Consider a one-way authentication technique based on asymmetric encryption: A --> B: IDA B --> A: R1 A --> B: E(PRa, R1) Explain the protocol. What type of attack is this protocol susceptible to?

Answer:

Explanation:

Well here you only have to add 2 new horses .... so simply what you can do is find 2 slots where the gap is maximum ... lets take the given example in detail to understand the same.... here 10001001000010 here you can observe that the difference between the first two filled slots is 3 then 2 then 4 then 1....... make it a separate array ...... now as i said that we just need to place two new horses so we just need two of the biggest numbers from this array and place the new horses at the center of the corresponding slots ... like here 3 and 4 are greatest ... so one horse is placed in between the gap corresponding to 3  while other at 4 ..... after making the new array answer can be find in linear time itself... I hope this makes sense ....

Now lets understand how to program it easily so firstly after scanning all stuff ..... we will make a gap array .... then just think now what we will do is find the greatest two numbers then again putting the horses and then find our answers .... so now an easy approach is that instead of making the string again and again searching for answer .... what we do is just do half of the two greatest ones and again find the greatest one in array....

Answer:

1. The average turn around time is 21.2 minute using round robin scheduling algorithm with a quantum time of 1 minute.

2. The average turn around time is 20 minute using Priority scheduling algorithm.

3. The average turn around time is 19.2 minute using First Come First Serve scheduling algorithm.

4. The average turn around time is 14 minute using Shortest Job First scheduling algorithm.

Explanation:

Gantt chart was used to solve the scheduling problem.

Image showing the Gantt chart and working solution is attached.

Round Robin scheduling algorithm is a type of scheduling that uses a time slice for each process/job. Once the time slice is complete, the process leave to join the queue again if it has not finished execution.

Priority scheduling algorithm is a type of scheduling algorithm that allows a process/job to execute/complete based on the order of priority. In this case, the higher the number, the higher the priority. With 5 having the highest priority.

First Come First Serve scheduling algorithm is a type of scheduling algorithm that allows execution of a process/job based on first come first server i.e based on order of arrival time. In this case the order is ABCDE.

Shortest Job First scheduling algorithm is a type of scheduling algorithm that allocate the CPU to the process/job having the shortest completion time first. In this case, Job C has the shortest completion time.

Answer:

Explanation:

Firstly, let's create a method swapPairs that take one ArrayList (Line 18). In the method, use a for-loop to traverse through each item in the ArrayList and swap the items between the current items at index-i and at index-i+1 (Line 19-22). The index-i is incremented by two in next loop and therefore the next swapping will proceed with third and fourth items and so forth.

In the main program, create a sample ArrayList (Line 5-12) and then test the method (Line 14) and print the output (Line 15). We shall get  [be, to, not, or, be, to, hamlet].

Answer:

168

Explanation:

=(109-103)+(120-109)+(123-120)+(151-123)+(198-151)+(87-74)+(74-32)+(32-23)+(23-19)+(19-14)+(14-10)+(10-5)+(1-0)

=6+11+3+28+47+3+42+9+4+5+4+5+1

=168

Answer:

Check the explanation

Explanation:

import java.util.Scanner;

public class DuplicateElimination {

   public static void main(String[] args) {

       Scanner scanner = new Scanner(System.in);

       int arr[] = new int[5];

       int size = 0, n;

       for(int i = 0;i<5;i++){

           System.out.print("Enter an integer between 10 and 100:");

           n = scanner.nextInt();

           if(size==0 || elemanVarmı(arr,size,n)){

               System.out.println("This is the first time "+n+" has been entered");

               arr[size] = n;

               size++;

           }

       }

       System.out.println("The complete set of unique values entered is");

       for(int i = 0;i<size;i++){

           System.out.println("Unique Value "+(i+1)+": is "+arr[i]);

       }

   }

   private static boolean elemanVarmı(int dizi[], int size, int n)

   {

       for (int i = 0; i < size; i++)

       {

           if(dizi[i]==n)

           {

               return false;

           }

       }

       return true;

   }

}

Final answer:

The question is about designing a programming application that takes in up to five integers between 10 and 100 and displays each unique number entered by the user. The program maintains a small array to store the unique numbers and possibly includes functionality to determine the minimum and maximum values from the input, as well as error handling for invalid inputs.

Explanation:

Programming Unique Values Application

The application you're describing involves creating a program that accepts five integers between 10 and 100, inclusive. Each number entered by the user should be checked against the existing list of input numbers to determine if it is a duplicate. If it is not a duplicate, it should then be displayed and added to the list of unique values. To handle this scenario, the program could use an array with a length that corresponds to the maximum number of unique integers, which in the worst case would be five. As the user inputs each number, the program should check if the number is already present in the array. If it isn't, the number should be added to the array, and the current list of unique numbers should be displayed. This helps in avoiding the storage of duplicate numbers and efficiently manages the memory by using the smallest array necessary.

