Functional side effects: 15) Given the following Java/C++ like code. Assuming function calls (operands) are evaluated from left-to-right, one at a time. a) What outputs are printed to the screen? int a = 4; // global variable int fun1() { a = 2; return 3; } int fun2() { return a; } void main() { int x = fun1() + fun2(); System.out.println("x = " + x); } Answer: x = b) What outputs are printed to the screen? int a = 4; // global variable int fun1() { a = 2; return 3; } int fun2() { return a; } void main() { int x = fun2() + fun1(); System.out.println("x = " + x); } Answer: x =

Answer:

Check the explanation

Explanation:

* Filename: ShadeDesigner.java

* Programmer: Jamin A. Bishop

* Date: 3/31/2012

* version 1.00 2012/4/2

*/

import java. awt.*;

import java. awt. event.*;

import java. text. DecimalFormat;

import javax. swing.*;

import javax. swing. event. ListSelectionEvent;

import javax. swing. event. ListSelectionListener;

public class ShadeDesigner extends JFrame

{

private String[] styles = {"Regular Shades", "Folding Shades", "Roman Shades"};

private String[] size = {"25 Inches Wide", "27 Inches Wide",

"32 Inches Wide", "40 Inches Wide"};

private String[] colors = {"Natural", "Blue", "Teal",

"Red", "Green"};

private JLabel banner;// To display a banner

private JPanel bannerPanel;// To hold the banner

private JPanel stylesPanel;//

private JPanel sizePanel;//

private JPanel colorPanel;

private JPanel buttonPanel;//

private JList stylesList;

private JList sizeList;

private JList colorList;

private JTextField Styles;

private JTextField Size;

private JTextField Color;

private JButton calcButton;

private JButton ExitButton;

private double totalCharges = 50.00;

//Constants

private final int ROWS = 5;

private final double regularCost = 0.00;//base price for the blinds

private final double foldingCost = 10.00;//extra cost for folding blinds

private final double romanCost = 15.00;//extra cost for roman blinds

private final double twentyfiveInCost = 0.00; //extra cost for 25" blinds

private final double twentySevenInCost = 2.00;//extra cost for 27" blinds

private final double thirtyTwoInCost = 4.00;//extra cost for 32" blinds

private final double fourtyInCost = 6.00;//extra cost for 40" blinds

private final double naturalColorCost = 5.00;//extra cost for color

public ShadeDesigner()

{

//display a title

setTitle("Shade Designer");

// Specify what happens when the close button is clicked.

setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

// Create the banner on a panel and add it to the North region.

buildBannerPanel();

add(bannerPanel, BorderLayout. NORTH);

stylesPanel();

add(stylesPanel, BorderLayout. WEST);

sizePanel();

add(sizePanel, BorderLayout. CENTER);

colorPanel();

add(colorPanel, BorderLayout. EAST);

buttonPanel();

add(buttonPanel, BorderLayout. SOUTH);

pack();

setVisible(true);

}

//build the bannerpanel

private void buildBannerPanel()

{

bannerPanel = new JPanel();

bannerPanel.setLayout(new FlowLayout(FlowLayout.CENTER));

banner = new JLabel("Shade Designer");

banner.setFont(new Font("SanSerif", Font.BOLD, 24));

bannerPanel.add(banner);

}

//stylepanel

private void stylesPanel()

{

JLabel styleTitle = new JLabel("Select a Style.");

stylesPanel = new JPanel();

stylesPanel. setBorder(BorderFactory. createEmptyBorder(5,5,5,5));

stylesList = new JList (styles);

stylesList.setVisibleRowCount(ROWS);

JScrollPane stylesScrollPane = new JScrollPane(stylesList);

stylesList. setSelectionMode(ListSelectionModel.SINGLE_SELECTION);

stylesList. addListSelectionListener(new stylesListListener());

stylesPanel. setLayout(new BorderLayout());

stylesPanel. add(styleTitle, BorderLayout. NORTH);

stylesPanel. add(stylesScrollPane, BorderLayout. CENTER);

Styles = new JTextField (5);

Styles. setEditable(false);

//stylesPanel. add(StylesLabel, BorderLayout. CENTER);

stylesPanel. add(Styles, BorderLayout. SOUTH);

