You're asked to implement classes to manage personnels at a hospital. Create a class diagram that includes classes (and some variables and methods) that manage employees and patients in a hospital management program. Use interfaces and inheritance to create a hierarchy of all employees including doctors, nurses, and patients. Then, write the code outline of these classes. You don't have to implement any of the methods or a main method. Just add a simple comment or println statement describing what the method would do.

Python Code with Explanation:

# create a function named func to implement the required logic

def func():

# get the input from the user and store it in a variable named number

   number = int(input("Please enter an integer between 1 to 10 inclusive\n"))

# if the input number is equal or greater than 1 and equal to 10 or less then the input is correct    

   if number>=1 and number<=10:

# print correct input

       return print("Correct input")

# else the input is wrong

   else:

# print wrong input

       return print("Wrong input")

# call the function func

func()

Output:

Test 1:

Please enter an integer between 1 to 10 inclusive

4

Correct input

Test 2:

Please enter an integer between 1 to 10 inclusive

0

Wrong input

Test 3:

Please enter an integer between 1 to 10 inclusive

11

Wrong input

Answer:

dont know if it works!

Explanation:

input=(input("enter number")

if input > 1

print("correct input")

if input <10

print("correct input")

else

print("wrong input")

Answer:

The software can't locate the device

Explanation:

The device wasn't properly setup in order for the software to locate it.

Answer:

Water holing is the correct answer to this question.

Explanation:

Waterholing:-

It is a kind of attack in which the attacker detects the sites that the targets of the group frequently access and then afflicts the sites with the ransomware. Which afflicts selected representatives of the target group.

The watering hole assault is a data breach wherein the individual attempts to infiltrate a particular demographic of end-users by harming sites reported to be visited by team members. The aim is to compromise a specific target data and gain network access at the perpetrator's place of work.

Answer:

See explaination

Explanation:

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

Answer:

Explanation:

Program:

#include<iostream>

#include <bits/stdc++.h>

using namespace std;

//check for operator

bool isOperator(char c)

{

switch(c)

{

case '+': case '-': case '/': case '*': case '^':

return true;

}

return false;

}

//Converter class

class Converter

{

private:

string str;

public:

//constructor

Converter(string s):str(s){}

//convert from infix to postfix expression

string toPostFix(string str)

{

stack <char> as;

int i, pre1, pre2;

string result="";

as.push('(');

str = str + ")";

for (i = 0; i < str.size(); i++)

{

char ch = str[i];

if(ch==' ') continue;

if (ch == '(')

as.push(ch);

else if (ch == ')')

{

while (as.size() != 0 && as.top() != '('){

result = result + as.top() + " ";

as.pop();

}

as.pop();

}

else if(isOperator(ch))

{

while (as.size() != 0 && as.top() != '(')

{

pre1 = precedence(ch);

pre2 = precedence(as.top());

if (pre2 >= pre1){

result = result + as.top() + " ";

as.pop();

}

else break;

}

as.push(ch);

}

else

{

result = result + ch;

}

}

while(as.size() != 0 && as.top() != '(') {

result += as.top() + " ";

as.pop();

}

return result;

}

//return the precedence of an operator

int precedence(char ch)

{

int choice = 0;

switch (ch) {

case '+':

choice = 0;

break;

case '-':

choice = 0;

break;

case '*':

choice = 1;

break;

case '/':

choice = 1;

break;

case '^':

choice = 2;

default:

choice = -999;

}

return choice;

}

};

//Node class

class Node

{

public:

string element;

Node *leftChild;

Node *rightChild;

//constructors

Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}

Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}

};

//ExpressionTree class

class ExpressionTree

{

public:

//expression tree construction

Node* covert(string postfix)

{

stack <Node*> stk;

Node *t = nullptr;

for(int i=0; i<postfix.size(); i++)

{

if(postfix[i]==' ') continue;

string s(1, postfix[i]);

t = new Node(s);

if(!isOperator(postfix[i]))

{

stk.push(t);

}

else

{

Node *r = nullptr, *l = nullptr;

if(!stk.empty()){

r = stk.top();

stk.pop();

