#define DIRECTN 100
#define INDIRECT1 20
#define INDIRECT2 5
#define PTRBLOCKS 200

typedef struct {
filename[MAXFILELEN];
attributesType attributes; // file attributes
uint32 reference_count; // Number of hard links
uint64 size; // size of file
uint64 direct[DIRECTN]; // direct data blocks
uint64 indirect[INDIRECT1]; // single indirect blocks
uint64 indirect2[INDIRECT2]; // double indirect

} InodeType;

Single and double indirect inodes have the following structure:

typedef struct
{
uint64 block_ptr[PTRBLOCKS];
}
IndirectNodeType;

Required:
Assuming a block size of 0x1000 bytes, write pseudocode to return the block number associated with an offset of N bytes into the file.

Answer:

Check the explanation

Explanation:

SDLC

SDLC stands for Software Development Life Cycle

It provides steps for developing a software product

It also checks for the quality and correctness of a software

The main aim of SDLC is to develop a software, which meets a customer requirements

SDLC involves various phases to develop a high-quality product for its end user

Phases of SDLC

           There are seven phases in a software development life cycle as follows,

Requirement Analysis   Feasibility Study   Design    Coding    Testing Deployment      Maintenance

kindly check the attached image below

Requirement analysis

This is the first stage in a SDLC.

In this phase, a detailed and precise requirements are collected from various teams by senior team members

It gives a clear idea of the entire project scope, opportunities and upcoming issues in a project

This phase also involves, planning assurance requirements and analyze risk involved in the project

Here, the timeline to finish the project is finalized

Feasibility Study

In this stage, the SRS document, that is “Software Requirement Specification” document is defined, which includes everything to be designed and developed during the life cycle of a project

Design

In this phase as the name indicates, software and system design documents are prepared, based on the requirement specification document

This enable us to define the whole system architecture

Two types of design documents are prepared in this phase as follows,

High-Level Design

Low-Level Design

Coding

In this phase, the coding of project is done in a selected programming language

Tasks are divided into many units and given to various developers

It is the longest phase in the SDLC

Testing

In this stage, the developed coding is tested by the testing team

The testing team, checks the coding in the user point of view

Installation/Deployment

In this stage, a product is released by the company to the client

Maintenance

Once the product is delivered to the client, the maintenance phase work will start by fixing errors, modifying and enhancing the product.

Employee Payroll System

Requirement analysis

Gather information from various team for requirement

Analyze the number of employees to handle the project

Timeline to finish the project, eg.1 month

Feasibility Study

The system requirements such as how many systems are required

Decide which software to be installed to handle the project.

For example, if we are going to develop the software using C language, then software for that has to be installed

The SRS document has to be prepared

Design

In this the HLD and LLD is prepared, the overall software architecture is prepared

The modules involved in coding are decided

Here the modules can be employee, payroll manager, Employee Log time, account details

The input and output are designed, such as employee name is the input and output is the salary for him, based on working hours

Coding

Here the coding is divided into various sections and given to 2 or more employees

One may code employee detail, one will code working hours of employees and one may code the banking details of employee

Testing

The coding is tested for syntax, declaration, data types and various kinds of error

The testing team will test the coding with various possible values

They may even check with wrong values and analyze the output for that

Installation

Now the software is installed in the client system and the client is calculating the payroll for an employee based on working hours in a month

Maintenance

If any error occurs, the team will clear the issue.

When a new employee joins, then the employee data will added to the database and the database is updated

Also if the client asks for any new features, it will done in this phase.

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

Answer:

The formula used in this question is called the probability of combinations or combination formula.

Explanation:

Solution

Given that:

Formula applied is stated as follows:

nCr = no of ways to choose r objects from n objects

  = n!/(r!*(n-r)!)

The Data given :

Menu A : 5 appetizers and 3 main dishes

Menu B : 3 appetizers and 4 main dishes

Total appetizers - 6

Total main dishes - 5

Now,

Part A :

Total ways = No of ways to select menu A + no of ways to select menu B

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = 6C5*5C3 + 6C3*5C4

          = 6*10 + 20*5

          = 160

Part B :

Since, we can select the same number of appetizers/main dish again so the number of ways to select appetizers/main dishes will be = (total appetizers/main dishes)^(no of appetizers/main dishes to be selected)

Total ways = No of ways to select menu A + no of ways to select menu B  

          = (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

          = (6^5)*(5^3) + (6^3)*(5^4)

          = 7776*125 + 216*625

          = 1107000

Part C :

No of ways to select same appetizers and main dish for all the options

= No of ways to select menu A + no of ways to select menu B  

= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)