}

private class stylesListListener implements ListSelectionListener

{

public void valueChanged (ListSelectionEvent e)

{

String selection = (String) stylesList. getSelectedValue();

Styles. setText(selection);

}

}

//size panel

private void sizePanel()

{

JLabel sizeTitle = new JLabel("Select a Size.");

sizePanel = new JPanel();

sizePanel. setBorder(BorderFactory. createEmptyBorder(5,5,5,5));

sizeList = new JList (size);

sizeList.setVisibleRowCount(ROWS);

JScrollPane stylesScrollPane = new JScrollPane(sizeList);

sizeList.setSelectionMode(ListSelectionModel.SINGLE_SELECTION);

sizeList.addListSelectionListener(new sizeListListener());

sizePanel. setLayout(new BorderLayout());

sizePanel. add(sizeTitle, BorderLayout. NORTH);

sizePanel. add(stylesScrollPane, BorderLayout. CENTER);

//sizeLabel = new JLabel("Style Selected: ");

Size = new JTextField (5);

Size.setEditable(false);

//stylesPanel. add(StylesLabel, BorderLayout. CENTER);

sizePanel. add(Size, BorderLayout. SOUTH);

}

private class sizeListListener implements ListSelectionListener

{

public void valueChanged (ListSelectionEvent e)

{

String selection = (String) sizeList. getSelectedValue();

Size. setText(selection);

}

}

Final answer:

A test plan typically details 'the test data to be used,' which includes sets of inputs designed to test software functionality and validate outcomes.

Explanation:

The question asks which component is specified by the detailed procedures in a test plan. Of the options provided, the one that is typically detailed in a test plan is the test data to be used. A test plan contains comprehensive information about the testing strategy, test objectives, resources needed for testing, test environment, test limitations, and the schedule of testing activities. Importantly, it specifies the test data which are sets of inputs given to a software program during testing to ascertain an expected outcome and validate the software's functionality.

Final answer:

To design an application in which the number of days for each month in the year is stored in an array, create an array of integers representing the number of days in each month. Use a loop to display 12 sentences, each containing the month number and the corresponding number of days.

Explanation:

To design an application in which the number of days for each month in the year is stored in an array, you can create an array of integers representing the number of days in each month. Here's an example:

int[] daysInMonth = new int[]{31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

Then, you can use a loop to display 12 sentences, each containing the month number and the corresponding number of days. Here's an example:

for (int i = 0; i < 12; i++) {
  System.out.println("Month " + (i + 1) + " has " + daysInMonth[i] + " days.");
}

Final answer:

To design the application, create an array with the number of days corresponding to each month and iterate through the array to display sentences for each month using the provided mnemonic or the knuckle mnemonic to remember the days.

Explanation:

To design an application that stores the number of days for each month in an array, we'll assume a non-leap year. We have the mnemonics "Thirty days hath September, April, June, and November. All the rest have thirty-one, save the second one alone, Which has four and twenty-four, till leap year gives it one day more". So, we'll create an array with the following values: 31 (January), 28 (February), 31 (March), 30 (April), 31 (May), 30 (June), 31 (July), 31 (August), 30 (September), 31 (October), 30 (November), 31 (December). Then, we can iterate through this array and display the sentences accordingly.

Print "Month (index + 1) has (value at current index) days"

The knuckle mnemonic is a physical way to remember the number of days in each month, where months with 31 days are represented by the protruding knuckles and shorter months fall in the spots between the knuckles.

Answer:

See explaination for the program code

Explanation:

code:

public static int[] countLastDigits(int[] list) {

int[] count = new int[10];

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

int digit = list[i] % 10;

count[digit]++;

}

return count;

}

The program which counts the number of elements with ends with a certain digit is given below with, elements ending with 0 as index, 1 and so on.

public static int[] countLastDigits(int[] list) {

#initializes a function named countLastDigits

int[] count = new int[10];

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

#a for loop to iterate over the array created

int digit = list[i] % 10;

count[digit]++;

#appends to counts variable

}

return count;

}

Learn more : https://brainly.com/question/18478903

Answer:

c. You use separator.join(a_list) where a_list is a list of strings.