To extend the functionality, as indicated in Exercise 5.2, the same logical structure can be applied to also keep track and display the maximum and minimum value entered by the user at the end of the input sequence. We could also implement error handling to provide feedback if invalid input is given, similar to the approach outlined in Exercise 5.1 for detecting user input mistakes. This kind of functionality often involves iterative loops, array manipulation, and conditional statements to handle the logic.

Answer:

Explanation:

check below for the answer in d attached file

Answer:

import java.util.Scanner;

public class num11 {

   public static void main(String[] args) {

   Scanner in = new Scanner(System.in);

       System.out.println("Enter a word or phrase");

       String word = in.nextLine();

       //Calling the methods

       System.out.println("Total vowels in "+word+" are "+numVowels(word));

       System.out.println("Total consonants in "+word+" are "+numConsonants(word));

   }

   public static int numVowels(String word){

       //Remove spaces and convert to lowercase

       //Assume that only correct character a-z are entered

       String newWord = word.toLowerCase().replaceAll(" ","");

       int vowelCount = 0;

       for(int i = 0; i <= newWord.length()-1; i++){

           if(newWord.charAt(i)=='a'||newWord.charAt(i)=='e'||newWord.charAt(i)=='i'

                   ||newWord.charAt(i)=='o'||newWord.charAt(i)=='u'){

               vowelCount++;

           }

       }

       return vowelCount;

   }

   public static int numConsonants(String word){

       //Remove spaces and convert to lowercase

       //Assume that only correct character a-z are entered

       String newWord = word.toLowerCase().replaceAll(" ","");

       int consonantCount = 0;

       //Substract total vowels from the length of the word to get the consonants

       consonantCount = newWord.length()-numVowels(word);

       return consonantCount;

   }

}

Explanation:

Answer:

Is in the provided screenshot

Explanation:

Get the last two characters and concatenate.

Final answer:

To solve the task, a method in Java is written that appends the last two characters of the input string three additional times to the end, using the substring method.

Explanation:

The task is to write a method in Java that takes a given string and appends the last two characters of that string three more times onto the end of it. This operation modifies the original string by repeatedly adding a substring that consists of its last two characters.

Method Implementation:

Here is how you can write the method:

public static String repeatTwo(String word) {
   if (word == null || word.length() < 2) {
       return word; // handle null and short strings
   }
   String lastTwoChars = word.substring(word.length() - 2);
   return word + lastTwoChars + lastTwoChars + lastTwoChars;
}

This method first checks if the input string is null or shorter than two characters. If it isn't, it proceeds to use the substring method to extract the last two characters and concatenates this substring three times to the end of the original string.

Answer:

Here is the program:

public class DoubleValue

{ public static void main(String[] args) { //start of the main function

 double sum = 0.0;  // initialize double type sum by 0.0

//for statement to sum the double values passed by the command-line //arguments

       for (String str : args)        {  

         sum += Double.parseDouble(str);  

//returns double representation of the passed str argument and takes the //sum of these double values         }

       System.out.printf("The sum of the double values passed in from the command line is %.1f\n", sum);  }  }      

//prints the above message with output 0.0                                  

Explanation:

The above program has a static method parseDouble of class Double to convert a string str to a double value and the for statement is used to sum the double values passed by command line arguments. This sum is stored in sum variable and is displayed in the output. The program along with its output is attached.

If you want to get the input from the user you can use the following code. Just use the Scanner class to take input from the user.

import java.util.Scanner;

public class DoubleValue

{ public static void main(String[] args) {

//two string type variables

 String str1;

               String str2;

//scans and reads input from user

Scanner input = new Scanner(System.in);

System.out.println("Enter value of str1: ");

str1 = input.next();

System.out.println("Enter value of str2: ");

str2 = input.next();

double sum = 0.0;

//sum double values and parseDouble function is used to convert string to //double value

sum += Double.parseDouble(str1) + Double.parseDouble(str2);

System.out.printf("The sum of the double values passed in from the command line is %.1f\n", sum); } }  //prints result of the sum

Final answer:

The code provided illustrates how to sum double values passed through command-line arguments in Java, using Double.parseDouble and an enhanced for loop, demonstrating important concepts in parsing and arithmetic operations in programming.