=(6*5) + (6*5)

= 60

Total ways = Part B - (same appetizers and main dish selected)      

= 1107000 - 60

= 1106940

Answer:

See explaination

Explanation:

//ReadFile.java

import java.io.File;

import java.io.FileNotFoundException;

import java.util.Scanner;

public class ReadFile

{

public static void main(String[] args)

{

int elementsCount=0;

//Create a File class object

File file=null;

Scanner fileScanner=null;

String fileName="sample.txt";

try

{

//Create an instance of File class

file=new File(fileName);

//create a scanner class object

fileScanner=new Scanner(file);

//read file elements until end of file

while (fileScanner.hasNext())

{

double value=fileScanner.nextInt();

elementsCount++;

}

//print smallest value

System.out.println(elementsCount+" data values are read");

}

//Catch the exception if file not found

catch (FileNotFoundException e)

{

System.out.println("File Not Found");

}

}

}

Sample output:

sample.txt

10

20

30

40

50

output:

5 data values are read

Answer:

Check the explanation

Explanation:

#include <stdio.h>

int inversions(int a[], int low, int high)

{

int mid= (high+low)/2;

if(low>=high)return 0 ;

else

{

int l= inversions(a,low,mid);

int r=inversions(a,mid+1,high);

int total= 0 ;

for(int i = low;i<=mid;i++)

{

for(int j=mid+1;j<=high;j++)

if(a[i]>a[j])total++;

}

return total+ l+r ;

}

}

int main() {

int a[]={5,4,3,2,1};

printf("%d",inversions(a,0,4));

  return 0;

}

Check the output in the below attached image.

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

Answer:

A Program was written to carry out some set activities. below is the code program in C++ in the explanation section

Explanation:

Solution

CODE

#include <iostream>

using namespace std;

int main() {

string name; // variables

int number;

cin >> name >> number; // taking user input

while(number != 0)

{

// printing output

cout << "Eating " << number << " " << name << " a day keeps the doctor away." << endl;

// taking user input again

cin >> name >> number;

}

}

Note: Kindly find an attached copy of the compiled program output to this question.

In this exercise we have to use the knowledge in computational language in C++ to describe a code that best suits, so we have:

The code can be found in the attached image.

In a very summarized way, we can describe the loop or looping in software as an instruction that keeps repeating itself until a certain condition is met.

To make it simpler we can write this code as:

#include <iostream>

using namespace std;

int main() {

string name; // variables

int number;

cin >> name >> number; // taking user input

while(number != 0)

{

// printing output

cout << "Eating " << number << " " << name << " a day keeps the doctor away." << endl;

// taking user input again

cin >> name >> number;

}

}

See more about C++ at brainly.com/question/19705654

Answer: I would suggest you consider your audience and how you can connect to them. Is your presentation, well, presentable? Is whatever you're presenting reliable and true? Also, no more than 6 lines on each slide. Use colors that contrast and compliment. Images, use images. That pulls whoever you are presenting to more into your presentation.

Explanation:

Answer:

Solid state drive

Explanation:

The term solid-state drive is used for the electronic circuitry made entirely from semiconductors. This highlights the fact that the main storage form, in terms of a solid-state drive, is via semiconductors instead of a magnetic media for example a hard disk.  In lieu of a more conventional hard drive, SSD is built to live inside the device. SSDs are usually more resistant to physical shock in comparison to the electro-mechanical drives and it functions quietly and has faster response time. Therefore,  SSD will be best suitable for the Dave.

Answer:

Explanation:

The solution code is written in Python 3.

Firstly, we can create a Library class with one constructor (Line 2). This constructor won't take any input parameter value. There is only one instance variable, books, in the class (Line 3). This instance variable is an empty list.

To test our class, we can create an object lib1 (Line 5).  Next use that object to add the book item to the books list in the object (Line 6-8).  

Answer:

Following are the code to this question:

def pairSum(a,x): #find size of list

   s=len(a) # use length function to calculte length of list and store in s variable

   for x1 in range(0, s):  #outer loop to count all list value

       for x2 in range(x1+1, s):    #inner loop

           if a[x1] + a[x2] == x:    #condition check

               print(x1," ",x2) #print value of loop x1 and x2  

pairSum([12,7,8,6,1,13],13) #calling pairSum method

Output:

0   4

1   3

Explanation:

Description of the above python can be described as follows:

