Defeating authentication follows the method–opportunity–motive paradigm.
1. Discuss how these three factors apply to an attack on authentication?

Answer:

Spyware

Explanation:

Internet is a type of computer network that allow device communicate with each other world wide. So, it is not the correct option.

Worm: this is a standalone malware computer program that replicates itself in order to spread to other computers. It spreads copies of itself from computer to computer. A worm can replicate itself without any human interaction, and it does not need to attach itself to a software program in order to cause damage. The major feature of a worm is ability to replicate on it own. So, it is not the correct option.

Bot: this is a software application that runs automated tasks. So, it is not the correct option.

Middleware: this is software that lies between an operating system and the applications running on it, enabling communication and data management. It provides services to software applications beyond those available from the operating system. So, it is not the correct option.

Spyware: this is the correct answer. Spyware is similar to trojan horse because it hides itself in a system and a user may not know that it exist on the system. Spyware is a form of malware that hides on your device, monitors your activity, and steals sensitive information without knowledge of the user.

Answer:

Spyware

Explanation:

Computer network systems are intercommunication of connected devices, to share resources within and to other remote networks. Network security policies and technical control system technologies are used by a network or system administrator to prevent unauthorized and unauthenticated access to user account in the network by cyber attackers.

Cyber attackers are individuals or group of individuals with the traditional hacking skills to exploit network and system vulnerabilities, to steal information. Malwares are used by hackers to covertly access networks. The spyware is a monitoring Trojan horse malware, used to monitor the activities of a user, steal information for fraudulent purposes.

Answer:

There are four options for these questions: and d is correct answer

a. signature

b. priority file

c. moles

d. honeytoken

Explanation:

Honeytokens are words or fictitious records implemented in a database, with this option we can track data in a complex situation, for example, in cloud computing, if someone tries to steal data from the database we can detect, who was the thief, and we can add a honeytoken in every record of our database.

Answer:

d. Analyzing the unstructured content of Web pages

Explanation:

Web page mining is referred to that mining that  is used to extract the data or information from the web page. The main focus behind the web page mining is to discover all useful information from web pages.

The main objective of web page mining the unstructured data or content of the web pages.

All the other three option are not related with web content mining, so the correct option is D

Answer:

The Local, Remote and Upstream repository can each have multiple push /pull requests. When  a Participant in a collaborative group on Github is ready to have their work used by the group,  the participants makes a pull request.

Explanation:

These are simple blanks which are answer here

The Local, Remote and Upstream repository can each have multiple push /pull requests. When  a Participant in a collaborative group on Github is ready to have their work used by the group,  the participants makes a pull request.

The code is implemented based on the given operations.

Explanation:

#include <iostream>

#include <string>

using namespace std;

class dayType

{ private:

 string day[7];

 string presentDay;

 int numofDays;

public:

 void setDay(string freshDay);

 void printDay() const;

 int showDay(int &day);

 int nextDay(int day);

 int prevDay(int day) const;

 int calcDay(int day, int numofDays);    

 dayType()

 {

  day[0] = "Sunday";

  day[1] = "Monday";

  day[2] = "Tuesday";

  day[3] = "Wednesday";

  day[4] = "Thursday";

  day[5] = "Friday";

  day[6] = "Saturday";

  presentDay = day[0];

  numofDays = 0;

 };

 ~dayType();

};

#endif

#include "dayType.h"

void dayType::setDay(string freshDay)

{

  presentDay = freshDay;

}

void dayType::printDay()

{

  cout << "Day chosen is " << presentDay << endl;

}

int dayType::showDay(int& day)

{

  return day;

}

int dayType::nextDay(int day)

{

day = day++;

if (day > 6)

 day = day % 7;

switch (day)

{

case 0: cout << "The successive day is Sunday";

 break;

case 1: cout << "The successive day is Monday";

 break;

case 2: cout << "The successive day is Tuesday";

 break;

case 3: cout << "The successive day is Wednesday";

 break;

case 4: cout << "The successive day is Thursday";

 break;

case 5: cout << "The successive day is Friday";

 break;

case 6: cout << "The successive day is Saturday";

 break;

}

cout << endl;

return day;

}

int dayType::prevDay(int day)

{

day = day--;

switch (day)

{

case -1: cout << "The before day is Saturday.";

 break;

case 0: cout << "The before day is Saturday.";

 break;

case 1: cout << "The before day is Saturday.";

 break;

case 2: cout << "The before day is Saturday.";

 break;

case 3: cout << "The before day is Saturday.";

 break;

case 4: cout << "The before day is Saturday.";

 break;

case 5: cout << "The before day is Saturday.";

 break;

default: cout << "The before day is Saturday.";

}

cout << endl;

return day;

}

int dayType::calcDay(int addDays, int numofDays)

{

addDay = addDays + numofDays;

if (addDay > 6)

 addDay = addDay % 7;

switch(addDay)

{

case 0: cout << "The processed day is Sunday.";

 break;

case 1: cout << "The processedday is Monday.";

 break;

case 2: cout << "The processedday is Tuesday.";

 break;

case 3: cout << "The processedday is Wednesday.";

 break;

case 4: cout << "The processedday is Thursday.";

 break;

case 5: cout << "The processedday is Friday.";

 break;

case 6: cout << "The processedday is Saturday.";

 break;

default: cout << "Not valid choice.";

}

cout << endl;

return addDays;

}

To design and implement the dayType class, define the instance variable 'day' to store the day of the week. Create methods to set the day, print the day, return the day, return the next day, return the previous day, and calculate the day by adding certain days. Test the class by creating an object and calling the methods.