Explanation:

The question involves writing a Java application to sum double values passed through command-line arguments using an enhanced for loop, illustrating a practical application in parsing and arithmetic operations in programming. This task requires knowledge of Java's Double.parseDouble method and the enhanced for loop. To accomplish this, a main method is created where the command-line arguments are accessed. Each argument, which is initially a String, is converted to a double using Double.parseDouble, and then summed in an enhanced for loop. Here is a simple implementation:

public class SumDoubles {
   public static void main(String[] args) {
       double sum = 0.0;
       for (String arg : args) {
           sum += Double.parseDouble(arg);
       }
       System.out.println("The sum of the double values passed in from the command line is " + sum);
   }
}

This code snippet effectively addresses the task by converting command-line arguments from String to double and summing them. It demonstrates the use of command-line arguments, Double.parseDouble, and an enhanced for loop in Java.

Answer:

Explanation:

Note: Alphabets refers to respective problem in question

Program:

#include<stdio.h>

#define SIZE 25

int main()

{

  double numfrc[SIZE];                                   // (a)

  for(int i=0;i<SIZE;i++)                                   // (b)

      numfrc[i]=0;

  numfrc[13]=6.666;                                       // (c)

  numfrc[14]=-6.666;                                       // (d)

  numfrc[9]=1.667;                                       // (e)

  numfrc[6]=3.333;                                       // (f)

  for(int i=14;i>=9;i--)

      printf("%2d ---> %.2lf\n",i,numfrc[i]);               // (g)

  printf("\n");

  for(int i=0;i<SIZE;i++)

      printf("%2d ---> %.2lf\n",i,numfrc[i]);               // (h)

}

Output:

Answer:

Kindly check explaination for code

Explanation:

/* Java program for Merge Sort */

class MergeSort

{

// Merges two subarrays of arr[].

// First subarray is arr[l..m]

// Second subarray is arr[m+1..r]

void merge(int arr[], int l, int m, int r)

{

// Find sizes of two subarrays to be merged

int n1 = m - l + 1;

int n2 = r - m;

/* Create temp arrays */

int L[] = new int [n1];

int R[] = new int [n2];

/*Copy data to temp arrays*/

for (int i=0; i<n1; ++i)

L[i] = arr[l + i];

for (int j=0; j<n2; ++j)

R[j] = arr[m + 1+ j];

/* Merge the temp arrays */

// Initial indexes of first and second subarrays

int i = 0, j = 0;

// Initial index of merged subarry array

int k = l;

while (i < n1 && j < n2)

{

if (L[i] <= R[j])

{

arr[k] = L[i];

i++;

}

else

{

arr[k] = R[j];

j++;

}

k++;

}

/* Copy remaining elements of L[] if any */

while (i < n1)

{

arr[k] = L[i];

i++;

k++;

}

/* Copy remaining elements of R[] if any */

while (j < n2)

{

arr[k] = R[j];

j++;

k++;

}

}

// Main function that sorts arr[l..r] using

// merge()

void sort(int arr[], int l, int r)

{

if (l < r)

{

// Find the middle point

int m = (l+r)/2;

// Sort first and second halves

sort(arr, l, m);

sort(arr , m+1, r);

// Merge the sorted halves

merge(arr, l, m, r);

}

}

/* A utility function to print array of size n */

static void printArray(int arr[])

{

int n = arr.length;

for (int i=0; i<n; ++i)

System.out.print(arr[i] + " ");

System.out.println();

}

// Driver method

public static void main(String args[])

{

int arr[] = {12, 11, 13, 5, 6, 7};

System.out.println("Given Array");

printArray(arr);

MergeSort ob = new MergeSort();

ob.sort(arr, 0, arr.length-1);

System.out.println("\nSorted array");

printArray(arr);

}

}

Answer:

Check the explanation

Explanation:

/* Java program for Merge Sort */

class MergeSort

{

// Merges two subarrays of arr[].

// First subarray is arr[l..m]

// Second subarray is arr[m+1..r]

void merge(int arr[], int l, int m, int r)

{

// Find sizes of two subarrays to be merged

int n1 = m - l + 1;

int n2 = r - m;

/* Create temp arrays */

int L[] = new int [n1];

int R[] = new int [n2];

/*Copy data to temp arrays*/

for (int i=0; i<n1; ++i)

L[i] = arr[l + i];

for (int j=0; j<n2; ++j)

R[j] = arr[m + 1+ j];

/* Merge the temp arrays */

// Initial indexes of first and second subarrays

int i = 0, j = 0;

// Initial index of merged subarry array

int k = l;

while (i < n1 && j < n2)

{

if (L[i] <= R[j])

{

arr[k] = L[i];

i++;

}

else

{

arr[k] = R[j];

j++;

}

k++;