Answer:

public class SwitchCase {

   public static void main(String[] args) {

       int num = 0;

       int a = 10, b = 20, c = 20, d = 30, x = 40;

       switch (num){

           case 102: a += 1;

           case 103: a += 1;

           case 104: a += 1;

           case 105: a += 1;

           break;

           case 208: b += 1; x = 8;

           break;

           case 209: c = c * 3;

           case 210: c = c * 3;

           break;

           default: d += 1004;

       }

   }

}

Explanation:

Answer: Provided in the explanation segment

Explanation:

Below is the code to carry out this program;

/* C++ program helps prompts user to enter the size of the array. To display the array elements, sorts the data from highest to lowest, print the lowest, highest and average value. */

//main.cpp

//include header files

#include<iostream>

#include<iomanip>

using namespace std;

//function prototypes

void displayArray(int arr[], int size);

void selectionSort(int arr[], int size);

int findMax(int arr[], int size);

int findMin(int arr[], int size);

double findAvg(int arr[], int size) ;

//main function

int main()

{

  const int max=50;

  int size;

  int data[max];

  cout<<"Enter # of scores :";

  //Read size

  cin>>size;

  /*Read user data values from user*/

  for(int index=0;index<size;index++)

  {

      cout<<"Score ["<<(index+1)<<"]: ";

      cin>>data[index];

  }

  cout<<"(1) original order"<<endl;

  displayArray(data,size);

  cout<<"(2) sorted from high to low"<<endl;

  selectionSort(data,size);

  displayArray(data,size);

  cout<<"(3) Highest score : ";

  cout<<findMax(data,size)<<endl;

  cout<<"(4) Lowest score : ";

  cout<<findMin(data,size)<<endl;

  cout<<"(5) Lowest scoreAverage score : ";

  cout<<findAvg(data,size)<<endl;

  //pause program on console output

  system("pause");

  return 0;

}

/*Function findAvg that takes array and size and returns the average of the array.*/

double findAvg(int arr[], int size)

{

  double total=0;

  for(int index=0;index<size;index++)

  {

      total=total+arr[index];

  }

  return total/size;

}

/*Function that sorts the array from high to low order*/

void selectionSort(int arr[], int size)

{

  int n = size;

  for (int i = 0; i < n-1; i++)

  {

      int minIndex = i;

      for (int j = i+1; j < n; j++)

          if (arr[j] > arr[minIndex])

              minIndex = j;

      int temp = arr[minIndex];

      arr[minIndex] = arr[i];

      arr[i] = temp;

  }

}

/*Function that display the array values */

void displayArray(int arr[], int size)

{

  for(int index=0;index<size;index++)

  {

      cout<<setw(4)<<arr[index];

  }

  cout<<endl;

}

/*Function that finds the maximum array elements */

int findMax(int arr[], int size)

{

  int max=arr[0];

  for(int index=1;index<size;index++)

      if(arr[index]>max)

          max=arr[index];

  return max;

}

/*Function that finds the minimum array elements */

int findMin(int arr[], int size)

{

  int min=arr[0];

  for(int index=1;index<size;index++)

      if(arr[index]<min)

          min=arr[index];

  return min;

}

cheers i hope this help!!!

Answer:

Bits

Explanation:

Answer:

All the classes are mentioned with code and screenshots. . That class is shown at last.

Explanation:

Solution

Class.Java:

Code

**

* atauthor your_name

* This class denotes Class at the college.

*

*/

public class Class {

  private String className,classID;

  private Professor professor;

  /**

  * atparam className

  * atparam classID

  * atparam professor

  */

  public Class(String className, String classID, Professor professor) {

      this.classID=classID;

      this.className=className;

      this.professor=professor;

  }

     /**

  * at return classID of the class

  */

  public String getClassID() {

      return classID;

  }

    /**

  * Override toString() from Object Class

  */

  public String toString() {    

      return classID+" "+className+" "+professor.getName()+" "+professor.getEducation();        

  }  

}

Person.Java:

Code:

/**

* atauthor your_name

* This class represents Person

*

*/

public class Person {    

  protected String name;  

  /**method to fetch name

  * at return

  */

  public String getName() {

      return name;

  }

  /**method to set name

  * at param name

  */

  public void setName(String name) {

      this.name = name;

  }

Professor.java:

Code:

import java.util.ArrayList;

import java.util.List;  

/**

* at author your_name

*

*This class represents professors

*

*/

public class Professor extends Person{    

  private String professorID, education;

  private List<Class> classes=new ArrayList<Class>();    

  /**

  * at param name

  * at param professorID

  * at param education

  */

  public Professor(String name,String professorID,String education) {        

      this.name=name;

      this.professorID=professorID;

      this.education=education;        

  }

  /**

  * at return

  */

  public String getEducation() {

      return this.education;