To design and implement the dayType class, we can start by defining the instance variable 'day' to store the day of the week. We can then create a setDay method to set the value of 'day', and a printDay method to print the current day. The returnDay method can be used to return the current day, and the nextDay and previousDay methods can be implemented to calculate and return the next and previous days based on the current day. To calculate the day after adding a certain number of days, we can create a calculateDay method that adds the given number of days to the current day and returns the resulting day. Lastly, appropriate constructors can be added to initialize the object with a specific day.

Here is an example implementation of the dayType class in Python:

To test the class, you can create an object of the dayType class with a specific day, and call the different methods to perform the required operations.

https://brainly.com/question/34680737

#SPJ3

Answer:

A)

Explanation:

Enable the list to be pinned in the console. This allows users to view the list alongside the case view in the console

Bookstore and BookSearch are the two entities for the given scenario.

Explanation:

Answer:

public static double drivingCost(int drivenMiles, double milesPerGallon,double dollarsPerGallon){

       return (drivenMiles/milesPerGallon)*(dollarsPerGallon);

   }

Find a complete java program that prompts the user to enter this values and the calls this Method in the explanation section.

Explanation:

import java.util.Scanner;

public class ANot {

   public static void main(String[] args) {

   Scanner in = new Scanner(System.in);

       System.out.println("Please Enter the Driven Miles");

       int drivenMiles = in.nextInt();

       System.out.println("Please Enter the Your fuel consumption in miles/gallon");

       double milesPerGallon = in.nextDouble();

       System.out.println("Please Enter the cost of fuel in Dollars/gallon");

       double dollarsPerGallon = in.nextDouble();

       double dollarCost = drivingCost(drivenMiles,milesPerGallon,dollarsPerGallon);

       System.out.println("The Total Cost in Dollars is "+dollarCost);

       }

   public static double drivingCost(int drivenMiles, double milesPerGallon,double dollarsPerGallon){

       return (drivenMiles/milesPerGallon)*(dollarsPerGallon);

   }

}

To ensure a loop executes as long as the user inputs a non-negative number, you can utilize a while loop in your code. The loop should continue if the condition is true, where the input number is greater than or equal to zero. Here's a sample code snippet:

number = int(input('Enter a number: '))

while number >= 0:

   print('Body')

   number = int(input('Enter a number: '))

print('Done')

This loop begins by prompting the user to enter a number. The while condition checks if the entered number is non-negative (i.e., greater than or equal to zero). If the condition is met, the loop 'Body' is executed, followed by another prompt for the user to input a number. This process repeats until the user inputs a negative number, which causes the while loop to end, and the program prints 'Done' after exiting the loop. Using a break statement isn't necessary here because the loop condition naturally terminates when a negative number is entered.

Answer:

The code is given below with appropriate comments for better understanding

Explanation:

#include <bits/stdc++.h>

using namespace std;

enum Digits {

zero,

one

} ;

struct CellType

{

  Digits bit;

  CellType* next;

};

class Minimum

{

public:

  struct CellType* head,*temp; // head to point MSB

  Minimum(string s)

  {

      int sz = s.size(); // binary size as per question it indicates n.

      for(int i=0;i<sz;i++)

      {

          if(s[i] == '0') // if the bit is zero , we add zero at the end of stream

              addAtEnd(zero);

          else // if the bit is one , we one zero at the end of stream

              addAtEnd(one);

      }

      addOne();

      showlist();

  }

  CellType* create(Digits x) // to create a node of CellType*

  {

      CellType* t = new CellType();

      t->bit = x;

      t->next = NULL;

      return t;

  }

  void addAtEnd(Digits x)

  {

      if(head == NULL) // if list is empty , then that will be the only node

      {

          CellType* t;

          t = create(x);

          head = temp = t;

      }

      else

      { // other wise we add the node at end indicated by the temp variable

          CellType* t ;

          t = create(x);

          temp->next = t;

          temp=temp->next;

      }

  }

  void showlist()

  {

      // this is just a normla method to show the list

      CellType* t = head;

      while(t!=NULL)

      {

          cout<<t->bit;

          t=t->next;

      }

  }

  void addOne()

  {

      /*

          here since we need to add from the end and it is a singly linked list we take a stack

          and store in last in ,first out format .

          Then we keep on changing all ones to zeroes until we find a zero int he list,

          The moment a zero is found it should be changed to one.

          If everything is one in the sequence , then we add a new zero digit node at the beginning of the list.

      */

      stack<CellType*> st;

      CellType* t = head;

      while(t!=NULL)

      {

          st.push(t);

          t=t->next;

      }

      while(st.size()>0 && (st.top())->bit == one )

      {

          CellType* f = st.top();

          f->bit = zero ;

          st.pop();

      }

      if(st.size())

      {

          CellType* f = st.top();

          f->bit = one ;

      }

      else

      {

          t = create(one);

          t->next = head;

          head = t;

      }

  }

};

int main()

{

  /*

      Here i am taking an integer as input and then converting it to binary using a string varaible s

      if you want to directly take the binary stream as input , remove the comment from "cin>>s" line.