Explanation:

The join() is an in-built string method which returns a string concatenated with the elements of an iterable. It concatenates each element of an iterable (such as list, string and tuple) to the string and returns the concatenated string.

The syntax of join() is:

string.join(iterable)

From the above syntax, the string usually mean a separator and the iterable will be a string or list or tuple.

The answer is C.

c. You use separator.join(a_list) where a_list is a list of strings.

It is not A because the iterable could be a string. It is not D because the separator is outside not in the bracket.

Final answer:

To display the invoice number, line numbers, product SKUs, product descriptions, and brand ID for sales of sealer and top coat products of the same brand on the same invoice, you can use a SQL query with multiple joins and sorting conditions.

Explanation:

To display the invoice number, line numbers, product SKUs, product descriptions, and brand ID for sales of sealer and top coat products of the same brand on the same invoice, you can use SQL query with multiple joins and sorting conditions. Here's an example:

This query assumes that the tables for invoices are named 'invoices' and the tables for lines are named 'lines'. Modify the table names and column names according to your database schema.

Final answer:

The SQL query should select invoice numbers, line numbers, product SKUs, product descriptions, and brand IDs where product descriptions are 'sealer' or 'top coat' and group by invoice number and brand, ensuring there are at least two such products on the same invoice. The query should sort results as specified.

Explanation:

Writing a SQL query involves selecting specific columns from a table and applying the conditions that match the data retrieval requirements. For this scenario, we want to display invoice numbers, line numbers, product SKUs, product descriptions, and brand IDs for sales of 'sealer' and 'top coat' products of the same brand on the same invoice. Additionally, we are asked to ensure the results are sorted by invoice number in ascending order, first line number in ascending order, and then by second line number in descending order.

Here is an example of what the SQL query might look like:

SELECT InvoiceNumber, LineNumber1, LineNumber2, SKU, ProductDescription, BrandID FROM Sales WHERE (ProductDescription = 'sealer' OR ProductDescription = 'top coat') AND BrandID IS NOT NULL GROUP BY InvoiceNumber, BrandID HAVING COUNT(*) > 1 ORDER BY InvoiceNumber ASC, LineNumber1 ASC, LineNumber2 DESC;

Note that database structure is assumed, and actual table and column names may vary.

Answer:

Check the explanation

Explanation:

public static int countMatching(String s, char c) {

   int count = 0;

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

       if (s.charAt(i) == c)

           ++count;

   }

   return count;

}

Method in a complete Java program

public class FizzBuzz {

/* sample run:

    * z appears 2 time(s) in FIZZbuzz

    */

   public static void main(String[] args) {

       String s = "FIZZbuzz";

       char c = 'z';

       int count = countMatching(s, c);

       System.out.printf("%c appears %d time(s) in %s%n", c, count, s);

   }

   // Put your countMatching() method here:

   public static int countMatching(String s, char c) {

       int count = 0;

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

           if (s.charAt(i) == c)

               ++count;

       }

       return count;

   }

}

z appears 2 time(s) in FIZZbuzz Process finished with exit code

Answer:

A. Implement a mobile data loss agent on the devices to prevent any user manipulation with the contact information

Explanation:

Given that, the task is to provide Security controls to lower the risk

Hence, one should undertand the various purpose of troubleshooting, which are:

1. Before the security breach, preventive measures are designed to stop or avoid security breach from occurrence.

2. During the security breach, detective actions are designed to establish and characterize a security breach.

3. After the event, corrective actions are purposely designed to stop the level of any damage caused by the security breach.

Hence, in this case, the MOST efficient security control to implement to lower this risk is to implement a mobile data loss agent on the devices to prevent any user manipulation with the contact information

This is because, Mobile Data Loss Prevention (DLP) is a security function that is provided through email or data security solutions. Thus, through its policies application, to the ActiveSync agent, sensitive information can be restricted from being sent to any ActiveSync-enabled mobile device

Answer:

