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Dimas [21]
2 years ago
10

What are the arguments for writing efficient programs even though hardware is relatively inexpensive?

Computers and Technology
1 answer:
Ainat [17]2 years ago
7 0

Answer: Even though the hardware is inexpensive the writing of program is not efficient through this method as proper development of program is necessary for the clear execution due to factors like:-

  • The facility of writing program even the cost of hardware is less but it is not a free facility.
  • It also has a slower processing for the execution of the program
  • The construction of the efficient program is necessary for the compilation and execution of it rather than poorly constructed program is worthless and inefficient in working.

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python A year in the modern Gregorian Calendar consists of 365 days. In reality, the earth takes longer to rotate around the sun
e-lub [12.9K]

Answer:

// program in Python.

#read year

i_year=int(input("Please Enter a year:"))

#check leap year

if((i_year % 4 == 0 and i_year % 100 != 0) or (i_year % 400 == 0)):

   print("{} is a leap year.".format(i_year))

else:

   print("{} is not a leap year.".format(i_year))

Explanation:

Read year from user and assign it to variable "year".If year is completely divisible by 4 and not divisible by 100 or year is completely divisible by 400 then year is leap year otherwise year is not a leap year.

Output:

Please Enter a year:2003                                                                                                          

2003 is not a leap year.

3 0
2 years ago
Read 2 more answers
Assume that getPlayer2Move works as specified, regardless of what you wrote in part (a) . You must use getPlayer1Move and getPla
kakasveta [241]

Answer:

(1)

public int getPlayer2Move(int round)

{

  int result = 0;

 

  //If round is divided by 3

  if(round%3 == 0) {

      result= 3;

  }

  //if round is not divided by 3 and is divided by 2

  else if(round%3 != 0 && round%2 == 0) {

      result = 2;

  }

  //if round is not divided by 3 or 2

  else {

      result = 1;

  }

 

  return result;

}

(2)

public void playGame()

{

 

  //Initializing player 1 coins

  int player1Coins = startingCoins;

 

  //Initializing player 2 coins

  int player2Coins = startingCoins;

 

 

  for ( int round = 1 ; round <= maxRounds ; round++) {

     

      //if the player 1 or player 2 coins are less than 3

      if(player1Coins < 3 || player2Coins < 3) {

          break;

      }

     

      //The number of coins player 1 spends

      int player1Spends = getPlayer1Move();

     

      //The number of coins player 2 spends

      int player2Spends = getPlayer2Move(round);

     

      //Remaining coins of player 1

      player1Coins -= player1Spends;

     

      //Remaining coins of player 2

      player2Coins -= player2Spends;

     

      //If player 2 spends the same number of coins as player 2 spends

      if ( player1Spends == player2Spends) {

          player2Coins += 1;

          continue;

      }

     

      //positive difference between the number of coins spent by the two players

      int difference = Math.abs(player1Spends - player2Spends) ;

     

      //if difference is 1

      if( difference == 1) {

          player2Coins += 1;

          continue;

      }

     

      //If difference is 2

      if(difference == 2) {

          player1Coins += 2;

          continue;

      }

     

     

  }

 

  // At the end of the game

  //If player 1 coins is equal to player two coins

  if(player1Coins == player2Coins) {

      System.out.println("tie game");

  }

  //If player 1 coins are greater than player 2 coins

  else if(player1Coins > player2Coins) {

      System.out.println("player 1 wins");

  }

  //If player 2 coins is grater than player 2 coins

  else if(player1Coins < player2Coins) {

      System.out.println("player 2 wins");

  }

}

3 0
2 years ago
"So far this month you have achieved $100.00 in sales, which is 50% of your total monthly goal. With only 5 more workdays this m
Rus_ich [418]

Answer:

$20 per day for next 5 working days.

Explanation:

if 50% is $100.00 another $100 need to achieve in 5 days.

100/5=20

3 0
2 years ago
Read 2 more answers
Design an application for Bob's E-Z Loans. The application accepts a client's loan amount and monthly payment amount. Output the
zmey [24]

Answer:

part (a).

The program in cpp is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   std::cout << "Loan balance: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>0)

   {

       if(balance<payment)    

         { payment = balance;}

       else

       {  

           std::cout << balance <<"\t\t\t"<< payment << std::endl;

           balance = balance - payment;

       }

   }

   return 0;

}

part (b).

The modified program from part (a), is given below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold balance and monthly payment amounts

   double balance;

   double payment;

   double charge=0.01;

   //user enters balance and monthly payment amounts

   std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;

   std::cout << "Enter the balance amount: ";

   cin>>balance;

   std::cout << "Enter the monthly payment: ";

   cin>>payment;

   balance = balance +( balance*charge );

   std::cout << "Loan balance with 1% finance charge: " <<" "<< "Monthly payment: "<< std::endl;

   //balance amount and monthly payment computed

   while(balance>payment)

   {

           std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

           balance = balance +( balance*charge );

           balance = balance - payment;

       }

   if(balance<payment)    

         { payment = balance;}          

   std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;

   return 0;

}

Explanation:

1. The variables to hold the loan balance and the monthly payment are declared as double.

2. The program asks the user to enter the loan balance and monthly payment respectively which are stored in the declared variables.  

3. Inside a while loop, the loan balance and monthly payment for each month is computed with and without finance charges in part (a) and part (b) respectively.

4. The computed values are displayed for each month till the loan balance becomes zero.

5. The output for both part (a) and part (b) are attached as images.

4 0
1 year ago
PLEASE HELP PROGRAMMING WILL GIVE BRAINLIEST
JulijaS [17]
The fourth choice is correct.
5 0
2 years ago
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