  */

  long long int n,k;

  cin>>n;

  string s;

  k = n;

  while(k>0)

  {

      s = s + (char)(k%2 +48);

      k=k/2;

  }

  reverse(s.begin(),s.end());

  //cin>>s;

  Minimum* g = new Minimum(s);

  return 0;

}

Answer:

   public static int[] listLengthOfAllWords(String [] wordArray){

       int[] intArray = new int[wordArray.length];

       for (int i=0; i<intArray.length; i++){

           int lenOfWord = wordArray[i].length();

           intArray[i]=lenOfWord;

       }

       return intArray;

   }

Explanation:

A Complete Java program with a call to the method is given below

import java.util.Arrays;

import java.util.Scanner;

public class ANot {

   public static void main(String[] args) {

      String []wordArray = {"John", "James", "David", "Peter", "Davidson"};

       System.out.println(Arrays.toString(listLengthOfAllWords(wordArray)));

       }

   public static int[] listLengthOfAllWords(String [] wordArray){

       int[] intArray = new int[wordArray.length];

       for (int i=0; i<wordArray.length; i++){

           int lenOfWord = wordArray[i].length();

           intArray[i]=lenOfWord;

       }

       return intArray;

   }

}

This program gives the following array as output: [4, 5, 5, 5, 8]

Answer:

(a) What is the best case time complexity of the algorithm (assuming n > 1)?

Answer: O(1)

(b) What is the worst case time complexity of the algorithm?

Answer: O(n^4)

Explanation:

(a) In the best case, the if condition will be true, the program will only run once and return so complexity of the algorithm is O(1) .

(b) In the worst case, the program will run n^4 times so complexity of the algorithm is O(n^4).

Answer:

Option(b) i.e "connection " is the correct answer for the given question.

Explanation:

In the visual basics of .Net when we want to establish the connection with the database we have to use the data configuration wizard. A connection is the string contacts of the data source that establishes a connection with the database.

Option(a),Option(c) and Option(d) are not relate the connection of the database.

So the "connection" is the correct answer.

Answer:

import java.io.FileNotFoundException;

import java.io.FileReader;

import java.util.HashMap;

import java.util.Scanner;

import java.util.concurrent.SynchronousQueue;

import java.util.regex.Matcher;

import java.util.regex.Pattern;

public lass Test

{

public static void main(String[] args) {

FileReader fr;

try {

fr = new FileReader("expression.txt");

Scanner sc=new Scanner(fr);

while(sc.hasNextLine())

{

String line=sc.nextLine();

pareseString(line);

pareseString(line);

}

} catch (FileNotFoundException e) {

e.printStackTrace();

}

pareseString("6 * 3+2i");

pareseString("2 - 3");

}

public static ComplexNumber add(ComplexNumber c1, ComplexNumber c2) {

return new ComplexNumber(c1.getRealNumber() + c2.getRealNumber(), c1.getImaginaryNumber() + c2.getImaginaryNumber());

}

public static ComplexNumber substract(ComplexNumber c1, ComplexNumber c2) {

return new ComplexNumber(c1.getRealNumber() - c2.getRealNumber(), c1.getImaginaryNumber() - c2.getImaginaryNumber());

}

public static ComplexNumber multiply(ComplexNumber c1,ComplexNumber c2) {

ComplexNumber c3 = new ComplexNumber();

c3.setRealNumber( c1.getRealNumber() * c2.getRealNumber() - c1.getImaginaryNumber() * c2.getImaginaryNumber());

c3.setImaginaryNumber( c1.getRealNumber() * c2.getImaginaryNumber() - c1.getImaginaryNumber() * c2.getRealNumber());;

return c3;

}

public static ComplexNumber divide(ComplexNumber c1,ComplexNumber c2) {

ComplexNumber c3 = new ComplexNumber();

c3.setRealNumber( c1.getRealNumber() / c2.getRealNumber() - c1.getImaginaryNumber() / c2.getImaginaryNumber());

c3.setImaginaryNumber( c1.getRealNumber() / c2.getImaginaryNumber() - c1.getImaginaryNumber() / c2.getRealNumber());;

return c3;

}

public static void pareseString(String line)

{

String [] strArr=line.split(" ");

String cn1=strArr[0];

String operation=strArr[1];

String cn2=strArr[2];

ComplexNumber c1=validation(cn1);

ComplexNumber c2=validation(cn2);

if(c1!=null && c2!=null)

{

switch (operation) {

case "+":

System.out.println(add(c1, c2));

break;

case "-":

System.out.println(substract(c1, c2));

break;

case "*":

System.out.println(multiply(c1, c2));

break;

case "/":

System.out.println(divide(c1, c2));

break;

}

}

}

private static ComplexNumber validation(String comp) {

String numberNoWhiteSpace = comp.replaceAll("\\s","");

Pattern patternA = Pattern.compile("([-]?[0-9]+\\.?[0-9]?)([-|+]+[0-9]+\\.?[0-9]?)[i$]+");

Pattern patternB = Pattern.compile("([-]?[0-9]+\\.?[0-9]?)$");

Pattern patternC = Pattern.compile("([-]?[0-9]+\\.?[0-9]?)[i$]");