}

/* Copy remaining elements of L[] if any */

while (i < n1)

{

arr[k] = L[i];

i++;

k++;

}

/* Copy remaining elements of R[] if any */

while (j < n2)

{

arr[k] = R[j];

j++;

k++;

}

}

// Main function that sorts arr[l..r] using

// merge()

void sort(int arr[], int l, int r)

{

if (l < r)

{

// Find the middle point

int m = (l+r)/2;

// Sort first and second halves

sort(arr, l, m);

sort(arr , m+1, r);

// Merge the sorted halves

merge(arr, l, m, r);

}

}

/* A utility function to print array of size n */

static void printArray(int arr[])

{

int n = arr.length;

for (int i=0; i<n; ++i)

System.out.print(arr[i] + " ");

System.out.println();

}

// Driver method

public static void main(String args[])

{

int arr[] = {12, 11, 13, 5, 6, 7};

System.out.println("Given Array");

printArray(arr);

MergeSort ob = new MergeSort();

ob.sort(arr, 0, arr.length-1);

System.out.println("\nSorted array");

printArray(arr);

}

}

/* This code is contributed by Rajat Maurya */

Answer:

FCFS Scheduling Algorithm:-

// C++ program for implementation of FCFS

// scheduling

#include<iostream>

using namespace std;

// Function to find the waiting time for all

// processes

void findWaitingTime(int processes[], int n,

int bt[], int wt[])

{

// waiting time for first process is 0

wt[0] = 0;

// calculating waiting time

for (int i = 1; i < n ; i++ )

wt[i] = bt[i-1] + wt[i-1] ;

}

// Function to calculate turn around time

void findTurnAroundTime( int processes[], int n,

int bt[], int wt[], int tat[])

{

// calculating turnaround time by adding

// bt[i] + wt[i]

for (int i = 0; i < n ; i++)

tat[i] = bt[i] + wt[i];

}

//Function to calculate average time

void findavgTime( int processes[], int n, int bt[])

{

int wt[n], tat[n], total_wt = 0, total_tat = 0;

//Function to find waiting time of all processes

findWaitingTime(processes, n, bt, wt);

//Function to find turn around time for all processes

findTurnAroundTime(processes, n, bt, wt, tat);

//Display processes along with all details

cout << "Processes "<< " Burst time "

<< " Waiting time " << " Turn around time\n";

// Calculate total waiting time and total turn

// around time

for (int i=0; i<n; i++)

{

total_wt = total_wt + wt[i];

total_tat = total_tat + tat[i];

cout << " " << i+1 << "\t\t" << bt[i] <<"\t "

<< wt[i] <<"\t\t " << tat[i] <<endl;

}

cout << "Average waiting time = "

<< (float)total_wt / (float)n;

cout << "\nAverage turn around time = "

<< (float)total_tat / (float)n;

}

// Driver code

int main()

{

//process id's

int processes[] = { 1, 2, 3};

int n = sizeof processes / sizeof processes[0];

//Burst time of all processes

int burst_time[] = {10, 5, 8};

findavgTime(processes, n, burst_time);

return 0;

}

Explanation:

See output

Answer:

see explaination and attachment

Explanation:

5.

To convert any byte address, By, to word address, Wo, first divide By by No, the no. of bytes/word, and ignores the remainder. To calculate a byte offset, O, in word, calculate the remainder of By divided by No.

i) 0 : word address = 0/8 = 0 and offset, O = 0 mod 8 = 0

ii) 9 : word address = 9/8 = 1 and offset, O = 9 mod 8 = 1

iii) 27 : word address = 27/8 = 3 and offset, O = 27 mod 8 = 3

iv) 31 : word address = 31/8 = 3 and offset, O = 31 mod 8 = 7

v) 120 : word address = 120/8 = 15 and offset, O = 120 mod 8 = 0

vi) 256 :word address = 256/8 = 32 and offset, O = 256 mod 8 = 0

6. see attachment for the python programming screen shot and output

The word address is the integer division of the physical address by 8, while the offset is the remainder of the physical address divided by 8.

(a) Calculate the word address and offset

The number of bit is given as 64.

So, we have:

[tex]2^n = 64[/tex]

Express 64 as a base of 2

[tex]2^n = 2^8[/tex]

Cancel out the base of 2

[tex]n = 8[/tex]

So, we have:

(i) Physical address: 0

(ii) Physical address: 9

(iii) Physical address: 27

(iv) Physical address: 31

(v) Physical address: 120

(v) Physical address: 256

(b) The program

The program written in Python, where comments are used to explain each line is as follows:

#This creates a list of physical address

pA = [0, 9, 27, 31,120, 256]

#This iterates through the list

for i in range(len(pA)):

   #This prints the word address

   print("Word Address:",(pA[i]//8))

   #This prints the offset

   print("Offset:",(pA[i]%8))

Read more about word address at:

https://brainly.com/question/17493537

Answer:

Answer is provided in the attached screenshot

Explanation:

The character in the 2nd position will be always be the one we need to check. We then check if that character is a vowel, and replace the string as required.

