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Julli [10]
4 years ago
9

Which of the following statements are true?

Computers and Technology
1 answer:
AleksAgata [21]4 years ago
8 0

Answer:

c. All of these

Explanation:

Generic code by replacing generic types with Object, or with with upper bounds.

The compiler has to insert type casts into code that results from translating generic types do not exist at the byte code level.

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Define a void function that calculates the sum (+), difference (-), product (*), quotient (/), and modulus (%) of two integer nu
BabaBlast [244]

Answer:

#include <iostream>

using namespace std;

void multipurpose(int &num1,int &num2,int &add,int &subt,int &multi,int &divi,int &modulo )

{

   add=num1+num2;//adding two numbers...

   subt=abs(num1-num2);//subtracting two numbers...

   multi=num1*num2;//multiplying two numbers...

   divi=num1/num2;//Finding quotient of two numbers...

   modulo=num1%num2;//Finding modulo of two numbers...

}

void print(int add,int sub,int divi,int multi,int modulo) //function to print the values.

{

   cout<<"The addition is "<<add<<endl<<"The subtraction is "<<sub<<endl

   <<"The quotient is "<<divi<<endl<<"The multiplication is "<<multi<<endl

   <<"The modulus is "<<modulo<<endl;

}

int main() {

   int a,b,sum=0,mult=0,divi=0,sub=0,modulo=0;

   cin>>a>>b;

   multipurpose(a,b,sum,sub,mult,divi,modulo);

   print(sum,sub,divi,mult,modulo);

return 0;

}

Enter the numbers

12 12

The addition is 24

The subtraction is 0

The quotient is 1

The multiplication is 144

The modulus is 0

Explanation:

I have created a function multipurpose that has a argument list of two numbers ,and variables for addition,subtraction,multiplication,Division,Modulus and these all are passed by reference.Since the function is of type void so it cannot return anything so the only way to store the result is by passing the variables to store the results by  reference.

The other function print has the arguments of all the results here you can pass them by value or by reference because you only need to print the results.

7 0
4 years ago
A Java main method uses the parameter (String[ ] variable) so that a user can run the program and supply "command-line" paramete
Firdavs [7]

Answer:

The answer is "Option a"

Explanation:

  • In java, the main function is the point of entry of every java program. Its syntax always starts "public static void main" with (String args[]), in which it can also be modified by the name of the string array name.
  • It also known as an entry point is the key process. In any program, it is the first method, that executes whenever you run a program. There is one main feature in a regular app that uses instances of certain classes to operate.
4 0
3 years ago
Create a program that contains 4 methods / functions... main(), getTestScores(), calcAverage(), and displayAverage(). The main()
jeyben [28]

Answer:

import java.util.Scanner;

public class  num6{

   static int getTestScores(){

       System.out.println("Enter the Score");

       Scanner in = new Scanner(System.in);

       int score = in.nextInt();

       return score;

   }

   static double calcAverage(int score1, int score2, int score3){

       return (score1+score2+score3)/3;

   }

   static void displayAverage(double ave){

       System.out.println("The Average is "+ave);

   }

   public static void main(String[] args) {

       int num1 = getTestScores();

       int num2 = getTestScores();

       int num3 = getTestScores();

       //Calling Calculate Average

       double average = calcAverage(num1,num2,num3);

       //Calling displayAverage

       displayAverage(average);

   }

}

Explanation:

  • Using Java programming Language
  • Create the four methods
  • getTestScores() Uses the Scanner Class to receive an in variable and return it
  • calcAverage() accepts three ints as parameter calculates their average and return it
  • displayAverage() Accepts a double and prints it out with a concatenated string as message
  • In the Main Method, getTestScores is called three times to obtain three numbers from the user
  • calAverage is called and handed the three numbers
  • printAverage is called to output the calculated average
8 0
3 years ago
(20 points) Given a collection of n nuts and a collection of n bolts, arranged in an increasing order of size, give an O(n) time
Blababa [14]

Answer:

Check the explanation

Explanation:

Keep two iterators, i (for nuts array) and j (for bolts array).

while(i < n and j < n) {

if nuts[i] == bolts[j] {

We have a case where sizes match, output/return

}

else if nuts[i] < bolts[j] {

what this means is that the size of nut is lesser than that of bolt and we should go to the next bigger nut, i.e., i+=1

}

else {

what this means is that the size of bolt is lesser than that of nut and we should go to the next bigger bolt, i.e., j+=1

}

}

Since we go to each index in both the array only once, the algorithm take O(n) time.

6 0
4 years ago
Richard uses Microsoft outlook as his email client. He wants to check whether he has received any new mail. Identify which key R
Kazeer [188]
Refresh...... I mean, that is what I hit
4 0
4 years ago
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