}

if(!stk.empty()){

l = stk.top();

stk.pop();

}

t->leftChild = l;

t->rightChild = r;

stk.push(t);

}

}

return stk.top();

}

//inorder traversal

void infix(Node *root)

{

if(root!=nullptr)

{

cout<< "(";

infix(root->leftChild);

cout<<root->element;

infix(root->rightChild);

cout<<")";

}

}

//postorder traversal

void postfix(Node *root)

{

if(root!=nullptr)

{

postfix(root->leftChild);

postfix(root->rightChild);

cout << root->element << " ";

}

}

//preorder traversal

void prefix(Node *root)

{

if(root!=nullptr)

{

cout<< root->element << " ";

prefix(root->leftChild);

prefix(root->rightChild);

}

}

};

//main method

int main()

{

string infix;

cout<<"Enter the expression: ";

cin >> infix;

Converter conv(infix);

string postfix = conv.toPostFix(infix);

cout<<"Postfix Expression: " << postfix<<endl;

if(postfix == "")

{

cout<<"Invalid expression";

return 1;

}

ExpressionTree etree;

Node *root = etree.covert(postfix);

cout<<"Infix: ";

etree.infix(root);

cout<<endl;

cout<<"Prefix: ";

etree.prefix(root);

cout<<endl;

cout<< "Postfix: ";

etree.postfix(root);

cout<<endl;

return 0;

}

Answer:

A) 22 ⅓

B) 111>18

C) There is an error in the expression

D) 25

E) 62500

Question:

Evaluate the following expressions

A) 5 + 2 * (x + 4) / 3, where x has a value of 18

B) 7/2 * 2 + 4 * (5 + 7 * 3) & gt; 18

C) 2 <= x AND x<= 7, where x has value 23

D) 50% 10 * 5 =

F) 31250/5/5 * 2 =

Explanation:

A) 5 + 2 * (x + 4) / 3

x = 18

First we would insert the value of x

5 + 2 * (x + 4) / 3

5 + 2(18 + 8) / 3

Then we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.

= 5 + 2(26) / 3

= 5 + 52/3

= 5 + 17 ⅓

= 22 ⅓

B) 7/2 * 2 + 4 * (5 + 7 * 3) > 18

we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.

7/2 * 2 + 4 * (5 + 7 * 3) >18

= 7/2 × 2 + 4× (5 + 7 × 3)>18

= (7×2)/2 + 4× (5+21) >18

= 14/2 + 4(26) >18

= 7 + 104 >18

= 111>18

C) 2 <= x AND x<= 7, where x has value 23

There is an error in the expression

D) 50% of 10 * 5

we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.

The 'of' expression means multiplication

= 50% × 10×5

= 50% × 50

50% = 50/100

=50/100 × 50

= 1/2 × 50

= 25

F) 31250/5/5 * 2

The expression has no demarcation. Depending on how it is broken up, we would arrive at different answers. Let's consider:

31250/(5/5 × 2)

Apply BODMAS

= 31250/[5/(5 × 2)]

= 31250/(5/10)

= 31250/(1/2)

Multiply by the inverse of 1/2 = 2/1

= 31250 × (2/1)

= 62500

Answer:

There are no limitation, policies or regulations that limit these tools for use on privately owned machines or home networks. In most businesses networks, intranets or internets the use of them is illegal if used for malicious intent. Penetration testing teams sign rules of engagement before using these tools.

Explanation:

There are no limitation, limitations, policies, or regulations in their use on local machines, privately owned machines or home networks, or small business networks.

In most businesses networks, intranets or internets the use of them is often and mostly illegal especially in a situation where they are been used for malicious intent.

Penetration testing teams would often or always make sure that they sign rules of engagement before using these tools.

