correct answer of modulo

double percentage = correctAns * 25;   //computes percentage

System.out.println("You got " + correctAns + " correct answers.");   //display number of correct answers given by user

System.out.println("That's " + percentage + "%!");         }  } //displays percentage

Explanation:

The program is well explained in the comments mentioned with each line of code. The screenshot of the output is attached.

3 0
3 years ago
You have been asked to create a query that will join the Production.Products table with the Production.Categories table. From th
Flauer [41]

Answer:

The Fastest car in the world would be between 1,184 Horse Power-1,106 Torque

Explanation:

6 0
3 years ago
____ are designed to be used with everyday objects, such as home appliances, gaming consoles, digital cameras, e-readers, digita
Fudgin [204]

Embedded Operating Systems are designed to be used with everyday objects, such as home appliances, gaming consoles, digital cameras, e-readers, digital photo frames, ATMs, toys, watches, GPS systems, home medical devices, voting terminals, and cars.

6 0
3 years ago
Write a C++ program to count even and odd numbers in array. The array size is 50. The array elements will be entered by the user
vlabodo [156]

Answer:

The program in C++ is as follows:

#include <iostream>

using namespace std;

int main(){

   int numbers[50];

   int evekount = 0, odkount = 0;

   for(int i = 0; i<50;i++){

       cin>>numbers[i];

       if(numbers[i]%2==0){            evekount++;        }

       else{            odkount++;        }

   }

   cout<<"Even Count: "<<evekount<<endl;

   cout<<"Odd Count: "<<odkount<<endl;

   return 0;

}

Explanation:

This declares the integer array of number

   int numbers[50];

This initializes the even count and odd count to 0

   int evekount = 0, odkount = 0;

This iterates from 1 to 50

   for(int i = 0; i<50;i++){

This gets input for the array

       cin>>numbers[i];

This checks for even

<em>        if(numbers[i]%2==0){            evekount++;        }</em>

This checks for odd

<em>        else{            odkount++;        }</em>

   }

This prints the even count

   cout<<"Even Count: "<<evekount<<endl;

This prints the odd count

   cout<<"Odd Count: "<<odkount<<endl;

3 0
3 years ago
Select the correct answer.
Dvinal [7]

Answer:

A

Explanation:

because when you make a presentation so you are already use technology

4 0
3 years ago
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