Why were the practitioners of alternative software development methods not satisfied with the traditional waterfall method?

Answer:

Below is the JAVA program:

import java.util.Scanner;   //to take input from user

public class Main {  //name of the class

  public static void main(String[] args) {  //beginning of main method

      Scanner input = new Scanner(System.in);  //creates Scanner instance

      int integer = input.nextInt();  //defines and collects an integer for the number of words

      String stringArray[] = new String[integer]; //initializes an array to hold strings

      for (int i = 0; i < integer; i++) {  //iterates through the array

          stringArray[i] = input.next();         }  //collects strings

      int frequency[] = new int[integer];  //creates an array for holding frequencies

      int count;         //initializes variable to calculate frequency of each word

      for (int i = 0; i < frequency.length; i++) {  //iterates through frequency array

          count = 0;  //sets count to zero

          for (int j = 0; j < frequency.length; j++) {  //iterates through array

              if (stringArray[i].equals(stringArray[j])) {  //checks if element at ith index matches jth index

                  count++;                 }            }  //increments count

          frequency[i] = count;      }  //stores count in the frequency array

      for (int i = 0; i < stringArray.length; i++) {  //iterates through the string array

          System.out.println(stringArray[i] + " " + frequency[i]);         }    }    } //displays each word and its frequency

Explanation:

To illustrate the program, consider:

let integer = 3

for (int i = 0; i < integer; i++) is used to input strings into stringArray.

On the first pass:

i = 0

0<3

stringArray[i] = input.next(); takes a word and saves it to the ith index of stringArray. For example, if the user types "hey", it will be in stringArray[0].

Then, i increments to i = 1

On the second pass:

i = 1

1<3

stringArray[i] = input.next(); takes another word and assigns it to stringArray[1]. If the user enters "hi", then hi will be in stringArray[1]

Next, i becomes i = 2

On the third pass:

i = 2

2<3

stringArray[i] = input.next(); captures a word and places it in stringArray[2]. If the user types "hi", it goes to stringArray[2]

Then, i increments to i = 3

The loop terminates since i<integers is false.

The next outer loop for (int i = 0; i < frequency.length; i++) and inner loop for (int j = 0; j < frequency.length; will check each word and if (stringArray[i].equals(stringArray[j])) identifies any duplicates. Words that repeat will have their count incremented and the frequency array will record these values. For example, hey appears once while hi shows up twice, thus resulting in the final outputs:

hey 1

hi 2

hi 2

The visual representation of the program and its outcome based on the example provided is attached.

Answer:

The following C++ program is displayed below. No output is produced as per the requirements stated in the question.

#include <iostream>

using namespace std;

int main() {    

   // declaration and initialization of integer variables

   int k, memberID = 12, nMembers=5;

   bool isAMember;    

   // declaration and initialization of integer array

   int currentMembers[] = {12, 34, 56, 78, 90};    

   for(k=0; k<nMembers; k++)

   {

       if(memberID == currentMembers[k])

       {

           // if the member is found in the array, the loop exits using the break statement

           isAMember = true;

           break;

       }

       else

           isAMember = false;

   }    

   return 0;

}

Explanation:

The program starts with the declaration and initialization of integer variables, followed by the definition of an array that holds the identifiers of all members.

The Boolean variable is declared, although it isn't initialized at this point.

int k, memberID = 12, nMembers=5;

bool isAMember;

int currentMembers[] = {12, 34, 56, 78, 90};

Subsequently, the array containing the members' IDs is searched for the specific member ID. This action is performed using a for loop along with an if-else condition within that loop.

If the member ID exists in the array, the variable isAMember is set to true; otherwise, it is marked as false.

If isAMember is set to true, the break statement allows for the loop to be exited early.

for(k=0; k<nMembers; k++)

   {

       if(memberID == currentMembers[k])

       {

           isAMember = true;

           break;

       }

       else

           isAMember = false;

 }

The break statement is utilized since any other IDs in the array won't match the given member ID and thus, the variable isAMember would incorrectly be initialized to false even if the member ID is present in the array. Therefore, it's crucial to terminate the loop as soon as the matching member ID is located.

This program can be tested by utilizing different member ID values and varying the size of the array.

Answer:

c. The pivot could either be 7 or 9.

Explanation:

When sorting an array of eight integers through quicksort, the first partitioning indicates that either 7 or 9 may serve as the pivot. Observing the array, it is specifically 7 and 9 that occupy their correct positions within the ordered array. All integers preceding 7 and 9 are lesser, and all numbers following them are greater. Therefore, it suggests that the pivot is located between 7 and 9.

Answer:

import random

def name_change( name: list ) -> list :

   choice = random.choice( name )

   choice_index = name.index( choice )

   name.pop( choice_index )

   return name

Explanation:

The provided Python code utilizes the random library to randomly select an item from a list within the function "name_change". The index of the selected item is then retrieved from the list and removed from the original list called "name", with the modified name subsequently returned.

To activate the function, assign it to a variable, supply a list (it must be a list) as its argument, and then express the result.