Final answer:

To modify a phrase that starts with 'A' followed by a word that starts with a vowel by adding an 'n', one must write a Java function that checks the following character and adjusts the string accordingly.

Explanation:

The student's question pertains to a programming exercise in Java where the user needs to modify a string based on whether the word following an 'A' begins with a vowel or not. To solve this, you would need to check the character after the 'A ' for a vowel. If it's a vowel, an 'n' should be added to form 'An'. The Java function would use conditionals to check this and then return the modified or original phrase accordingly.

Here is an example of how the grammar function can be implemented:

public String grammar(String phrase) {
   if(phrase.length() > 2 && phrase.substring(0, 2).equalsIgnoreCase("A ")) {
       char nextChar = phrase.charAt(2);
       if(nextChar == 'a' || nextChar == 'e' || nextChar == 'i' || nextChar == 'o' || nextChar == 'u') {
           return "An " + phrase.substring(2);
       }
   }
   return phrase;
}

Answer:

c. The pivot could be either 7 or 9.

Explanation:

Since we are trying to sort an array of eight integers using quick sort, from the first partitioning it shows that the pivot or the central point can either be 7 or 9. When you look at the array, it is only 7 and 9 that are placed correctly in the sorted array. Every element to the left of 7 and 9 are smaller and every element on the right of 7 and 9 are integers higher than them. Hence this shows that the pivot lies between 7 or 9.

The pivot element of an array is the middle element of the array.

The pivot element could be either 7 or 9

From the question, we have:

[tex]\mathbf{n = 8}[/tex]

So, the pivot element is:

[tex]\mathbf{Pivot = \frac{n +1}{2}}[/tex]

So, we have:

[tex]\mathbf{Pivot = \frac{8 +1}{2}}[/tex]

[tex]\mathbf{Pivot = \frac{9}{2}}[/tex]

Divide 9 by 2

[tex]\mathbf{Pivot = 4.5}[/tex]

So, the pivot element is the 4.5th element.

The 4.5th element is between the 4th and 5th element (i.e. 7 or 9)

This means that, the pivot element could be either 7 or 9

Hence, the correct option is (c).

Read more about pivot elements at:

https://brainly.com/question/16260051

Answer:

See explaination

Explanation:

/File: ValidateCheckDigits.java

import java.io.BufferedWriter;

import java.io.File;

import java.io.FileWriter;

import java.io.IOException;

import java.util.Scanner;

public class ValidateCheckDigits {

public static void main(String[] args) {

String fileName = "numbers.txt"; //input file name

String validFileName = "valid_numbers.txt"; // file name of the output file

//Open the input file for reading and output file for writing

try(Scanner fileScanner = new Scanner( new File(fileName));

BufferedWriter fileWriter = new BufferedWriter( new FileWriter(validFileName))) {

//Until we have lines to be read in the input file

while(fileScanner.hasNextLine()) {

//Read each line

String line = fileScanner.nextLine().trim();

//calculate the sum of first 5 digits

int sum = 0;

for(int i = 0; i < 5; i++) {

sum += Integer.parseInt( line.charAt(i) + "");

}

//Get the last digit

int lastDigit = Integer.parseInt(line.charAt(5)+"");

//Check if the remainder of the sum when divided by 10 is equal to last digit

if(sum%10 == lastDigit) {

//write the number in each line

fileWriter.write(line);

fileWriter.newLine();

System.out.println("Writing valid number to file: " + line);

}

}

System.out.println("Verifying valid numbers completed.");

} catch(IOException ex ) {

System.err.println("Error: Unable to read or write to the file. Check if the input file exists in the same directory.");

}

}

}

Check attachment for output and screenshot

Final answer:

The process involves reading six-digit account numbers, summing the first five digits, and validating if the last digit matches the remainder of the sum modulo 10, outputting only the verified numbers to a file.

Explanation:

The goal is to validate six-digit account numbers to ensure their authenticity. The last digit of each account number must match the remainder when the sum of the first five digits is divided by 10. This concept utilizes simple mathematical operations like addition and modulo in conjunction with programming to accomplish the task of validation. Here's an outline of the procedure:

This process highlights the importance of validating data and using programming to automate otherwise time-consuming tasks.

