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Artist 52 [7]
3 years ago
13

The term ________ refers to the use of a single unifying device that handles media, internet, entertainment, and telephone needs

.
Computers and Technology
1 answer:
dezoksy [38]3 years ago
5 0
The correct answer that would best complete the given statement above would be DIGITAL CONVERGENCE. The term Digital Convergence <span>refers to the use of a single unifying device that handles media, internet, entertainment, and telephone needs. Thanks for posting your question. Hope this helps. </span>
You might be interested in
Write a program that will input the names, ages and weights of three siblings and display the lightest followed by the youngest
jekas [21]

Answer:

Here is the JAVA program:

Sibling.java class:

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

public class Sibling {  //class name

//private data members of class

   private String name;  // String type variable to hold the name of sibling

   private int age;  // int type variable to hold the age of sibling

   private int weight;  // int type variable to hold the weight of sibling

   public Sibling() {}     //constructor of class Sibling

   public Sibling (String n, int a, int w)  {  //parameterized constructor

       name = n;  // refers to name field

       age = a;  // refers to age field

       weight = w;    }   // refers to weight field

   public String getName ()    {   //accessor method to get the current name of sibling

       return name;    }   //returns the name field of the sibling

   public int getAge ()    {   //accessor method to get the current age of sibling

       return age;    }   //returns the age field of the sibling

   public int getWeight (){ //accessor method to get the current weight of sibling

    return weight;    } //returns the weight field of the sibling

   public void getInput() {    //to take input name,age and weight from user

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

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

 name = input.nextLine();  //scans and reads the input name from user

 System.out.print("Enter the age:  ");  //prompts user to enter the age of sibling

 age = input.nextInt();   //scans and reads the age from user

 System.out.print("Enter the weight:  ");  //prompts user to enter the weight of sibling

 weight = input.nextInt(); } }   //scans and reads the input weight from user

Explanation:

Here is the TestSibling class that contains main method:

public class TestSibling {   //class name

public static void main (String[] args) {   //main method

String name;  // to hold name of sibling

int age, weight;   // to hold age and weight of sibling

Sibling sib1, sib2, sib3;   // objects of Sibling class

sib1 = new Sibling ();  // creates object of class Sibling for sibling 1 and calls constructor

sib1.getInput();   // calls getInput method using sib1 object to get the name, age and weight of sibling 1

sib2 = new Sibling ();  // creates object of class Sibling for sibling 2 and calls constructor

sib2.getInput();   //calls getInput method using sib2 object to get the name, age and weight of sibling 2

sib3 = new Sibling ();  //creates object of class Sibling for sibling 3 and calls constructor

sib3.getInput();   //calls getInput method using sib3 object to get the name, age and weight of sibling 3

Sibling youngest=null, lightest=null;    //holds the youngest age and lightest weight of siblings

if (sib1.getAge( )<= sib2.getAge( ) && sib1.getAge( ) <= sib3.getAge( ) )   /*if condition checks if age of sibling 1 is less than or equals to that of sibling 2 and sibling 3,  using object of each sibling and getAge method to access age. the logical operator && AND is used so that if condition evaluates to true if sib1 is younger than both sib2 and sib3 */

{ youngest=sib1;}   //if the above condition is true then sets sib1 as youngest

else if (sib2.getAge( ) <= sib1.getAge( ) && sib2.getAge( ) <= sib3.getAge( ) )   // else if condition checks if age of sibling 2 is less than or equals to that of sibling 1 and sibling 3

{youngest=sib2;}   // if above condition is true then sets sib2 as the youngest

else   //if none of the above condition is true then this means that the third has the lowest age

{            youngest=sib3;    }    //sets sib3 as the youngest

if (sib1.getWeight( ) <= sib2.getWeight( ) && sib1.getWeight( ) <= sib3.getWeight( ) )   //if condition checks if weight of sibling 1 is less than or equals to that of sibling 2 and sibling 3,  using object of each sibling and getAge method to access weight of each sibling

          { lightest=sib1; }   //if the above condition is true then sets sib1 as having the lightest weight

else if (sib2.getWeight( ) <= sib1.getWeight( ) && sib2.getWeight( ) <= sib3.getWeight( ) )  // else if condition checks if weight of sibling 2 is less than or equals to that of sibling 1 and sibling 3

{lightest=sib2; }  //if the above condition is true then sets sib2 as the lightest

else  //if none of the above condition is true then this means that the third has the lightest weight

{ lightest=sib3;   }  //sets sib3 as the lightest

System.out.println("The lightest sibling: " + lightest.getName() +" " + lightest.getAge()+" "+ lightest.getWeight());  } } //calls the getName() getAge() and getWeight() method using object lightest to print the lightest of the siblings

System.out.println("The youngest sibling: " + youngest.getName() +" " + youngest.getAge()+" "+ youngest.getWeight());  //calls the getName() getAge() and getWeight() method using object youngest to print the youngest of the siblings

The screenshot of the output is attached.