Answer: Provided in the explanation segment

Explanation:

The following code provides answers to the given problem.

import java.util.*;

public class untitled{

public static void main(String[] args)

{int tens,fives,ones,quarters,dimes,nickles,pennies,amt;

double amount;

Scanner in = new Scanner(System.in);

System.out.println("What do you need to find the change for?");

amount=in.nextDouble();

System.out.println("Your change is");

amt=(int)(amount*100);

tens=amt/1000;

amt%=1000;

fives=amt/500;

amt%=500;

ones=amt/100;

amt%=100;

quarters=amt/25;

amt%=25;

dimes=amt/10;

amt%=10;

nickles=amt/5;

pennies=amt%5;

System.out.println(tens +" ten dollar bills");

System.out.println(fives +" five dollar bills");

System.out.println(ones +" one dollar bills");

System.out.println(quarters +" quarters");

System.out.println(dimes +" dimes");

System.out.println(nickles +" nickles");

System.out.println(pennies +" pennies");

}

}

cheers i hope this helped !!

Answer:

You should name it honest and sincere intention

Explanation:

That is a good name for what you are going to talk about in your slideshow.

Answer:

See explaination

Explanation:

clc;

clear all;

close all;

x=input(' choose abc or fgh or mno:','s');

to=input('enter intial fluid temperature in celcius:');

t1=input('enter time in minutes:');

ta=input('enter ambient temperature in celcius:');

abc=1;

fgh=2;

mno=3;

if x==1

k=0.01;

elseif x==2

k=0.03;

else

k=0.02;

end

t=ta+((to-ta)*exp((-k)*t1));

X = sprintf('%s has temp %f degrees celcius after %d minutes.',x,t,t1);

disp(X);

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:

Is there a picture of the question?

Explanation:

Answer:

Check the explanation

Explanation:

PSEUDO CODE:

for(int i=0,j=0;i<phone.length,j<company.length;i++,j++)

{

print(phone[i]," - ",company[j]);

}

this for loop will print each phone number associated with the company names.

Answer:

Check the explanation

Explanation:

solution a:

def Rotate(string) :

n = len(string)

temp = string + string

for i in range(n) :

for j in range(n) :

print(temp[i + j], end = "")

print()

string = ("abcde")

Rotate(string)

solution b:

nums = [3,9,5,8,2,4,7]

res = list(filter(lambda n : n < 5, nums))

print(res)

Answer:

The answer is 1646

Explanation:

The original code requires 1024 bytes and is called 70 times ,it requires 7 byte and its size is 132 bytes

1024 + (70*7) + 132 = 1024 + 490 +132

= 1646

Answer:

The output of the code:

Enter a 4 digit integer : 1 2 3 4

The decrypted number is : 0 1 8 9

The original number is : 1 2 3 4

Explanation:

Okay, the code will be written in Java(programming language) and the file must be saved as "Encryption.java."

Here is the code, you can just copy and paste the code;

import java.util.Scanner;

public class Encryption {

public static String encrypt(String number) {

int arr[] = new int[4];

for(int i=0;i<4;i++) {

char ch = number.charAt(i);

arr[i] = Character.getNumericValue(ch);

}

for(int i=0;i<4;i++) {

int temp = arr[i] ;

temp += 7 ;

temp = temp % 10 ;

arr[i] = temp ;

}

int temp = arr[0];

arr[0] = arr[2];

arr[2]= temp ;

temp = arr[1];

arr[1] =arr[3];

arr[3] = temp ;

int newNumber = 0 ;

for(int i=0;i<4;i++)

newNumber = newNumber * 10 + arr[i];

String output = Integer.toString(newNumber);

if(arr[0]==0)

output = "0"+output;

return output;

}

public static String decrypt(String number) {

int arr[] = new int[4];

for(int i=0;i<4;i++) {

char ch = number.charAt(i);

arr[i] = Character.getNumericValue(ch);

}

int temp = arr[0];

arr[0]=arr[2];

arr[2]=temp;

temp = arr[1];

arr[1]=arr[3];

arr[3]=temp;

for(int i=0;i<4;i++) {

int digit = arr[i];

switch(digit) {

case 0:

arr[i] = 3;

break;

case 1:

arr[i] = 4;

break;

case 2:

arr[i] = 5;

break;

case 3:

arr[i] = 6;

break;

case 4:

arr[i] = 7;

break;

case 5:

arr[i] = 8;

break;

case 6:

arr[i] = 9;

break;

case 7:

arr[i] = 0;

break;

case 8:

arr[i] = 1;

break;

case 9:

arr[i] = 2;

break;

