Explain how you would assess the operating system requirements for virtualization if the organization wanted to virtualize 25% of physical stand-alone servers.

Answer:

d) selector patterns

Explanation:

To apply and create rules to elements in the same class, selector patterns are a simple solution to make the task easier.

Selector patterns allow you to have a set of rules, tools already tested, which allow you to solve most problems directly.

To direct a style rule to specific elements, selector patterns are used. Selectors can be a tag, class, or ID and are specified at the start of a CSS rule followed by curly braces containing the style declarations.

Using Selectors in CSS

To direct a style rule to specific elements, selector patterns can be used to match only those page elements that correspond to a specified pattern. The selector is the first part of a CSS rule that identifies the HTML elements the subsequent styles will apply. It can reference an HTML tag directly, or it can be a custom class, ID or even a combination of selectors for more specificity. Once you specify a selector, you can define styles inside the curly braces {} that follow.

For example, using a descendant selector like .completed ul, you can target all <ul> elements within an element with the class .completed, and apply a specific style such as text-decoration: line-through; to them. This type of specificity ensures that styles are applied precisely to the elements intended. The colon (:) is used in CSS to assign values rather than an equal sign, which is typical in many programming languages.

When defining classes in our CSS file, we precede the name with a period to indicate it is a class and use the class attribute within the HTML document to apply it to elements. Classes allow us to apply different styles to the same type of elements, enhancing the page's design and user experience.

Answer:

A. Utilize the native VLAN only on trunk ports to reduce the risk of a Double-Tagged 802 1q VLAN hopping attack

C. Utilize DHCP snooping on a per VLAN basis an apply IP DHCP snooping entrusted on all ports.

D. Utilize the ARP inspection feature to help prevent the misuse of garpe Utilize private VLANs an ensure that all ports are part of the isolated port group.

Explanation:

We must configure the native VLAN only on trunk ports because in this way we are not to receive fake VLAN, and can steal information.

We can use DHCP snooping to configure trusted ports and entrusted ports, in this case, the trusted port can accept trust messages, only with those messages we can connect with the server.

We could ask the MAC address with ARP inspections, MAC address is unique value for a physical hardware in the world.

Answer:

Option A is the correct answer to the following question.

Explanation:

Because multitasking is the process of an operating system in which provides users to accomplish more than one jobs at the same time, for example, we can listen music with creating documents in word such as we can do multiple jobs at a time by which we can save much more time to do other works. So, that's why the following option is correct.

A radiologist’s computer monitor has higher resolution than most computer screens. Resolution refers to the: number of horizontal and vertical pixels on the screen.

Explanation:

The resolution of a screen refers to the size and number of pixels that are displayed on a screen. The more pixels a screen has, the more information is displayed on it without the user having to scroll for it.

More pixels also define and enhance the clarity of the pictures and content displayed on the screen. Pixels combine in different ratios to provide different viewing experiences for users.

Answer:

              .

Explanation:

Answer:

The function solution is in JavaScript

Explanation:

function isRoundedNumberEven(decimal) {

   return Math.round(decimal) % 2 === 0;

/* the modulus 2 operation is used for numbers divisible by 2 */

/* This indicates even numbers */

}

console.log(isRoundedNumberEven(2.2));

console.log(isRoundedNumberEven(2.8));

Answer:

(a) - True

(b)- False

(c)-  False

(d) - False

(e) - False

(f) - False

(g)- True

(h)-  False

(i) - False

(j) - False

Explanation:

(a)-  In an adjacency list the usage of space depends upon the number of edges and vertices in the graph while in the case of the matrix efficiency depends upon the square of the vertices.

(b) The maximum number of edges is n*(n-1).

(c) The number of edges with a spanning tree of n vertices contains n-1 edges.

(d) Dijkstra's Theorem simply refers the adjacent vertices of a vertex and is independent of the fact that a graph is directed or in directed.