Matcher matcherA = patternA.matcher(numberNoWhiteSpace);

Matcher matcherB = patternB.matcher(numberNoWhiteSpace);

Matcher matcherC = patternC.matcher(numberNoWhiteSpace);

double realNumber=0.0;

double imaginaryNumber=0.0;

ComplexNumber cn=null;

if (matcherA.find()) {

realNumber = Double.parseDouble(matcherA.group(1));

imaginaryNumber = Double.parseDouble(matcherA.group(2));

cn=new ComplexNumber(realNumber, imaginaryNumber);

} else if (matcherB.find()) {

realNumber = Double.parseDouble(matcherB.group(1));

imaginaryNumber = 0;

cn=new ComplexNumber(realNumber, imaginaryNumber);

} else if (matcherC.find()) {

realNumber = 0;

imaginaryNumber = Double.parseDouble(matcherC.group(1));

cn=new ComplexNumber(realNumber, imaginaryNumber);

}

return cn;

}

}

class Number

{

private double realNumber;

public Number(double realNumber) {

this.realNumber= realNumber;

}

public Number() {

this.realNumber= realNumber;

}

public double getRealNumber() {

return realNumber;

}

public void setRealNumber(double realNumnber) {

this.realNumber = realNumnber;

}

@Override

public String toString() {

return this.getRealNumber()+"";

}

@Override

public boolean equals(Object obj) {

if(obj instanceof Number)

{

Number cn=(Number)obj;

return this.getRealNumber()==cn.getRealNumber();

}

return false;

}

}

class ComplexNumber extends Number

{

public double getImaginaryNumber() {

return imaginaryNumber;

}

public void setImaginaryNumber(double imaginaryNumber) {

this.imaginaryNumber = imaginaryNumber;

}

double imaginaryNumber;

public ComplexNumber(double realNumnber,double imaginaryNumber) {

super(realNumnber);

this.imaginaryNumber=imaginaryNumber;

}

public ComplexNumber() {

super();

}

@Override  

public String toString() {

return this.getRealNumber()+"+"+this.getImaginaryNumber()+"i";

}

@Override

public boolean equals(Object obj) {

if(obj instanceof ComplexNumber)

{

ComplexNumber cn=(ComplexNumber)obj;

return this.getRealNumber()==cn.getRealNumber() && cn.getImaginaryNumber()==this.getImaginaryNumber();

}

return false;

}

}

Explanation:

Create the add method, that takes object c2 as parameter.

Create the subtract method, followed by the methods to multiply and divide.

Create a regular expression that matches complex number with BOTH real AND imaginary parts.

Answer:

tell him to do it himself and call him lazy

(i know im a genius)

Answer:

B. "This new software system has been giving us problems for a while now."

Explanation:

Of all the given, answer B is the only impersonal statement. Passive voice is used and it is highly effective in remaining professional while communicating from a managerial role. By using an impersonal statement, the employee (Jack Blair) won't get offended by any means. Although he isn't personally mentioned in the answer E, he may feel guilt because there is someone else that is able to resolve the problem.

Answer:

a) 56789₁₀ = 11011110111010101₂ (unsigned integer)

b) 56789₁₀ = 0000000011011110111010101₂ (Two's complement)

c) 56789₁₀ = 01010110011110001001 (BCD)

d) 56789₁₀ = ÝÕ (ASCII)

e) 56789₁₀ = 0 - 1000 1110 - 101 1101 1101 0101 0000 0000 (IEEE single precision)

Explanation:

a) 56789₁₀ = (1 × 2¹⁵) + (1 × 2¹⁴) + (0 × 2¹³) + (1 × 2¹²) + (1 × 2¹¹) + (1 × 2¹⁰) + (0 × 2⁹) + (1 × 2⁸) + (1 × 2⁷) + (1 × 2⁶) + (0 × 2⁵) + (1 × 2⁴) + (0 × 2³) + (1 × 2²) + (0 × 2¹) + (1 × 2⁰) = 11011110111010101₂

This requires 2 bytes - 16 bits and hexadecimal byte value of DDD5.

2) since the number is positive, the two's complement is just that same binary number with a signed 0 to indicate positive number in front.

56789₁₀ = 0000000011011110111010101₂

This requires 3 bytes - 24 bits and hexadecimal byte value of 11DDD5.

c) BCD

This converts each single bit in the base-10 to binary.

5 = 0101, 6 = 0110, 7 = 0111, 8 = 1000, 9 = 10001, then combined, we have

56789₁₀ = 01010110011110001001 (BCD)

It's an historic code.

This requires 3 bytes - 24 bits and hexadecimal byte value of 56789.

d) ASCII

This uses symbols to represent the numbers.

56789₁₀ = ÝÕ (ASCII)

This requires 1 byte - 8 bits.

e) IEEE single precision

Step 1, convert to base 2

56789₁₀ = 11011110111010101₂

Step 2, normalize the binary,

11011110111010101₂ =11011110111010101 × 2⁰ = 1.1011110111010101 × 2¹⁵

Sign = 0 (a positive number)

Exponent (unadjusted) = 15

Mantissa (not normalized) = 1.1011110111010101

Step 3, Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary

Exponent (adjusted) = Exponent (unadjusted) + 2⁽⁸⁻¹⁾ - 1 = 15 + 2⁽⁸⁻¹⁾ - 1 = (15 + 127)₁₀ = 142₁₀

Exponent (adjusted) = 142₁₀ = 1000 1110₂

Step 4, Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, by adding the necessary number of zeros to the right:

Mantissa (normalized) = 1.101 1101 1101 0101 0000 0000 = 101 1101 1101 0101 0000 0000

Therefore,

56789₁₀ = 0 - 1000 1110 - 101 1101 1101 0101 0000 0000

This requires 4 bytes - 32 bits and hexadecimal byte value of 8E5DD500.