  }    

  /**

  * at return

  */

  public String getprofessorID() {

      return this.professorID;

  }  

  /** to add classes

  * at param Class

  */

  public void addClass(Class c) {

      classes.add(c);

  }  

  /**

  * Override toString() from Object Class

  */

  public String toString() {

      String result=this.getName()+" - "+professorID+" - "+education;

      for(Class c:classes) {

          result+=c.toString()+"\n";

      }      

      return result;

  }

}

}

Student.java:

Code:

import java.util.ArrayList;

import java.util.List;  

/**

* This class represents students

*  at author your_Name

*

*/

public class Student extends Person{    

  private String studentID;

  private List<Class> classes=new ArrayList<Class>();    

  /**

  * atparam name

  * atparam studentID

  */

  public Student(String name,String studentID) {      

      this.name=name;

      this.studentID=studentID;      

  }  

  /**

  * atreturn

  */

  public String getStudentID() {

      return studentID;

  }

     /**

  * atparam c

  */

  public void addClass(Class c) {

      classes.add(c);

  }    

  /**

  * atreturn

  */

  public int getClassCount() {

      return classes.size();

  }

  /**

  * Override toString() from Object Class

  */

  public String toString() {

      String result=this.getName()+" - "+studentID+"\n";

      for(Class c:classes) {

          result+=c.toString()+"\n";

      }    

      return result;

  }  

}

NOTE: Kindly find an attached copy of screenshot of the output, which is a part of the solution to this question

Answer:

The java program for the given scenario is as follows.

import java.util.*;

//interface with method area

interface ObjectWithTwoParameters

{

double area (double d1, double d2);

}

class RectangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return d1*d2;

}

}

class TriangleClass implements ObjectWithTwoParameters

{

//overriding area()

public double area (double d1, double d2)

{

return (d1*d2)/2;

}  

}

class CylinderClass implements ObjectWithTwoParameters

{

public double area (double d1, double d2)

{

return ( 2*3.14*d1*d1 + d2*(2*3.14*d1) );

}

}

public class Test

{

public static void main(String[] args)

{

//area displayed for all three shapes

ObjectWithTwoParameters r = new RectangleClass();

double arear = r.area(2, 3);

System.out.println("Area of rectangle: "+arear);

ObjectWithTwoParameters t = new TriangleClass();

double areat = t.area(4,5);

System.out.println("Area of triangle: "+areat);

ObjectWithTwoParameters c = new CylinderClass();

double areac = c.area(6,7);

System.out.println("Area of cylinder: "+areac);

}

}

OUTPUT

Area of rectangle: 6.0

Area of triangle: 10.0

Area of cylinder: 489.84

Explanation:

1. The program fulfils all the mentioned requirements.

2. The program contains one interface, ObjectWithTwoParameters, three classes which implement that interface, RectangleClass, TriangleClass and CylinderClass, and one demo class, Test, containing the main method.

3. The method in the interface has no access specifier.

4. The overridden methods in the three classes have public access specifier.

5. No additional variables have been declared.

6. The test class having the main() method is declared public.

7. The area of the rectangle, triangle and the cylinder have been computed as per the respective formulae.

8. The interface is similar to a class which can have only declarations of both, variables and methods. No method can be defined inside an interface.

9. The other classes use the methods of the interface by implementing the interface using the keyword, implements.

10. The object is created using the name of the interface as shown.

ObjectWithTwoParameters r = new RectangleClass();

Answer:

A. 0.0450

B. 4

C. 0.25

D. 37.68

E. 6Hz

F. -0.523

G. 1.5m/s

H. vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

I. -1.67m/s.

Explanation:

Given the equation:

y(x,t) = 0.0450 sin(25.12x - 37.68t-0.523)

Standard wave equation:

y(x, t)=Asin(kx−ωt+ϕ)

a.) Amplitude = 0.0450

b.) Wave number = 1/ λ

λ=2π/k

From the equation k = 25.12

Wavelength(λ ) = 2π/25.12 = 0.25

Wave number (1/0.25) = 4

c.) Wavelength(λ ) = 2π/25.12 = 0.25

d.) Angular frequency(ω)

ωt = 37.68t

ω = 37.68

E.) Frequency (f)

ω = 2πf

f = ω/2π

f = 37.68/6.28

f = 6Hz

f.) Phase angle(ϕ) = -0.523

g.) Wave propagation speed :

ω/k=37.68/25.12=1.5m/s

h.) vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)

(i) vy(3.5m, 21s) = 0.045(-37.68) cos (25.12*3.5-37.68*21-0.523) = -1.67m/s.