(e)  Greedy Algorithm makes short sighted choices

(f) Consider the term overlapping sub-problems which is the technique used in Dynamic programming it uses previous calculations to find optimal solutions however in case of merge sort the idea is to break down array into smaller pieces that don't overlap unlike dynamic programming.

(g) Prim and Kruskal's algorithm are examples of greedy algorithm because the use the idea to pick the minimum weight edges at every step which is the approach of a greedy algorithm.

(h) the amount of flow on an edge cannot exceed its capacity.

(i) Dynamic Programming can be implemented using tabulation too.

(j) Different data structures have different time complexity.

Network scan and banner grab are the following tasks should be done FIRST.

A. Network scan

B. Banner grab

Explanation:

The penetration tester has been assigned a task to determine which ports are open on the given network.

To kick start the process, the tester should give a network scan in order to gather information about the various elements of the network and their details.

To follow up, the tester should proceed with banner grab. Banner grabbing is the process through which he/she would gain information about a computer system on the network and services running on its open ports.

Answer:

Programming language not stated.

I'll use python for this question

Explanation:

import random

# library that we use in order to choose

# on random words from a list of words

words = ['rain, 'computer', 'science', 'program, 'python', 'mathematics', 'player', 'condition','reverse', 'water', 'board', 'geeks','learn','school','days','scholar','collar','flood','house','flies']

# Function will choose one random word from this list of words

word = random.choice(words)

print("Guess the word")

guesses = ''"

# 5 turns

turns = 5

while turns > 0:

# counts the number of times a user fails

failed = 0

# all characters from the input word taking one at a time.

for char in word:

# comparing that character with the character in guesses

if char in guesses:

print(char)

else:

print("_")

# for every failure 1 will be incremented in failure

failed += 1

if failed == 0:

# user will win the game if failure is 0 and 'You Win' will be given as output

print("You Win")

# this print the correct word

print("The word is: ", word)

break

# if user has input the wrong alphabet then it will ask user to enter another alphabet

guess = input("guess a character:")

# every input character will be stored in guesses

guesses += guess

# check input with the character in word

if guess not in word:

turns -= 1

# if the character doesn’t match the word then “Wrong” will be given as output

print("Wrong")

# this will print the number of turns left for the user

print("You have", + turns, 'more guesses')

if turns == 0:

print("You Loose")

Answer:

a) Custom AppExchange-app for product-pricing

Explanation:

We can find a solution in AppExchange for this product-pricing, surfing in the option, we could get solutions like BoonPlus, and easy pricing Opportunity, this App has a free trial.

With this option is easy to choose a book price, add new products, select pricing rule, we can get an order's product, and calculate pricing, just we must download the app and install it.

Answer:

D. The port is removed from the EtherChannel group.

Explanation:

The configuration SPAN is when we copy traffic, and we transfer that traffic to another port, in this case, we can analyze that traffic in our benefit.

If a physical port belongs to the EtherChannel group, it can be configured as a SPAN source port and still be a part of the EtherChannel.

But a physical port cannot be configured as a SPAN destination and belong to the EtherChannel group.

This means we can be the SPAN source but not the SPAN destination.

Answer:

The answer is "Option 3".

Explanation:

In the given question the code (ii) is correct because in the class definition class "Student" defines a constructor that does not use any return type. and code (i) class definition the class "Student" defines a constructor that uses return type void which is not allowed in the constructor. and other options are not correct that can be described as follows:

Final answer:

In Excel, to calculate the time an item has been on sale, use the formula =($C$20-D4)/365 in cell E4 and copy it down to E5:E14. This formula subtracts the launch date from the Data Updated date and converts the result to years using an absolute reference for the Data Updated date.

Explanation:

To calculate how long each item has been on sale in an Excel spreadsheet, you would subtract the launch date in cell D4 from the Data Updated date in cell C20 and then divide by 365 to convert the difference to years. Since cell C20 should not change when you copy the formula, you need to use an absolute reference for this cell.