Hope this helps!

a) 10011101₂ = 9D₁₆ or 235₈

b) 00010101₂ = 15₁₆ or 025₈

c) 11100110₂ = E6₁₆ or 346₈

d) 01101001₂ = 69₁₆ or 151₈

An hexadecimal is a group of 4bits while an octal is a group of 3 bits. They are represented in the table below;

Table for conversion;

Octal   =>    binary

0         =>     000

1          =>     001

2          =>    010

3          =>    011

4          =>    100

5          =>    101

6          =>    110

7          =>    111

Hexadecimal   => binary

0                      =>     0000

1                       =>     0001

2                      =>     0010

3                      =>     0011

4                      =>     0100

5                      =>     0101

6                      =>     0110

7                      =>     0111

8                      =>     1000

9                      =>     1001

A                      =>    1010

B                      =>    1011

C                      =>    1100

D                      =>    1101

E                      =>    1110

F                      =>    1111

(a)

(i) Convert 10011101 to hexadecimal

Step 1: Starting from the right, split the number into groups of 4s as follows;

1001   1101

Step 2: Convert each of the groups into its equivalent hexadecimal using the table above;

1001 = 9

1101 = D

Step 3: Put them together;

1001 1101₂ = 9D₁₆

(ii) Convert 10011101 to octal

Step 1: Starting from the right, split the number into groups of 3s as follows;

10  011  101

Step 2: The last group (10) in the result of step 1 above has only 2 bits. Therefore, add zero to its left to make it 3 bits as follows;

010  011  101

Step 3: Convert each of the groups into its equivalent octal using the table above;

010 = 2

011 = 3

101 = 5

Step 4: Put them together;

10 011 101₂ = 235₈

(b)

(i) Convert 00010101 to hexadecimal

Step 1: Starting from the right, split the number into groups of 4s as follows;

0001   0101

Step 2: Convert each of the groups into its equivalent hexadecimal using the table above;

0001 = 1

0101 = 5

Step 3: Put them together;

0001 0101₂ = 15₁₆

(ii) Convert 00010101 to octal

Step 1: Starting from the right, split the number into groups of 3s as follows;

00  010  101

Step 2: The last group (00) in the result of step 1 above has only 2 bits. Therefore, add zero to its left to make it 3 bits as follows;

000  010  101

Step 3: Convert each of the groups into its equivalent octal using the table above;

000 = 0

010 = 2

101 = 5

Step 4: Put them together;

00 010 101₂ = 025₈

(c)

(i) Convert 11100110 to hexadecimal

Step 1: Starting from the right, split the number into groups of 4s as follows;

1110  0110

Step 2: Convert each of the groups into its equivalent hexadecimal using the table above;

1110 = E

0110 = 6

Step 3: Put them together;

1110 0110₂ = E6₁₆

(ii) Convert 11100110 to octal

Step 1: Starting from the right, split the number into groups of 3s as follows;

11 100 110

Step 2: The last group (11) in the result of step 1 above has only 2 bits. Therefore, add zero to its left to make it 3 bits as follows;

011 100 110

Step 3: Convert each of the groups into its equivalent octal using the table above;

011 = 3

100 = 4

110 = 6

Step 4: Put them together;

11 100 110₂ = 346₈

(d)

(i) Convert 01101001 to hexadecimal

Step 1: Starting from the right, split the number into groups of 4s as follows;

0110 1001

Step 2: Convert each of the groups into its equivalent hexadecimal using the table above;

0110 = 6

1001 = 9

Step 3: Put them together;

0110 1001₂ = 69₁₆

(ii) Convert 01101001 to octal

Step 1: Starting from the right, split the number into groups of 3s as follows;

01 101 001

Step 2: The last group (01) in the result of step 1 above has only 2 bits. Therefore, add zero to its left to make it 3 bits as follows;

001 101 001

Step 3: Convert each of the groups into its equivalent octal using the table above;

001 = 1

101 = 5

001 = 1

Step 4: Put them together;

01 101 001₂ = 151₈

Answer:

Answer is provided in the explanation section

Explanation:

1. For testing this query, first create a table:

CREATE TABLE STUDENT (NAME CHARACTER(25), ROLLNO int PRIMARY KEY, AREACODE int, REGD_YEAR date)  

2. Insert some data for checking the query

 insert into student values(101,'Mark',203,'03-12-1997')  

            insert into student values(106,'Zack',204,'06-18-1992')

 insert into student values(104,'Jess',203,'01-11-1995')

3. Select query for creating a list of student names from area code 203

SELECT NAME AS "Student Name", AREACODE, REGD_YEAR

FROM STUDENT

WHERE AREACODE LIKE '203%'

ORDER BY “REGD_YEAR“,”Student Name”;

Answer:

The answer is A.Absolute reference.