Explanation:

ef poker(hands):

   scores = [(i, score(hand.split())) for i, hand in enumerate(hands)]

   winner = sorted(scores , key=lambda x:x[1])[-1][0]

   return hands[winner]

def score(hand):

   ranks = '23456789TJQKA'

   rcounts = {ranks.find(r): ''.join(hand).count(r) for r, _ in hand}.items()

   score, ranks = zip(*sorted((cnt, rank) for rank, cnt in rcounts)[::-1])

   if len(score) == 5:

       if ranks[0:2] == (12, 3): #adjust if 5 high straight

           ranks = (3, 2, 1, 0, -1)

       straight = ranks[0] - ranks[4] == 4

       flush = len({suit for _, suit in hand}) == 1

       '''no pair, straight, flush, or straight flush'''

       score = ([1, (3,1,1,1)], [(3,1,1,2), (5,)])[flush][straight]

   return score, ranks

>>> poker(['8C TS KC 9H 4S', '7D 2S 5D 3S AC', '8C AD 8D AC 9C', '7C 5H 8D TD KS'])

'8C AD 8D AC 9C'

Answer:

d. Neither are more expensive than CDP nor Lose less data during a disaster than CDP

Explanation:

A hot site is considered a site that remain very much functional and permits immediate recovery when a disaster occurs. One great advantage of hot sites is that they can be used for operation before the occurrence of a disaster. Hot sites are usually cost effective and hence are less expensive than CDP. With hot sites, users are only faced with minimal downtime in the course of a disaster that affects one of the data centers, hence it is not that they lose less data during a disaster than CDP.

Final answer:

To merge the two customer files, you can use a simple algorithm that compares the customer numbers in each file and merges them accordingly.

Explanation:

To merge the two customer files, you can use a simple algorithm that compares the customer numbers in each file and merges them accordingly. Here's one possible logic for the program:

By following this logic, you can merge the customer files into one file containing all the customers.

To merge the customer files, initialize file pointers for each file, compare customer numbers, write the smaller entry to the output file, and continue until all entries are merged. Ensure no duplicates by using unique customer numbers. Consider the general format used in geographic data management.

Merging Customer Files for Geraldine's Landscaping Service and Gerard's Lawn Maintenance

To merge the customer files of both Geraldine's Landscaping Service and Gerard's Lawn Maintenance, you need to design a program that combines the two files into one. Each file is sorted by customer number, which makes the merging process more straightforward. Here's a step-by-step explanation of the logic:

This program ensures that all customers from both files are included in the merged output file without duplicates, thanks to the unique customer numbers. The same general format is applied when managing geographic data, where rows represent records and columns represent attributes.

By merging these files, businesses can geocode their customers to better understand their distribution and needs, leading to improved services.

Answer:

Detailed program code is written at explaination

Explanation:

Program:

import java.io.File;

import java.io.FileWriter;

import java.io.IOException;

import java.util.Scanner;

public class InputTextToOutputFile

{

public static void main(String[] args) throws IOException

{

Scanner in=new Scanner(System.in);//Scanner object to get user input

System.out.println("What is the name of your output file? ");

String fileName = in.nextLine();//get output file name

File file = new File(fileName);//File object with fileName as input

//Create the file by method file.createNewFile()

if (file.createNewFile())

{

System.out.println("File is created!");

} else {

System.out.println("File already exists.");

}

//FileWriter object with user file name given as input

FileWriter writer = new FileWriter(file);

System.out.println("Enter text to write to a file : ");

String line;//variable to store line content

do {

line=in.nextLine();//get line content from user

writer.write(line+"\n"); //write content to file by adding new line(\n) character

}while(!line.equals("STOP!"));//repeat a loop until user enters "STOP!" line

writer.close();//close the file object

}

Final answer:

To solve this problem, use the Scanner class to get user input and the FileWriter and BufferedWriter classes to write to a file. Prompt the user for the output file name and write everything the user types to the file until they enter "STOP!"

Explanation:

In order to solve this problem in the class named InputTextToOutputFile.java, you can use the Scanner class to get user input. First, create a Scanner object to read user input from the console. Then, prompt the user with the message "What is the name of your output file?" and store their response in a variable called 'fileName'.

Next, you can use the FileWriter and BufferedWriter classes to create and write to a file. Open the output file using the 'fileName' variable, and create a FileWriter and BufferedWriter object to write to the file.

Now, you can use a while loop to continuously read user input and write it to the file until the input contains the string "STOP!". Inside the while loop, read the user input using the Scanner object, and use the BufferedWriter object to write the input to the file. Finally, outside of the while loop, write the string "STOP!" to the file and close the BufferedWriter object to ensure that all the data is written to the file.

Answer:

2125 ns.

Explanation:

First of all Execution Time = Number of Instructions * CPI * Clock Cycle

Number of Instructions = 1000

CPI = 0.5

Clock Cycle = 1/clock rate = 1/4GHz = 0.25 * 10-9 s

So Execution Time = 1000 * 0.5 * 0.25 * 10-9

Execution Time = 125 * 10-9 s

Execution Time = 125 ns

Now 20% of the instructions take 10 ns extra time for remote communication.

1 instruction takes 10ns

so 20% of 1000 = 200 instructions will take 200*10 = 2000ns

So Total Execution Time = 125+2000 = 2125 ns.

Answer:

option C i.e. m^3 is the correct option.

Explanation:

The Multiplication of one m x m matrix with a different m x m matrix will take time m x m x m = m3

And given that this multiplication occurred for j iterations, therefore, the total number of multiplication will be j * m3 = jm3

Consequently, the complexity order will be jm3

Since 5< j < 100, hence j can be substituted by constant and for this reason among all the options mentioned, option C i.e. m^3 is correct.

Answer:

Answer is in the attached screenshot.

Explanation:

Using regex to split a given input string via whitespace, then returns the first and last element of the array.

Answer:

25, 40, 64, 16, 34

Explanation:

the size of the auxiliary file is 25

The first value in the auxiliary file is 40

the fourth value is 64

Largest value will be replaced with 16

The largest remaining value will be replaced with 34

Answer:

For this given problem as RXOR is a 1 bit left shift rotation performed on a current tash,it is true that this bit flips flips result in finding a message different than the original and that this new message "produces" the same hash.

Answer:

see explaination

Explanation:

The program code

lst = []

for i in range(5):

n = input("Enter an integer: ")

if(n.isdecimal()):

lst.append(int(n))

else:

print("Input value cannot be converted to integer")

print("Integer list:",lst)

see attachment for output

Answer:

Check the explanation

Explanation:

WriteData.java:

import java.io.FileWriter;

import java.io.PrintWriter;

class WriteData {

  String[][] data;

  public WriteData() {

      // Data to write to the file

      data = new String[][]{{ "Fiction", "Abraham Lincoln Vampire Hunter", "Grahame-Smith", "Wiley", "NY", "13.99", "222"},

          {"Fiction", "Frankenstein", "Shelley", "Prescott", "GA", "7.99", "321"},

          {"NonFiction", "Life of Kennedy", "Jones", "Pearson", "MT", "12.90", "biography"},

          {"Fiction", "Dracula", "Stoker", "Addison", "CA", "5.99", "145"},

          {"Fiction", "Curse of the Wolfman", "Hageman", "Wesley", "MA", "10.59", "876"},

          {"NonFiction", "How to Pass Java", "Willis", "Wiley", "NY", "1.99", "technology"},

          {"Fiction", "The Mummy", "Rice", "Addision", "CA", "7.99", "954"},

          {"NonFiction", "History of Texas", "Smith", "Prescott", "CA", "9.75", "history"}};

      }

  public void writeToFile() {

      try {

          PrintWriter pw = new PrintWriter(new FileWriter("books.txt"));   // Creating an output stream to write to file

          // Writing data to the file line by line

          for (int i = 0; i < 8; i++) {

              pw.println(data[i][0] + ";" + data[i][1] + ";" + data[i][2] + ";" + data[i][3] + ";" + data[i][4] + ";" + data[i][5] + ";" + data[i][6]);

          }

          pw.close();       // Closing the created output stream

      }

      catch(Exception e) {

          System.err.println(e);

      }

  }

}

Publisher.java:

class Publisher {

  private String name, state;

  public Publisher(String name, String state) {

      this.name = name;

      this.state = state;

  }

  // Getters and Setters

  public String getName() {

      return name;

  }

  public void setName(String name) {

      this.name = name;

  }

  public String getState() {

      return state;

  }

  public void setState(String state) {

      this.state = state;

  }

  public String toString() {

      return name + ";" + state;