}

}

int newNumber = 0 ;

for(int i=0;i<4;i++)

newNumber = newNumber * 10 + arr[i];

String output = Integer.toString(newNumber);

if(arr[0]==0)

output = "0"+output;

return output;

}

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

System.out.print("Enter a 4 digit integer:");

String number = sc.nextLine();

String encryptedNumber = encrypt(number);

System.out.println("The decrypted number is:"+encryptedNumber);

System.out.println("The original number is:"+decrypt(encryptedNumber));

}

}

Answer:

public class RectangleArea {

   public static void main(String[] args) {

       Rectangle myRectangle = new Rectangle(15.8, 7.9);

       double calArea = myRectangle.calcArea(15.8, 7.9);

       System.out.println("The area is "+ calArea);

   }

}

class Rectangle{

   private double length;

   private double width;

   //Defined Constructor

   public Rectangle(double length, double width) {

       this.length = length;

       this.width = width;

   }

   public double calcArea(double len, double wi){

       return len* wi;

   }

}

Explanation:

Answer:

See explaination

Explanation:

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

Answer:

WOW! that does not look easy!

Explanation:

I wish i could help but i have no idea how to do that lol

Answer:

99.6

Explanation:

The cloud provider  provides the certain modules of the cloud service like infrastructure as a service , software as a service ,etc to other companies .The cloud provider implements a new SaaS service that includes a cloud platform also the cloud provider intends to disclose the Service Level Agreement that is score based on its approximate accessibility during the next 12 months.

The Cloud providers predict that, depending on past cloud system data, it is a 25 % risk that a physical server holding the cloud platform will fail therefore such a failure will require 2 days to recover the cloud service so 99.6 is the Cloud storage accessibility ranking, to be released in the SLA.

Answer:

And user documentation consists of instructions and information to users who will interact with the system

Explanation:

user documentation will not allow user to prepare overall documentation because it is already prepared so not A.

user documentation does not need implementation, So not B.

user documentation may not be a printed one . So not C

And user documentation consists of instructions and information to users who will interact with the system, so D is the right option

Answer:

The following are the program in the C++ Programming Language.

//header files

#include <iostream>

#include <vector>

//using namespace

using namespace std;

//define a function

int find(vector<int> &vec, int n)  

{

   //set the for loop

   for(int i = 0; i < vec.size(); ++i)  

   {

       //check the elements of v is in the num

       if(vec[i] == n)  

       {

           //then, return the values of i

          return i;

       }

   }

  return -1;

}

//define the main method

int main()  

{

   //declare a integer vector type variable

   vector<int> vec;

   //print message

   cout << "Enter numbers: ";

   //declare integer type variable

   int n;

   //set the for loop

   for(int i = 0; i < 20; ++i)  

   {

       //get input in the 'num' from the user

       cin >> n;

       //check the following conditions

       if(n >= 10 && n <= 100 && find(vec, n) == -1)  

       {

           //then, assign the values of 'num' in the vector

          vec.push_back(n);

       }

   }

   //print the message

   cout << "User entered: ";

   //set the for loop

   for(int i = 0; i < vec.size(); ++i)  

   {

       //print the output

       cout << vec[i] << " ";

  }

  cout << "\n";

  return 0;  

}

Output:

Enter numbers: 14 18 96 75 23 65 47 12 58 74 76 92 34 32 65 48 46 28 75 56

User entered: 14 18 96 75 23 65 47 12 58 74 76 92 34 32 48 46 28 56

Explanation:

The following are the description of the program:

Answer:

Check the explanation

Explanation:

a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.

(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.

Explanation:

a) A compare validator: A compare validator can be described as a particular method that is utilized to control and also enabled to perform different types of validation tasks. It is being used to carry out as well as to execute a correct data type that determines the entered value is proper value into a form field or not. For instance when in an application registration it makes use of the confirm password after the first-password.

(b) Whenever it is compulsory to determine if the value that has been entered is the correct value into the form field then the usage of compare validator will be needed whereas a range validator is used to confirm the entered value is in a specific range.

Answer:

see explaination

Explanation:

a)Worst Case-time complexity=O(n)

def intersection_list(lst1, lst2):

lst3 = [value for value in lst1 if value in lst2]

return lst3

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele) # adding the element

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele) # adding the element

print(intersection_list(lst1, lst2))

b)Average case-time complexity=O(min(len(lst1), len(lst2))

def intersection_list(lst1, lst2):

return list(set(lst1) & set(lst2))

lst1 = []

lst2 = []

n1 = int(input("Enter number of elements for list1 : "))

for i in range(0, n1):

ele = int(input())

lst1.append(ele)

n2 = int(input("Enter number of elements for list2 : "))

for i in range(0, n2):

ele = int(input())

lst2.append(ele)

print(intersection_list(lst1, lst2))

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();

=======================================================

//Class header definition

public class DisArray {

   //First method with int array as parameter

   public static void convert2D(int[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the

       //length of  the one dimensional array,

       //then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       // number of rows

       int col = ((row * row) >= arraylength) ? row : row + 1;

       //3. Create a 2D int array with the number of rows and cols

       int [ ][ ] twoD = new int [row][col];

       //4. Place the elements in the one dimensional array into

       //the two dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       // two  dimensional array.

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

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

              }

              //Otherwise, just pad the array with zeros

               else {

                   twoD[i][j] = 0;

               }

           }

       }

       //You might want to create another pair of loop to print the elements

       //in the  populated two dimensional array as follows

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

           for (int j = 0; j < twoD[i].length; j++) {

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

           }

           System.out.println("");

       }

   }     //End of first method

   //Second method with String array as parameter

   public static void convert2D(String[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the length of  

       //the one dimensional array, then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       //number of rows.

       int col = (row * row >= arraylength) ? row : row + 1;

       //3. Create a 2D String array with the number of rows and cols

       String[][] twoD = new String[row][col];

       //4. Place the elements in the one dimensional array into the two

       // dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       //two  dimensional array.

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

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

               }

               //Otherwise, just pad the array with null values

               else {

                   twoD[i][j] = null;

               }

           }

       }

       //You might want to create another pair of loop to print the elements  

       //in the populated two dimensional array as follows:

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

           for (int j = 0; j < twoD[i].length; j++) {

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

           }

           System.out.println("");

       }

   }    // End of the second method

   //Create the main method

   public static void main(String[] args) {

       //1. Create an arbitrary one dimensional int array

       int[] x = {23, 3, 4, 3, 2, 4, 3, 3, 5, 6, 5, 3, 5, 5, 6, 3};

       //2. Create an arbitrary two dimensional String array

       String[] names = {"abc", "john", "dow", "doe", "xyz"};

       //Call the respective methods

       convert2D(x);

       System.out.println("");

       convert2D(names);

   }         // End of the main method

}            // End of class definition

=========================================================

==========================================================

23 3 4 3  

2 4 3 3  

5 6 5 3  

5 5 6 3  

abc john dow  

doe xyz null

==========================================================

The above code has been written in Java and it contains comments explaining each line of the code. Please go through the comments. The actual executable lines of code are written in bold-face to distinguish them from the comments.

Answer:

Explanation:

Please find attachment for the Python program

Answer:

Following are the step to this question:

i) search "intitle passwd" in google search engine.

ii) Open first link of the page.

iii) use commands to access data.  

Explanation:

Google Hacking operates as follows of web engines like google and yahoo to find web pages and sensitive information, that are insecure. It builds on the idea that web pages index a large number of public pages and files and make their discovery easy to create the correct query.

Explanation:

The most reliable hard drives are those whose most common size is 3.5 inches, their advantages are their storage capacity and their speed and a disadvantage is that they usually make more noise.

For greater speed, it is ideal to opt for two smaller hard disks, since large disks are slower, but are partitioned so that there are no problems with file loss.

For a person who needs to use content creation programs, it is ideal to opt for a hard drive that has reliability.