Explanation:

Absolute reference is a cell reference whose location remains constant when the formula is copied.

Answer:

Option 1 is the correct answer for the above question

Explanation:

In the provided code of 'SomeClass', the variable 'VALUE1' is a class constant because it is the only one declared as public, static, and final, indicating a constant value that cannot be changed.

In Java, a class constant is typically defined with the keywords public, static, and final. A class constant is a variable with a constant value that cannot be changed. In the given class SomeClass, the variable VALUE1 is declared with these keywords: public static final. Therefore, VALUE1 is a class constant. It is important to understand that class constants are useful for defining values that should not change throughout the execution of a program. Since it is declared with the final keyword, it must be assigned a value only once, and it cannot be modified afterwards.

Answer:

Hi there Collegebound! The implementation of the Triangle class and the Drive program is below. Copy the code below into the file Triangle.java and then compile it with the command: "javac Triangle.java". To run the program, type the command: "java Drive". You should see the following result if you test with the same inputs:

$java Drive

Enter side 1 of the triangle:

1

Enter side 2 of the triangle:

1

Enter side 3 of the triangle:

1

Enter color of the triangle:

orange

Enter if triangle is filled:

true

Area of Triangle is: 0.4330127018922193

Perimeter of Triangle is: 3.0

Color of Triangle is: orange

Triangle is filled: true

Explanation:

import java.lang.Math;

import java.util.Scanner;

public class Triangle extends GeometricObject {

 double side1, side2, side3;

 protected Triangle() {}

 protected Triangle(double s1, double s2, double s3) {

   this.side1 = s1;

   this.side2 = s2;

   this.side3 = s3;

 }

 public double getSide1() {

   return side1;

 }

 public double getSide2() {

   return side2;

 }

 public double getSide3() {

   return side3;

 }

 public double getArea() {

   /* use Heron's formula */

   double s = (this.side1 + this.side2 + this.side3) / 2;

   double area = Math.sqrt(s*((s-this.side1)*(s-this.side2)*(s-this.side3)));

   return area;

 }

 public double getPerimeter() {

   return side1+side2+side3;

 }

 @Override

 public String toString() {

   return "Triangle: side1 = " + side1 + " side2 = " + side2 + " side3 = " + side3;

 }

}

class Drive {

 public static void main(String args[]) {

   double side1, side2, side3;

   System.out.println("Enter side 1 of the triangle: ");

   Scanner scan = new Scanner(System.in);

   side1 = scan.nextDouble();

   System.out.println("Enter side 2 of the triangle: ");

   scan = new Scanner(System.in);

   side2 = scan.nextDouble();

   System.out.println("Enter side 3 of the triangle: ");

   scan = new Scanner(System.in);

   side3 = scan.nextDouble();

   System.out.println("Enter color of the triangle: ");

   scan = new Scanner(System.in);

   String color = scan.next();

   System.out.println("Enter if triangle is filled: ");

   scan = new Scanner(System.in);

   Boolean isFilled = scan.nextBoolean();

   Triangle triangle = new Triangle(side1, side2, side3);

   triangle.setColor(color);

   triangle.setFilled(isFilled);

   System.out.println("Area of Triangle is: " + triangle.getArea());

   System.out.println("Perimeter of Triangle is: " + triangle.getPerimeter());

   System.out.println("Color of Triangle is: " + triangle.getColor());

   System.out.println("Triangle is filled: " + triangle.isFilled());

 }

}

Final answer:

A Triangle class in Java entails creating a subclass of GeometricObject with fields for its sides, methods to calculate its area and perimeter, and an overridden toString method. Test the class with a program that gathers user inputs to instantiate a triangle and display its attributes.

Explanation:

The student's question pertains to the construction of a Triangle class in Java that extends a given GeometricObject abstract class. To design this class, one must include three data fields representing the sides of the triangle, constructors for default and specified values, accessor methods for the sides, a method to calculate the area of the triangle, a method to calculate the perimeter, and an override of the toString method to describe the triangle.

The test program will prompt the user for inputs regarding the sides, color, and fill of the triangle and will use this information to create a Triangle object and display its attributes.

Example Implementation:

public class Triangle extends GeometricObject {
   private double side1 = 1.0, side2 = 1.0, side3 = 1.0;
   public Triangle() {}
   public Triangle(double side1, double side2, double side3) {
       this.side1 = side1;
       this.side2 = side2;
       this.side3 = side3;
   }
   // Accessor methods
   public double getSide1() {
       return side1;
   }
   public double getSide2() {
       return side2;
   }
   public double getSide3() {
       return side3;
   }
   // Area calculation using Heron's formula
   public double getArea() {
       double s = (side1 + side2 + side3) / 2;
       return Math.sqrt(s * (s - side1) * (s - side2) * (s - side3));
   }
   // Perimeter calculation
   public double getPerimeter() {
       return side1 + side2 + side3;
   }
   // Description of Triangle
   Override
   public String toString() {
       return "Triangle: side1 = " + side1 + " side2 = " + side2 + " side3 = " + side3;
   }
}

To test the Triangle class, one can implement a Driver program within the same file that uses a Scanner to collect user input, initializes a Triangle object, sets its attributes, and finally prints the area, perimeter, color, and filled status.

Answer:

Hi there! The question is asking for a simple loop implementation that prints payment advice for Payroll.