Answer:

clc

clear

size = input('Enter the number of boxes in a row : ');

for i=1:size

for j=1:size

if i==size-1 || j==2

rectangle('Position',[i*size,j*size,size,size],'FaceColor',[0,1,1],'EdgeColor','black',...

'LineWidth',3)

else

rectangle('Position',[i*size,j*size,size,size],'FaceColor',[1 1 1],'EdgeColor','black',...

'LineWidth',3)

end

end

end

Explanation:

Answer:

-1.75

Explanation:

kindly refer to attachment for a detailed step by step solution to the problem.

Answer:

import java.util.Scanner;

public class num8 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter month's budget");

       double monthBudget = in.nextDouble();

       double totalExpenses = 0.0;

       double n;

       do{

           System.out.println("Enter expenses Enter zero to stop");

           n = in.nextDouble();

           totalExpenses += n;

       }while(n>0);

       System.out.println("Total expenses is "+totalExpenses);

System.out.println("The amount over your budget is "+ Math.abs(monthBudget-totalExpenses));

   }

}

Explanation:

Answer:

import java.util.Scanner;

public class num6 {

   static double balance = 10000.5; //Current Balance

   //The main Method

   public static void main(String[] args) {

       //Request User Input

       System.out.println("Enter 1 for deposit \nEnter 2 for Withdraw\nEnter 3 for Balance Enquiry\n0 to quit ");

       Scanner in = new Scanner(System.in);

       int op = in.nextInt();

       //Check to determine the Operation the user wants

       if(op == 1){

           System.out.println("Enter amount to deposit");

           double depositeAmount = in.nextDouble();

           deposit(depositeAmount); //Call the deposit method

       }

       if(op == 2){

           System.out.println("Enter amount to Withdraw");

           double withdrawAmount = in.nextDouble();

           //validate the withdraw amount

           if(withdrawAmount > balance){

               System.out.println("Insuffucient fund");

               System.out.println("Your current balance is "+balance);

           }

           else {

               withdraw(withdrawAmount);//Call withdraw Method

           }

           }

       if(op == 3){

           balanceEnquiry();

       }

   }

   //Deposit Method

   public static double deposit(double depositAmount){

       double newBalance = balance + depositAmount;

       System.out.println("You deposited "+depositAmount+"Your new balance is "+newBalance);

       return newBalance;

   }

   //Withdraw Method

   public static double withdraw(double withdrawAmount){

           double newBalance = balance - withdrawAmount;

       System.out.println("You Withdrew "+withdrawAmount+"Your new balance is "+newBalance);

           if(newBalance < 100){

           System.out.println("Your Balance is low");

       }

       return  newBalance;

   }

   //Balance Enquiry Method

   public static void balanceEnquiry(){

       System.out.println("Your current balance is "+balance);

   }

}

Explanation:

This is solved in Java programming language

See code explation provided as comments

Final answer:

Iris should ask Charlie and Kelvin thorough questions about the job's responsibilities, the team dynamics, and the company's future. Reflecting on past decisions with an understanding of cognitive bias leads to better outcomes. It's essential to commit to ongoing critical reflection to avoid future biases in decision-making.

Explanation:

Discussion Questions for Iris

When considering questions that Iris should ask Charlie, she should be inquisitive about the specifics of the new job, the dynamics within Kelvin's team, and the future of the company. This includes asking about the job responsibilities, the team's current projects and goals, the team's working style and culture, any potential for advancement, and the company's vision for the future and stability.

Questions for Kelvin

Regarding questions for Kelvin, Iris could probe deeper into how the role fits into the company's larger strategy, how success will be measured for the new position, what the expectations are during the first few months, and how Kelvin envisions the position evolving. These questions would provide Iris with a comprehensive understanding of how her potential role manifests within the company's framework.

Hypothetical Scenario as Kelvin

If placed in Kelvin's position, to prepare for the meeting, I would ensure that I clearly understood the objectives for the new role, prepared a structured plan outlining how I would integrate the new member into the team, and had a thorough understanding of the company's future plans to address any inquiries. Reflecting on past decisions through the lens of cognitive bias would help mitigate any irrational elements involved in decision-making.

Impact of Different Behaviors

Had a different approach been taken, such as being more critical or self-aware, the consequences may have included more informed decision-making and reduced chances of miscommunication. Understanding the cognitive bias and employing critical reflection and metacognition could have led to better outcomes and a healthier decision-making process.

Conclusion and Lessons Learned

In conclusion, reflecting on experiences and understanding cognitive biases can profoundly impact personal growth and decision-making. This level of self-awareness can lead to improved choices in the future and help avoid repeating the same mistakes. Committing to ongoing critical reflection and seeking diverse perspectives are steps one can take today to prevent cognitive bias from clouding judgment.

Answer:

See explaination for program code

Explanation:

Java code below:

Musician.java :

public class Musician {

private String name;

private int weeksInTop40;

private int albumsSold;

private boolean isPlatinum;

public Musician(String name, int weeksInTop40, int albumsSold) {

this.name = name;

this.weeksInTop40 = weeksInTop40;

this.albumsSold = albumsSold;

setPlatinum(albumsSold);

}

public void setPlatinum(int albumsSold) {

if(albumsSold >= 1000000) {

isPlatinum = true;

}else {

isPlatinum = false;

}

}

public int getWeeksInTop40() {

return this.weeksInTop40;

}

public String getName() {

return this.name;

}

public boolean getIsPlatinum() {

return this.isPlatinum;

}

public String toString() {

return this.name;

}

}

Billboard.java :

import java.util.ArrayList;

public class Billboard {

private ArrayList<Musician> top10 = new ArrayList<Musician>();

public void add(Musician newMusician) {

if(this.top10.size()<10 && newMusician.getIsPlatinum()==true) {

/**

* Less than 10 musician are present in the list currently

* And also the newMusician is platinum

* */

this.top10.add(newMusician);

}else {

if(newMusician.getIsPlatinum()==false) {

System.out.println("The new musician "+newMusician+" cannot be added because not enough records are sold");

}else {

/**

* We need to call the replace method

* */

replace(newMusician);

}

}

}

public void replace(Musician newMusician) {

/**

* First we need to find what is the lowest number of weeks

* an musician was there in top40.

* */

int lowestOnTop40 = Integer.MAX_VALUE;

int index=-1;

for(int i=0;i<this.top10.size();i++) {

if(this.top10.get(i).getWeeksInTop40()<lowestOnTop40) {

lowestOnTop40 = this.top10.get(i).getWeeksInTop40();

index = i;

}

}

/**

* We have at what position the lowest number of week on top value in variable

* lowestOnTop40 at index

* We need to compare this value with newMusician's number of weeks on top 40

* */

if(lowestOnTop40 < newMusician.getWeeksInTop40()) {

System.out.println("The old musician "+this.top10.get(index)+" has been replaced by new musician "+newMusician);

this.top10.set(index, newMusician);

}else {

System.out.println("The new musician "+newMusician+" cannot be added because not enough weeks on top40.");

}

}

public void printTop10() {

System.out.println(top10);

}

}

BillboardTester.java :

public class BillboardTester {

public static void main(String[] args) {

Billboard top10 = new Billboard();

top10.add(new Musician("Beyonce", 316, 100000000));

top10.add(new Musician("The Beatles", 365, 600000000));

top10.add(new Musician("Drake", 425, 150000000));

top10.add(new Musician("Pink Floyd", 34, 250000000));

top10.add(new Musician("Mariah Carey", 287, 200000000));

top10.add(new Musician("Rihanna", 688, 250000000));

top10.add(new Musician("Queen", 327, 170000000));

top10.add(new Musician("Ed Sheeran", 536, 150000000));

top10.add(new Musician("Katy Perry", 317, 143000000));

top10.add(new Musician("Justin Bieber", 398, 140000000));

//This musician should not be added to the top10 because they don't have enough records sold

top10.add(new Musician("Karel the Dog", 332, 60));

//This musician should replace the artist

top10.add(new Musician("Tracy the Turtle", 332, 150000000));

//This musician should not replace an artist, but is a Platinum artist

top10.add(new Musician("Alex Eacker", 100, 23400000));

top10.printTop10();

}

}

Answer:

read a FallStudents record

Explanation:

When data is input from a file, the data is processed and then placed in the output file. In order to verify the validity of this process, the output file is checked.

After comparing data in the FallStudents and SpringStudents files, and placing in sorted order by StudentID. The output data is placed in file FallStudents record. After the complete process is done, the output/file record which is the FallStudents record is read.

Answer:

Check the explanation

Explanation:

# List to store text values

grid=list()

# LOOP FOR 3 ROWS

for i in range(3):

   # LIST TO STORE COLUMN

   row = list()

   # LOOP FOR 3 COLUMNS

   for j in range(3):

       # CREATING TEXT WITH

       # COORDINATES FORMAT

       text = "({},{})".format(i,j)

       # APPEND TEXT TO ROW

       row.append(text)

   # APPEND ROW TO GRID

   grid.append(row)

# AFTER THE ABOVE LOOP

# GRID WILL BE CREATED IN TO

# 3 X 3 GRID

# PRINTING GRID

# FOR ALL ROWS

for i in grid:

   # FOR ALL COLUMNS

   for j in i:

       # PRINTING VALUE AT

       # ROW I, COLUMN J

       print(j,end=" ")

   print()

Kindly check the screenshot and output below.

