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Yuki888 [10]
2 years ago
8

A system of classifying and organizing online content into categories by the use of user-generated metadata such as keywords is

called a
Computers and Technology
1 answer:
SCORPION-xisa [38]2 years ago
6 0

Answer:

Folksonomy.

Explanation:

A system of classifying and organizing online content into categories by the use of user-generated metadata such as keywords is called a folksonomy.

This ultimately implies that, folksonomy is a user-generated system which is typically used for classifying and organizing online content into various categories through the use of metadata such as keywords, electronic tags and public tags in order to make it easier to find in the future.

Hence, folksonomy is highly beneficial in areas such as collaborative learning, teacher resource repository, collaborative research, educational platforms, e-commerce etc.

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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
3 years ago
100 POINTS NEED THIS BEFORE 11:59 TODAY!!!!!!!!!!!!!!!
AVprozaik [17]

Answer:ok be how hobrhkihfehgdhdj fuiufiisefif jfkijfjfhhfhfhfhf

Explanation:

5 0
3 years ago
What character separates the file extension and the filename?
PtichkaEL [24]
A period, Filenamehere.png or .pdf or any other file extension
5 0
3 years ago
Read 2 more answers
Write a for loop that iterates from 1 to numbersamples to double any element's value in datasamples that is less than minvalue.
CaHeK987 [17]

Answer:

function dataSamples=AdjustMinValue(numberSamples, userSamples, minValue)

dataSamples=userSamples;

%for loop

for i=1:numberSamples

%checking if dataSamples value at index,i

%is less than minValue

if dataSamples(i)<minValue

%set double of dataSamples value

dataSamples(i)= 2*dataSamples(i);

end

end

end

Explanation:

The given code is in MATLAB.

4 0
3 years ago
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astraxan [27]

Answer:

no

Explanation:

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