Explanation:

Assuming that the payment method is “mailed check” for hours worked less than 30 and “direct deposit” for hours worked greater than 30, we can implement the function as below.  

hours_worked = [45, 15, 63, 23, 39]

payment_method = “mailed check”

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

   if hours_worked[index] > 30 {

      payment_method = “direct deposit”

   }

   print(“Paying “ + (hours_worked[index] * pay_rate) + “by “ + payment_method)

}

To calculate the payments for a list of employees with given worked hours, a for loop can be used to multiply each hour count by the fixed rate of pay and print alternating payment method statements.

Using a definite loop, such as a for loop in a programming language, we can easily calculate the individual paychecks for a list of employees and their respective hours worked. Assuming a fixed rate of pay of $10 per hour, we can loop through a list with the values 45, 15, 63, 23, and 39 to calculate the gross pay for each employee. Since our task is to print a statement for the method of payment, alternating between direct deposit and mailed check, we would end up with statements that reflect the total pay multiplied by the rate_of_pay. Below is a potential implementation of such a loop:

This example takes the given hours, multiplies them by the rate of pay, and prints out a statement that indicates the payment amount and method.

Answer and explanation:

When buying a computer, there are a few factors that sould be taken into account. Those could be the following ones:

When deciding which computer to buy, important factors to consider are price, performance, operating system, usage, portability, and brand and support. Hardware refers to physical components, while software refers to programs and applications.

When deciding which computer to buy, there are several factors to consider. These include:

After selecting the system you would like to buy, you should identify which terms refer to hardware and which refer to software. Hardware refers to the physical components of a computer, such as the hard drive, processor, memory, and motherboard. Software refers to the intangible programs or applications that run on the computer, such as operating systems, utilities, and applications.

Answer:

A. 8-pin PCI-E connector.

F. 6-pin PCI-E connector.

Explanation:

The video card is a peripheral hardware component in a computer system that is used to run videos and graphic files, providing the required memory, runtime and bandwidth.

The PCI-e or peripheral component interconnect express is a connector or expansion slot used specifically for adding and powering video cards on a computer system.

Answer:

Answer explained below

Explanation:

Solution:

Some of the PHP functions used to interpret common HTML and URL encodings are as follows:

urlencode:

This function is used for encoding a string to be used in a query part of a URL and this is used as a convenient way to pass variables to the next page of a web form.

urldecode(): Same as urlencode() but in a reverse way. It is used to decode URL-encoded string

htmlentities(): This PHP function is used to convert all applicable characters to HTML entities

html_entity_decode(): This function converts HTML entities to characters and it is the reverse form of htmlentities() function.

Answer:

Transport layer:

- data packets are segment to smaller chunks.

- gives sequence number to segment.

- identifies the source and destination port number.

- initiates data transmission between nodes.

- rearrange and identifies the application, the transmitted data is meant for.

Explanation:

The transport layer is the fourth layer in the OSI network model. Protocols like TCP and UDP are found in this layer. It segment data packets and for a connection oriented protocol like TCP, it creates an established session between source and destination host (the session layer can also do this, but it is more defined in the transport layer).

The network and data-link layer can also transmit data packets.

Question continuation

Determine the following probabilities:

a. P(A)

b. P(B)

c. P(A ∩ B)

d. P(A ∪ B)

Answer:

a. P(A) = 0.1389

b. P(B) = 0.1389

c. P(AnB) = 0.0193

d. P(AuB) = 0.2585

Explanation:

Given

Password length = 6

Letters (a-z) = 26

Integers (0-9) = 10

Total usable characters = 26 + 10 = 36

a. P(A) = Probability that a password begins with vowel (a,e,i,o,u)

Probability = Number of required outcomes/ Number of possible outcomes

Number of required outcomes = Number of vowels = 5

Number of possible outcomes = Total usable characters = 36

P(A) = 5/36

P(A) = 0.13888888888

P(A) = 0.1389

b. P(B) = Probability that the password ends with an even number (0,2,4,6,8)

Probability = Number of required outcomes/ Number of possible outcomes

Number of required outcomes = Number of even numbers = 5

Number of possible outcomes = Total usable characters = 36

P(B) = 5/36

P(B) = 0.13888888888

P(B) = 0.1389

c. P(AnB)

This means that the probability that a password starts with a vowel and ends with an even number

P(AnB) = P(A) and P(B)

P(AnB) = P(A) * P(B)

P(AnB) = 5/36 * 5/36

P(AnB) = 25/1296

P(AnB) = 0.01929012345

P(AnB) = 0.0193 ----_---- Approximately

d. P(AuB)

This means that the probability that a password either starts with a vowel or ends with an even number

P(AuB) = P(A) or P(B)

P(AuB) = P(A) + P(B) - P(AnB)

P(AuB) = 5/36 + 5/36 - 25/1296

P(AuB) = 335/1296

P(AuB) = 0.25848765432

P(AuB) = 0.2585 ----_---- Approximately

Final answer:

Calculating the probability of events A and B for passwords satisfying specific conditions.

Explanation:

A denote the event that a password begins with a vowel and B denote the event that a password ends with an even number. The total number of possible passwords is 36^6 (26 letters + 10 integers). To determine the probability of A, we calculate the number of passwords that start with a vowel (5 vowels) followed by any character (36 options) for the remaining 5 characters. Similarly, to find the probability of B, we consider passwords that end with an even number (5 options) and any character for the other 5 places.

