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Rashid [163]
2 years ago
3

What would be suggested way to share and sustain knowledge with members in a team ?

Computers and Technology
1 answer:
Elis [28]2 years ago
4 0

Answer: D) Sharing best practices, lessons learned and other topics in a central place where the team can collaborate

Explanation: Knowledge sharing within a team is usually an important aspect of building a strong and efficient team. Acquiring knowledge usually comes from the experience gathered from previous projects, learning, or tips learned from various sources. Therefore, sharing knowledge is always a good way to ensure that what is gained is sustained by letting others know. Sharing and sustaining knowledge is most effective when transmitted or aired to a group of people usually a team gathered in an interactive setting which allows members to freely share experience and get feedback, views, or thoughts from other collaborators.

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(NumberFormatException)Write the bin2Dec(String binaryString) method to convert a binary string into a decimal number. Implement
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The following program will be used that prompts the user to enter a binary number as a string and displays decimal equivalent of the string.

<u>Explanation:</u>

NumberFormat.java

import java.util.Scanner;

public class NumberFormat{

public static void main(String[] args) {

      Scanner scan = new Scanner(System.in);

      try{

      System.out.println("Enter a binary string:");

      String s = scan.next();

      int integerValue = bin2Dec(Integer.parseInt(s));

      System.out.println("The decimal value of "+s+" is "+integerValue);

      }

      catch(NumberFormatException e){

          System.out.println(e);

      }

  }  

  public static int bin2Dec(int binaryNumber) throws NumberFormatException{

  int decimal = 0;

  int p = 0;

  try{

  while(true){

  if(binaryNumber == 0){

  break;

  } else {

  int temp = binaryNumber%10;

  decimal += temp*Math.pow(2, p);

  binaryNumber = binaryNumber/10;

  p++;

  }

  }

  }

  catch(NumberFormatException e){

      throw new NumberFormatException("Invalid binary number");

  }

  return decimal;

  }

}

Output:

Enter a binary string:

1011000101

The decimal value of 1011000101 is 709

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