The formula in cell E4 will be =($C$20-D4)/365. This formula uses the absolute reference $C$20 to ensure that as you copy the formula from cell E4 to cells   E5:E14, the reference to C20 remains constant.

After entering the formula in E4, you would then copy it down to the cells E5 through E14. The result will show you the time each item has been on sale in years.

Answer:

In order to ensure that only authorized users can send print jobs to the printer, the option of enabling the user authentication on shared printer can be used.

Explanation:

Answer:

1. Performance and uptime: In mission critical IT systems like those in the military performance and uptime are so important that the institutions might be reluctant to surrender their local infrastructure for cloud based systems, even if the cloud systems advertise better performance than what the institution's deployment can boast of. Even many corporates are still worried about this aspect of cloud computing. It’s no secret that a fraction of a second on load time can lead to loss of customers.

2. Security: Along with loss of control, security is arguably the biggest concern many large organizations like the military, academic institutions and hospitals have. Can other customers access our data? Have all the security patches been kept up to date? These question create a reluctance to migrate to the cloud among executives and IT directors in these institutions.

3. Data Protection: Closely tied in with security, institutions are concerned about data protection. Many governments place strict data protection requirements on large companies and standards audit schemes such as ISO-9001 place additional restrictions on firms.

For example, the UK’s Data Protection Act requires that personal data not to be transferred to any country outside the EU, unless that country “ensures an adequate level of protection for the rights and freedoms of data subjects in relation to the processing of personal data.” Practically, that means it’s often easier for servers to be physically located in the EU for many European institutions.

4. Losing control: When systems go down, it’s invariably the IT Director who takes the blame, so it’s no surprise that they are nervous about handing over responsibility for infrastructure to somebody they don’t even know.

Traditionally, if a hard drive fails or a CPU overheats, it’s the IT department who get the 3am call to fix it. It’s up to them to order the replacement hardware, power down the machine and get it back into production as soon as possible.

Even if you’re renting space in a hosting centre and paying for support, chances are you are physically allowed into the building and have the peace of mind that comes from knowing that, if all else fails, you can go in and fix the hardware yourself.

By moving to cloud computing, the IT Director or CIO can feel like they’re handing over control to a 3rd-party. What happens if one of the drives goes or there’s a networking issue? Who decides what hardware to buy?

Answer:

string1="There is a tide in the affairs of men, Which taken at the flood, leads on to fortune. Omitted, all the voyage of their life is bound in shallows and in miseries. On such a full sea are we now afloat. And we must take the current when it serves, or lose our ventures."

string1=string1.lower()   #covert whole string to lowercase

letter_counts={}

for ch in string1:

   letter_counts[ch]=letter_counts.get(ch,0)+1   #updates counter for each letter

letter_counts['o']

OUTPUT :

17

Explanation:

lower() is used to change the string to lowercase characters so that uppercase and lowercase is not differentiated for the same letter. After that, letter_counts is declared as a dictionary which is empty. for loop is used in which every letter of string will be stored ch variable. Dictionary in python has a get method in which 2 arguments can be passed, first one is the key whose value to be retrieved and the other argument would be default value to be set if not found and ch in string1 will increment the value of key ch in letter_counts.

Answer:

b. Standard Lightning component.

c. Custom Lightning component.

d. Global actions

These three components are best used in a ligthening app, for custom object i.e standard, Custom and Global Actions.

Adding Visualforce components to a Lightning Application is impossible, but one can add Lightning components into Visualforce pages.

Answer:

The Java code is given below

Explanation:

import java.io.*;

import java.util.Scanner;

import java.util.ArrayList;

public class StringCheck{

//There are several implementation approaches. This is one example framework/outline which might be helpful. Please feel free to try other approaches.

public static void checkWord(String word) throws Exception {

// Uses charAt method to test if the first letter of string variable word

// is a character. If not, throw new exception

if(Character.isLetter(word.charAt(0))){

 return;

}else{

 throw new Exception("This is not a word");

}

}

public static String getWord() {

// Declare a local scanner

// Prompt the user to enter a word

// Think about using a loop to give the user multiple opportunities to correctly enter a string

// Read into a string

// Call checkWord method passing the string as a parameter

// checkWord can throw an exception; call checkWord in a try/catch block

// Return the string to main if a valid string

Scanner sc = new Scanner(System.in);

boolean st = true;

String word = null;

while(st){

 System.out.println("Enter a Word : ");

 word = sc.next();

 try{

  checkWord(word);

  st = false;

 }catch(Exception e){

  System.out.println(e);

  st = true;

 }

}

sc.close();

return word;

}

public static void writeFile(String[] arrayToWrite, String filename) throws IOException {

// Example using FileWriter but PrintWriter could be used instead

// Create a FileWriter object

FileWriter fileWordStream = new FileWriter(filename);

// Use a loop to write string elements in arrayToWrite to fileWordStream

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

fileWordStream.write(arrayToWrite[i]+System.lineSeparator());

}

fileWordStream.flush();

fileWordStream.close();

// In the loop use the lineSeparator method to put each string on its own line

}

public static ArrayList readFile(String filename) throws FileNotFoundException, IOException {

// Declare local ArrayList

ArrayList<String> words = new ArrayList<>();

// Create a new File object using filename parameter

File file = new File(filename);

// Check if the file exists, if not throw a new exception

if(!file.exists()){

 throw new FileNotFoundException("File not Found");

}

// Create a new BufferedReader object      

Scanner fsc = new Scanner(file);

// use a loop and the readLine method to read each string (on its own line) from the file

while(fsc.hasNextLine()){

words.add(fsc.nextLine());

}

fsc.close();

// return the filled ArrayList to main

return words;

}

public static void main(String[] args) {

// create a string with literal values to write to the file

String[] testData = {"cat", "dog", "rabbit"};

// Create an ArrayList for reading the file

ArrayList<String> words = new ArrayList<>();

// Declare a string variable containing the file name "data.txt"

String file = "data.txt";

// Call getWord, assign the returned string to a variable and display it

String word = getWord();

System.out.println("The word is : "+word);

// Call writeFile and readFile methods in a try block

try{

writeFile(testData,file);

words = readFile(file);

// Printout the contents of the ArrayList after the call to readFile

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

 System.out.println(words.get(i));

}

}catch(FileNotFoundException e){

System.out.println(e);

}catch(IOException e){

System.out.println("IOException: error occured in input output");

}

// catch two types of exceptions

}

}

Final answer:

To create the 'CheckString' Java application, you need to write a method called 'checkWord' that throws an Exception if the word doesn't start with a letter, and another method called 'getWord' that prompts the user for input and handles any Exceptions thrown by 'checkWord'.

Explanation:

To create a new Java application called CheckString, you need to write two methods. The method checkWord will take a String parameter and throw an Exception if the word does not start with a letter. The method getWord will prompt the user for a word, call checkWord, and handle any Exceptions.

Method: checkWord

public void checkWord(String word) throws Exception {
  if (Character.isLetter(word.charAt(0)) == false) {
      throw new Exception("This is not a word.");
  }
  // otherwise, the input is considered a valid word
}

Method: getWord

public String getWord() {
  Scanner scanner = new Scanner(System.in);
  while (true) {
      try {
          System.out.print("Enter a word: ");
          String word = scanner.nextLine();
          checkWord(word);
          return word;
      } catch (Exception e) {
          System.out.println(e.getMessage());
      }
  }
}

Answer:

The algorithm is very similar to that of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Explanation:

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Correctness Proof: We will prove a stronger statement that on any input A our algorithm produces correctly the sorted version of A and the number of significant inversions in A. And, we will prove it by induction. Base Case: When n = 1, the algorithm returns 0, which is obviously correct.

Inductive Case: Suppose that the algorithm works correctly for 1, 2, . . . , n − 1. We have to show that the algorithm also works for n. By the induction hypothesis, B, x, C, y are correct. We need to show that the counting of significant split inversions and the normal merge-sort process are both correct. In other words, we are to show that z is the correct number of significant split inversions, and that D is correctly sorted.