Probability of A = (5 * 36^5) / 36^6
Probability of B = (5 * 36^5) / 36^6

Answer:

a. Microphone

d. Wi-Fi antenna

Explanation:

The wifi antenna , is present in the screen bezel , and the two cables are connected to the motherboard and the wifi adapter .

Hence , we need to be aware of the wifi antenna , before installing the webcam .

Microphone , as in most of the laptop , the microphone is present just near the screen bezel , and hence , need to be aware of before the installation process of the webcam .

Answer:

The correct option is WIFI antenna.

Answer:

Following is given the solution to the question. This question has two parts of source code:

The images are attached displaying code or each class.  Indentations are made clear so that the code get easier to understand.

Comments are given inside the code where necessary to make the logic clear.

Output for the code is also attached in the last image.

Explanation:

I hope it will help you!

Answer:

public class ConversionCalculator extends JFrame {

  private JLabel inch_Label, meter_Label, cm_Label, yard_Label;

  private JButton clear, calculate, exit;

  private JTextField inch_tf, cm_tf, meters_tf, yards_tf;

  public ConversionCalculator()

  {

      exitButtonHandler exitB;

      clearButtonHandler clearB;

      calcButtonHandler calcB;

      setTitle("Conversion Calculator");

      setSize(600,200);

      setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

      //The first layout we use will be 1 row with three columns

      setLayout(new GridLayout(1,3));

      //initialize all the components

      inch_Label = new JLabel("Inches");

      meter_Label = new JLabel("Meters");

      cm_Label = new JLabel("Centimeters");

      yard_Label = new JLabel("Yards");

      clear = new JButton("Clear");

      calculate = new JButton("Calculate");

      exit = new JButton("Exit");

      inch_tf = new JTextField("0.00");

      cm_tf = new JTextField("0.00");

      meters_tf = new JTextField("0.00");

      yards_tf = new JTextField("0.00");

      JPanel panel1 = new JPanel();

      panel1.setLayout(new GridLayout(2,2));

      JPanel panel2 = new JPanel();

      panel2.setLayout(new GridLayout(2,2));

      JPanel panel3 = new JPanel();

      panel3.setLayout(new GridLayout(3,1));

      panel1.add(cm_Label);

      panel1.add(cm_tf);

      panel1.add(meter_Label);

      panel1.add(meters_tf);

      panel2.add(inch_Label);

      panel2.add(inch_tf);

      panel2.add(yard_Label);

      panel2.add(yards_tf);

      panel3.add(clear);

      panel3.add(calculate);

      panel3.add(exit);

      add(panel1);

      add(panel2);

      add(panel3);

      exitB = new exitButtonHandler();

      exit.addActionListener(exitB);

      clearB = new clearButtonHandler();

      clear.addActionListener(clearB);

      calcB = new calcButtonHandler();

      calculate.addActionListener(calcB);

      setVisible(true);

  }

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

     // {

       // new ConversionCalculator();

      //}

  private class exitButtonHandler implements ActionListener

  {

      public void actionPerformed(ActionEvent e)

      {

          //exits program

          System.exit(0);

      }

  }

  private class clearButtonHandler implements ActionListener

  {

      public void actionPerformed(ActionEvent e)

      {

          //clear button sets all text fields to 0

          inch_tf.setText("0.00");

          meters_tf.setText("0.00");

          yards_tf.setText("0.00");

          cm_tf.setText("0.00");

      }

  }

  private class calcButtonHandler implements ActionListener

  {

      public void actionPerformed(ActionEvent e)

      {

          double inches, yards, meters, cms;

          DecimalFormat df = new DecimalFormat("0.00");

          //parse strings in textbox into doubles

          inches = Double.parseDouble(inch_tf.getText());

          yards = Double.parseDouble(yards_tf.getText());

          meters = Double.parseDouble(meters_tf.getText());

          cms = Double.parseDouble(cm_tf.getText());

          //we check which value has been tampered with and base our conversion off this

          //because of this it is important that the user clears, or else it will do inch conversion

          if(inches != 0.00)

          {

              cms = inches * 2.54;

              meters = cms / 100;

              yards = inches / 36;

              cm_tf.setText(df.format(cms));

              meters_tf.setText(df.format(meters));

              yards_tf.setText(df.format(yards));

          }

          else if(yards != 0.00)

          {

              inches = yards / 36;

              cms = inches * 2.54;

              meters = cms / 100;

              cm_tf.setText(df.format(cms));

              meters_tf.setText(df.format(meters));

              inch_tf.setText(df.format(inches));

          }

          else if(meters != 0.00)

          {

              cms = meters * 100;

              inches = cms / 2.54;

              yards = inches / 36;

              cm_tf.setText(df.format(cms));

              inch_tf.setText(df.format(inches));

              yards_tf.setText(df.format(yards));

          }

          else if(cms != 0.00)

          {

              inches = cms / 2.54;

              yards = inches / 36;

              meters = cms / 100;

              meters_tf.setText(df.format(meters));

              inch_tf.setText(df.format(inches));

              yards_tf.setText(df.format(yards));

          }

      }

  }

}

Explanation: