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Drupady [299]
3 years ago
9

What is the relationship between agile teams and project requirements

Computers and Technology
1 answer:
spin [16.1K]3 years ago
4 0

Answer:

Agile follows a non-linear process, unlike conventional project management, which focuses more on teamwork, cooperation, and versatility, as opposed to a strict sequence of activities.Agile project management takes an iterative approach to project management, which time-boxes tasks into fast sprints.

Explanation:

You might be interested in
The code size of 2-address instruction is________________.? 5 bytes? 7 bytes? 3 bytes? 2 bytes
Finger [1]

Answer:

7 bytes

Explanation:

<u>2 Address Instruction</u>

The 2 address instruction consist 3 components in the format.

One is opcode,other two are addresses of destination and source.

<u>Example-</u>

load b,c | Opcode   destination address,source address

add a,d  | Opcode   destination address,source address

sub c,f    | Opcode   destination address,source address

Opcode consists of 1 bytes whereas destination address and source address consist of 3 bytes each.

(1+3+3) bytes=7 bytes

5 0
3 years ago
Dawn is trying to find out how much weight she can push across the room. She is really trying to find her __________. A. flexibi
Marrrta [24]

<em>The correct answer is d. muscular strength.</em>

<em></em>

<em>Explanation :</em>

<em></em>

<em>The maximal force a muscle can create when it contracts is referred to as muscular strength. When compared to someone with lower physical strength, someone with better muscular strength can lift heavier weights. Lifting progressively larger weights over time and eating a diet rich in protein-based foods can help a person's physical strength gradually grow.</em>

<em></em>

<em>I hope this helps. Please mark "Brainliest" if you can.</em>

8 0
2 years ago
Let's implement a classic algorithm: binary search on an array. Implement a class named BinarySearcher that provides one static
yKpoI14uk [10]

Answer:

Hope this helped you, and if it did , do consider giving brainliest.

Explanation:

import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

 

//   main method

  public static void main(String[] args) {

     

//   create a list of Comparable type

     

      List<Comparable> list = new ArrayList<>();

     

//       add elements

     

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

     

//       print list

     

      System.out.println("\nList : "+list);

     

//       test search method

     

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

     

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

     

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

     

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

     

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

     

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

     

  }

 

 

 

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

     

//       if list is empty or parameter is null the throw IllegalArgumentException

     

      if(list.isEmpty() || par == null ) {

         

          throw new IllegalArgumentException();

         

      }

     

//       binary search

     

//       declare variables

     

      int start=0;

     

      int end =list.size()-1;

     

//       using while loop

     

      while(start<=end) {

         

//           mid element

         

          int mid =(start+end)/2;

         

//           if par equal to mid element then return

         

          if(list.get(mid).equals(par) )

          {

              return true ;

             

          }  

         

//           if mid is less than parameter

         

          else if (list.get(mid).compareTo(par) < 0 ) {

                 

              start=mid+1;

          }

         

//           if mid is greater than parameter

         

          else {

              end=mid-1;

          }

      }

     

//       if not found then retuen false

     

      return false;

     

  }

 

 

}import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

 

//   main method

  public static void main(String[] args) {

     

//   create a list of Comparable type

     

      List<Comparable> list = new ArrayList<>();

     

//       add elements

     

      list.add(1);

      list.add(2);

      list.add(3);

      list.add(4);

      list.add(5);

      list.add(6);

      list.add(7);

     

//       print list

     

      System.out.println("\nList : "+list);

     

//       test search method

     

      Comparable a = 7;

      System.out.println("\nSearch for 7 : "+search(list,a));

     

      Comparable b = 3;

      System.out.println("\nSearch for 3 : "+search(list,b));

     

      Comparable c = 9;

      System.out.println("\nSearch for 9 : "+search(list,c));

     

      Comparable d = 1;

      System.out.println("\nSearch for 1 : "+search(list,d));

     

      Comparable e = 12;

      System.out.println("\nSearch for 12 : "+search(list,e));

     

      Comparable f = 0;

      System.out.println("\nSearch for 0 : "+search(list,f));

     

  }

 

 

 

//   static method named search takes arguments Comparable list and Comparable parameter

  public static boolean search(List<Comparable> list, Comparable par) {

     

//       if list is empty or parameter is null the throw IllegalArgumentException

     

      if(list.isEmpty() || par == null ) {

         

          throw new IllegalArgumentException();

         

      }

     

//       binary search

     

//       declare variables

     

      int start=0;

     

      int end =list.size()-1;

     

//       using while loop

     

      while(start<=end) {

         

//           mid element

         

          int mid =(start+end)/2;

         

//           if par equal to mid element then return

         

          if(list.get(mid).equals(par) )

          {

              return true ;

             

          }  

         

//           if mid is less than parameter

         

          else if (list.get(mid).compareTo(par) < 0 ) {

                 

              start=mid+1;

          }

         

//           if mid is greater than parameter

         

          else {

              end=mid-1;

          }

      }

     

//       if not found then retuen false

     

      return false;

     

  }

 

 

}

7 0
2 years ago
Pendant Publishing edits multi-volume manuscripts for many authors. For each volume, they want a label that contains the author’
tigry1 [53]

Answer:

The answer to this question can be described as follows:

Explanation:

The Pseudocode to this question can be described as follows:

start  \\ start process

Declaration of variable

Input_File records.txt

Output_File result.txt

string author_Name , title

int number_of_volume

open Input_File

open Output_File

input author_Name, title, number_of_volume from records.txt

while not eof \\ start loop and eof is end of file  

output author_Name

output title

output number_of_volume

end while  \\end loop

close records.txt  \\close file

close results.txt   \\close file

END \\end the code

8 0
3 years ago
To provide both encryption and integrity protection, wpa2 uses aes encryption with:
lara [203]

WPA 2 stands for Wi-Fi Protected Access, version 2. It is the latest protocol that effectively protects 802.11 wireless traffic across a LAN. It belongs to the 802.11 protocol.

To provide both encryption and integrity protection, wpa2 uses aes encryption with Counter and CBC MAC (CCM) Mode.

3 0
2 years ago
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