First, we prove that z is correct. Due to the fact that B and C are both sorted by the induction hypothesis, for every C[j], where j = 1, . . . , length(C), we can always find the smallest i such that B[i] > 2C[j]. Therefore, B[1], B[2], . . . , B[i − 1] are all smaller than or equal to 2C[j], and B[i], B[i + 1], . . . , B[length(B)] are all greater than 2C[j]. Thus, length(B) − i + 1 is the correct significant split inversion count for pairs involving C[j]. After counting the number of correct significant split inversions for all C[j], where j = 1, . . . , length(C), z is the correct number of significant split inversions. Second, we show that D is the sorted version of A.

This is actually the proof of merge-sort. Each time, we append one element from B or C to D. At each iteration k, since B and C are both sorted by the induction hypothesis, min(B[i], C[j]), the one to be appended to D, is also the smallest number among all the numbers that have not been added to D yet.

As a result, every time we append the smallest remaining element (of B and C) to D, which makes D a sorted array

Answer:

The answer is "option B".

Explanation:

An OTB stands for an over-target baseline, it is a new management benchmark when an original target can't have coincided and new goals for management purposes are required.

Answer:

The main page of a website a visitor see while navigating, known as Home page.

Explanation:

The steps by which we can easily change our home page of Google Chrome, are as follows:

Step 1 :- Open Google Chrome.

Step 2 :- Click on the three vertical dots which appears next to the icon (profile).

Step 3 :- Click on settings, turn 'Show home button' to ON.

Step 4 :- Choose the new tab page you had like to use.

Answer:

First you would want to search in the bottom left search bar "settings", once you're there you would want to go to the top of the settings search bar and search up home page. Once you're there you can do whatever you want to do with your homepage.

Explanation:

got it right on edg.

Answer:

The answer is "Option (i)".

Explanation:

In the given java code a class is defined that name is "Rectangle" inside a class two global private double datatype variable is defined that is "length and width". A default constructor is defined that contains the value of private variables. In the next line, a parameterized constructor and a set function is defined that contain the parameter values. Then three function is defined that is "print, area, and perimeter" in which print function prints value and area, perimeter function calculate values.

Answer:

Vertical Scaling

Explanation:

Vertical scaling means the ability to increase the capacity of existing hardware or software by adding resources. In this case, more GB is added (8GB) to the server to address the problem. So vertical scaling is the resultant type john is expecting.

I could describe my leadership and service as being motivational and more a credit-giver.  

I do not own the credit for every successful task that my team has accomplished. In fact, I am giving them all the recognitions they deserved. In times of struggling and great pressure, I would motivate them. Most of their complaints about my leadership method is I do not give importance to time. I normally tend to finish everything before deadlines. With these attitudes, my family learn to adapt to this behavior. They value ideas and teamworks in their respective workplace and like me, they would also hate to submit their works late. They would sometimes get positive feedback from clients for finishing their assigned tasks on time.

Answer:

A. Questioning

Explanation:

Questioning is the skill of a scrummaster that is most closely alligned with coaching, inspite of others.

Final answer:

The correct answer is "Questioning". The ScrumMaster's skill of questioning is closely aligned with coaching, as it encourages discussion, reflection, and problem-solving within the team.

Explanation:

The role of a ScrumMaster in the Scrum methodology of Agile software development includes various responsibilities that contribute to the success of the project. When considering which characteristics or skills are closely aligned with coaching within the ScrumMaster's role, the most relevant option among those provided would be questioning. A ScrumMaster utilizes questioning to facilitate discussion, encourage team reflection, and promote problem-solving abilities, all of which are essential coaching techniques. This helps the team find solutions to their impediments and continuously improve their processes, aligning with the ScrumMaster's goal of coaching the team towards self-organization and leveraging their full potential.

Answer:

(a). IP routing.

(b). Multicasting.