A smart speaker has no screen or keypad for changing its settings.

The steps we will follow to configure the speaker are

Explanation:

A smart speaker has no screen or keypad for changing its settings.The steps we will follow to configure the speaker are

Thus above mentioned is the order of the three steps that we need to follow in order to configure the speaker.

To configure a smart speaker without a screen or keypad, you need to connect your smartphone to the speaker's Wi-Fi hotspot, download the speaker's app, and enter the password to your home Wi-Fi network.

To configure a smart speaker without a screen or keypad, follow these three steps:

Answer:

1. Get the number

2. Declare a variable to store the sum and set it to 0

3. Repeat the next two steps till the number is not 0

4. Get the rightmost digit of the number with help of remainder ‘%’ operator by dividing it with 10 and add it to sum.

5. Divide the number by 10 with help of ‘/’ operator

6. Print or return the sum

# include<iostream>

using namespace std;

/* Function to get sum of digits */

class gfg

{

   public:

   int getSum(float n)

   {

   float sum = 0 ;

   while (n != 0)

   {

    sum = sum + n % 10;

    n = n/10;

   }

return sum;

}

};

//driver code

int main()

{

gfg g;

float n = 687;  

cout<< g.getSum(n);

return 0;

}

Explanation:

Answer:

Each day, the professor observes whether the students sleep in class, and whether they have red eyes. The professor has the following domain theory: • The prior probability of getting enough sleep, with ... The probability of having red eyes if the student did not get enough sleep. • The probability of sleeping in class is if the ...

Explanation:

Answer:

.data

  str:   .asciiz   "Hello, world!\n"

  output:   .asciiz "\nTotal character numeric sum = "

.text

.globl main

main:

  la $t0,str       # Load the address of the string

  li $t2,0       # Initialize the total value to zero

loop:   lb $t1,($t0)       # Load one byte at a time from the string

  beqz $t1,end       # If byte is a null character end the operation

  add $t2,$t2,$t1       # Else add the numeric value of the byte to the running total

  addi $t0,$t0,1       # Increment the string pointer by 1

  b loop           # Repeat the loop

end:

  # Displaying the output

  li $v0,4       # Code to print the string

  la $a0,output       # Load the address of the string

  syscall           # Call to print the output string

  li $v0,1       # Code to print the integer

  move $a0,$t2       # Put the value to be printed in $a0

  syscall           # Call to print the integer

  # Exiting the program

  li $v0,10       # Code to exit program

  syscall           # Call to exit the program

Explanation:

Answer:

see explaination

Explanation:

Please find the screenprint and the VBA code. As shown in the screenprint, range B1 is used to enter the number of petals. If you want to use a different cell for petals input, then change the code accordingly.

Also, the VBA code is placed in the Sheet1 module. No new modules are inserted.

Screenprint: see attachment for screenshot

VBA Code:

Private Sub CommandButton1_Click()

Call daisyDecisions

End Sub

Private Sub daisyDecisions()

Dim remainder As Integer

Dim noOfPetals As Integer

noOfPetals = Sheet1.Range("B1").Value

remainder = noOfPetals Mod 2

If remainder <> 0 Then

MsgBox "He/She loves you!!!"

Else

MsgBox "He/She loves you not!!!"

End If

End Sub

Answer:

See the explanation for the answer.

Explanation:

Australian regulations makes extremely difficult for the enterprises to move organizations sensitive data to the cloud which is storing outside the Australian network. These are all managed by the Office of Australian Information Commissioner(OAIC) which provides oversight on the data privacy regulations designed to govern the dissemination of the sensitive information.

One rule they applied under this is The Australian National Privacy Act 1988 which tells how the organizations collect, use, secure, store the information. The National Privacy Principles in the Act tells how organizations should use the people's personal information. The NPP has a rule that An organization must take the responsibility to hold the information without misuse or modified by unauthorized access. They require enterprises to put security service level agreements with the cloud service providers that define audit rights, data location, access rights when there is cross border disclosure of information.

In later time they introduced a new principle called The Privacy Amendment Act 2012. This principle gives set of new rules along with the changes in the many rules in the previous act and also this is having a set of new principles those are called Australian Privacy Principles (APP).

In this there is principle for cross border disclosure of personal information which is APP8. This rule regulates the disclosure or transfer of personal information by an agency or company to a different location outside the country.

Before disclosure the information outside they have to take responsible steps that the company outside the Australia must not breach the APP 's.