4 0
3 years ago
A central issue of public sharing is:
Troyanec [42]
I think is b.how much time we spend alone
6 0
2 years ago
Read 2 more answers
The mathematical constant Pi is an irrational number with value approximately 3.1415928... The precise value of this constant ca
tatiyna

Answer:

I am writing a Python program:

def approxPIsquared(error):

   previous = 8

   new_sum =0

   num = 3

   while (True):

       new_sum = (previous + (8 / (num ** 2)))

       if (new_sum - previous <= error):

           return new_sum

       previous = new_sum

       num+=2    

print(approxPIsquared(0.0001))

Explanation:

I will explain the above function line by line.

def approxPIsquared(error):  

This is the function definition of approxPlsSquared() method that takes error as its parameter and approximates constant Pi to within error.

previous = 8     new_sum =0      num = 3

These are variables. According to this formula:

Pi^2 = 8+8/3^2+8/5^2+8/7^2+8/9^2+...

Value of previous is set to 8 as the first value in the above formula is 8. previous holds the value of the previous sum when the sum is taken term by term. Value of new_sum is initialized to 0 because this variable holds the new value of the sum term by term. num is set to 3 to set the number in the denominator. If you see the 2nd term in above formula 8/3^2, here num = 3. At every iteration this value is incremented by 2 to add 2 to the denominator number just as the above formula has 5, 7 and 9 in denominator.

while (True):  This while loop keeps repeating itself and calculates the sum of the series term by term, until the difference between the value of new_sum and the previous is less than error. (error value is specified as input).

new_sum = (previous + (8 / (num ** 2)))  This statement represents the above given formula. The result of the sum is stored in new_sum at every iteration. Here ** represents num to the power 2 or you can say square of value of num.

if (new_sum - previous <= error):  This if condition checks if the difference between the new and previous sum is less than error. If this condition evaluates to true then the value of new_sum is returned. Otherwise continue computing the, sum term by term.

return new_sum  returns the value of new_sum when above IF condition evaluates to true

previous = new_sum  This statement sets the computed value of new_sum to the previous.

For example if the value of error is 0.0001 and  previous= 8 and new_sum contains the sum of a new term i.e. the sum of 8+8/3^2 = 8.88888... Then IF condition checks if the

new_sum-previous <= error

8.888888 - 8 = 0.8888888

This statement does not evaluate to true because 0.8888888... is not less than or equal to 0.0001

So return new_sum statement will not execute.

previous = new_sum statement executes and now value of precious becomes 8.888888...

Next   num+=2  statement executes which adds 2 to the value of num. The value of num was 3 and now it becomes 3+2 = 5.

After this while loop execute again computing the sum of next term using       new_sum = (previous + (8 / (num ** 2)))  

new_sum = 8.888888.. + (8/(5**2)))

This process goes on until the difference between the new_sum and the previous is less than error.

screenshot of the program and its output is attached.

6 0
3 years ago
Match the following.
bija089 [108]

Answer:

#4 is quick access tool bar #7 is zoom bar #2ribbon #5 scroll bars #6 title bar #1contextual bars #3 ruler

6 0
3 years ago
Read 2 more answers
Write a method called swapPairs that switches the order of values in an ArrayList of strings in a pairwise fashion. Your method
arsen [322]

Answer:

  1. import java.util.ArrayList;
  2. public class Main {
  3.    public static void main(String[] args) {
  4.        ArrayList<String> strList =new ArrayList<String>();
  5.        strList.add("to");
  6.        strList.add("be");
  7.        strList.add("or");
  8.        strList.add("not");
  9.        strList.add("to");
  10.        strList.add("be");
  11.        strList.add("hamlet");
  12.        swapPairs(strList);
  13.        System.out.println(strList);
  14.    }
  15.    public static void swapPairs(ArrayList<String> list){
  16.        for(int i=0; i < list.size()-1; i+=2){
  17.            String temp = list.get(i);
  18.            list.set(i, list.get(i+1));
  19.            list.set(i+1, temp);
  20.        }
  21.    }
  22. }

Explanation:

Firstly, let's create a method swapPairs that take one ArrayList (Line 18). In the method, use a for-loop to traverse through each item in the ArrayList and swap the items between the current items at index-i and at index-i+1 (Line 19-22). The index-i is incremented by two in next loop and therefore the next swapping will proceed with third and fourth items and so forth.

In the main program, create a sample ArrayList (Line 5-12) and then test the method (Line 14) and print the output (Line 15). We shall get  [be, to, not, or, be, to, hamlet].

7 0
3 years ago
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