(c). converged.

(d). VoIP.

(e). VoIP.

(f). Plain text.

Explanation:

Ip routing is the method of transfer data from the one destination to another destination and it also send data from host to host across the network.

Multicasting is the type of distribution from one host to many hosts and also many hosts to many of them, it is used for the Internet Protocol Routing protocol in which the data is distributed to many receivers.

VoIP is that type of routing through which folks can use the internet services as the transmission mode.

Final answer:

Packet switching is based on travelling datagrams one hop at a time through internetworks. Multicast is group-based communication. A QoS network treats different data types differently.

Explanation:

a. **Packet switching** is based on travelling datagrams through internetworks one hop at a time. The entire route is unknown at the beginning of the journey.

b. **Multicast** is group-based communication. It can be one-to-many or many-to-many distribution.

c. The most important requirement of a **Quality of Service (QoS)** network is the ability to treat different data types differently.

d. **VoIP (Voice over IP)** process transmits telephone calls (voice) over the internet.

e. **G.729** is the standard recommended by the ITU for low bit-rate voice transmission over the Internet.

f. **Clear text** is an unencrypted message data.

Answer:

Trojan horse

Explanation:

A Trojan horse is malicious software that masquerades as a legitimate computer program such as a video game, image file, utility program, or even an antivirus so as to entice the user to run it, thereby allowing an attackers (the developer of the malicious program) to gain access to the data on the computer in an illegal manner.

Answer:

By unchecking the 'allow duplicate' box and checking the 'Block duplicate' box  in the SalesForce user interface. More also, including  a column in the import file that has other record name(SF ID)  will ensure no duplicate during import.

Explanation:

importing 1000 records is quiet lengthy and to avoid duplicate  it is important to include in the import file that has other record name(SF ID). By including a column that has another file ID every record will be referenced uniquely thereby easily identifying any duplicate.

It is also important to  check the ''Block duplicate' box in the salesForce user interface to ensure that records are imported without duplicate.

Answer:

Code given below

Explanation:

/*

* Class to hold the Passenger data

* Stores the name and classOfService

* */

public class Passenger {

  String name;

int classOfService;

  public Passenger(String string, int classOfService) {

      this.name = string;

      this.classOfService = classOfService;

  }

@Override

  public String toString() {

      return "Passenger [name=" + name + ", classOfService=" + classOfService

              + "]";

  }

@Override

  public int hashCode() {

      final int prime = 31;

      int result = 1;

      result = prime * result + classOfService;

      result = prime * result + ((name == null) ? 0 : name.hashCode());

      return result;

  }

@Override

  public boolean equals(Object obj) {

      if (this == obj)

          return true;

      if (obj == null)

          return false;

      if (getClass() != obj.getClass())

          return false;

      Passenger other = (Passenger) obj;

      if (classOfService != other.classOfService)

          return false;

      if (name == null) {

          if (other.name != null)

              return false;

      } else if (!name.equals(other.name))

          return false;

      return true;

  }

  public String getName() {

      return name;

  }

  public int getClassOfService() {

      return classOfService;

  }

}

----

import java.util.ArrayList;

import java.util.List;

/* Train class for holding the

* passengerList.

* The passengerList is the list of Passenger Object

*/

public class Train {

  List<Passenger> passengerList;

  public Train() {

      passengerList = new ArrayList<Passenger>();

  }

  public Passenger getPassenger(int index) {

      if (index <= 0 && getTotalNumberOfPassengersOnBoard() < index) {

          return null;

      } else {

          return passengerList.get(index - 1);

      }

  }

  public void addPassenger(Passenger p) {

      passengerList.add(p);

  }

  public int getTotalNumberOfPassengersOnBoard() {

      return passengerList.size();

  }

  public boolean isPassengerOnBoard(String name) {

      boolean flag= false;

      for (Passenger p : passengerList) {

          if (p.getName().equalsIgnoreCase(name)) {

              flag = true;

              break;

          } else {

              flag = false;

          }

      }

      return flag;

  }

public double getRevenue(double priceFirstClass, double priceSecondClass) {

      double total = 0.0;

      for (Passenger p : passengerList) {

          if (p.getClassOfService() == 1) {

              total += priceFirstClass;

          } else {

              total += priceSecondClass;

          }

      }

      return total;

  }

  public double getPercentageFirstClassTravellers() {

      double count = 0.0;

      for (Passenger p : passengerList) {

          if (p.getClassOfService() == 1) {

              count++;

          }

      }

      return count / getTotalNumberOfPassengersOnBoard() * 100;

  }

@Override

  public String toString() {

      return "Train [passengerList=" + passengerList + "]";

  }

@Override

  public int hashCode() {

      final int prime = 31;

      int result = 1;

      result = prime * result

              + ((passengerList == null) ? 0 : passengerList.hashCode());

      return result;

  }

@Override

  public boolean equals(Object obj) {

      if (this == obj)

          return true;

      if (obj == null)

          return false;

      if (getClass() != obj.getClass())

          return false;

      Train other = (Train) obj;

      if (passengerList == null) {

          if (other.passengerList != null)

              return false;

      } else if (!passengerList.equals(other.passengerList))

          return false;

      return true;

  }

}

-----------

public class TrainTest {

  public static void main(String[] args) {

      Train t = new Train();

      Passenger p1 = new Passenger("James", 1);

      Passenger p2 = new Passenger("Sarah", 2);

      Passenger p3 = new Passenger("Jhon", 1);

      Passenger p4 = new Passenger("Test", 2);

      // add other passengers Test1 .. Test6

      for (int j = 1; j < 7; j++) {

          t.addPassenger(new Passenger("Test" + j, j % 2));

      }

      t.addPassenger(p1);

      t.addPassenger(p2);

      t.addPassenger(p3);

      t.addPassenger(p4);

      System.out.println("total revenue $" + t.getRevenue(10.0, 25.0));

      System.out.println("total number of passengers on Board .. "

              + t.getTotalNumberOfPassengersOnBoard());

      System.out.println("percentage of first class traveller .. "

              + t.getPercentageFirstClassTravellers());

      System.out.println("getPassenger number # 2 .. " + t.getPassenger(2));

      System.out.println("is passenger Jhon on board ? "

              + t.isPassengerOnBoard("Jhon"));

      System.out.println("is passenger Dummy on board ? "

              + t.isPassengerOnBoard("Dummy"));

      System.out.println("Printing all the Passengers on Board ...");

      for (int i = 1; i <= t.getTotalNumberOfPassengersOnBoard(); i++) {

          System.out.println("" + t.getPassenger(i));

      }

      // using the the Train toString to print all the passengers from Train

      System.out.println(t.toString());

  }

}

Answer:

The code is given, paying attention to every column details

Explanation:

columns:

InvoiceNumber

InvoiceDate

InvoiceTotal

InvoiceLineItemDescription

InvoiceLineItemAmount

SELECT InvoiceNumber, InvoiceDate, InvoiceTotal, InvoiceLineItemDescription AS ItemDescription, InvoiceLineItemAmount AS ItemAmount

FROM Invoices AS Invoice JOIN InvoiceLineItems AS LineItem

ON Invoice.InvoiceID = LineItem.InvoiceID

WHERE Invoice.InvoiceID IN (

SELECT InvoiceID

FROM InvoiceLineItems

GROUP BY InvoiceID

HAVING COUNT(*) > 1)

ORDER BY InvoiceDate

FOR XML RAW, ELEMENTS;

Answer:

When admitting digital evidence at trial, the issue of Authenticity comes up when the evidence involves computer-generated records.

Explanation:

The digital evidence such as video, audio or some pictures have been presented before court as evidence. The first question that may arise is related to "authenticity of the material". To authenticate that evidence court may order the forensic audit of that particular evidence.

So, the issue of authenticity of evidence will arise, in case of digital evidence.