  }

}

Book.java:

class Book {

  private String title, author;

  private double price;

  private Publisher publisher;

  public Book(String title, String author, Publisher publisher, double price) {

      this.title = title;

      this.author = author;

      this.publisher = publisher;

      this.price = price;

  }

  public Double calculateCharge(int Qty){

      return getPrice()*Qty;

  }

  public String getTitle() {

      return title;

  }

  public void setTitle(String title) {

      this.title = title;

  }

  public String getAuthor() {

      return author;

  }

  public void setAuthor(String author) {

      this.author = author;

  }

  public String getPublisher() {

      return publisher.toString();

  }

  public void setPublisher(Publisher publisher) {

      this.publisher = publisher;

  }

  public double getPrice() {

      return price;

  }

  public void setPrice(double price) {

      this.price = price;

  }

}

FictionBook.java:

class FictionBook extends Book {

  String fictionCode;

  public FictionBook(String title, String author, Publisher publisher, double price, String fictionCode) {

      super(title, author, publisher, price);       // Calling the parent constructor to initialize values

      this.fictionCode = fictionCode;

  }

  // Getter and Setter

  public String getFictionCode() {

      return fictionCode;

  }

  public void setFictionCode(String fictionCode) {

      this.fictionCode = fictionCode;

  }

}

NonFictionBook.java:

class NonFictionBook extends Book {

  String categoryCode;

  public NonFictionBook(String title, String author, Publisher publisher, double price, String categoryCode) {

      super(title, author, publisher, price);       // Calling the parent constructor to initialize values

      this.categoryCode = categoryCode;

  }

  // Getter and Setter

  public String getCategoryCode() {

      return categoryCode;

  }

  public void setCategoryCode(String categoryCode) {

      this.categoryCode = categoryCode;

  }

}

BookTest.java:

import java.util.Scanner;

import java.util.*;

import java.io.File;

import java.io.FileNotFoundException;

public class BookTest {

  public static void main(String[] args) {

      // TODO code application logic here

      // Writing data to the file using WriteData.java program

      WriteData writeData = new WriteData();

      writeData.writeToFile();

      int[] BookQauntity = {12, 8, 3, 53, 7, 23, 14, 5, 6};

      Book[] book = new BookTest().buildInstances();

      Double GrandTotal= new BookTest().createCharges(BookQauntity, book);

      System.out.println("GrandTotal : "+ GrandTotal);

  }

      catch(FileNotFoundException e) {

          System.err.println("File not found");

      }

      return bk;

  };

  // ISBN info is removed

  Double createCharges(int[] BookQuantity, Book[] book){

      Double Gtotal =0.0, total=0.0;

      for(int i=0; i<8; i++){

          total = book[i].calculateCharge(BookQuantity[i]);

          System.out.println("Title : "+ book[i].getTitle() + ", Total Charge : "+ total);

          Gtotal +=total;

      }

      return Gtotal;

  };

}

Answer:

All of the given options apply and are correct.          

Explanation:

a) The first option is correct because the intruder can use denial-of-service attack and by flooding the site with traffic to get the voter server unavailable. Then he can direct the voters to a fraud vote server.

b) The second options correctly applies too because the thief also buys the pass codes from the people who want to sell their right to vote and this way the thief can purchase multiple pass code and cast multiple votes.

c) The hackers can set up fake vote servers and then direct the voters to these servers or websites in order to collect pass codes of voters. These pass codes can then be misused by the thief to cast multiple votes in an online election.

d) Social engineering techniques can also be used to get pass codes from certain voters. By manipulating the targeted voters psychologically they can get their pass codes to cast multiple votes illegally.

Answer:

1. Average of total Cost = Total Cost/n = O(1)

2. Average of total Cost = Total Cost/n = O(n)

Explanation:

1) Array has Initial size of 8

For adding 16 elements

1. So Cost for first 8 element its 8.

2. On adding 9th element cost is (8+1) = 9 and array is resized to size of 16

3. Then for 10th,11th,12th,.......,16th element Cost is 7.

So total cost is 8 +9 + 7 = 24 ..So for adding 16 elements Cost is 24 units.

For adding 32 elements

1. So Cost for first 8 element its 8.

2. On adding 9th element cost is (8+1) = 9 and array is resized to size of 16

3. Then for 10th,11th,12th,.......,16th element total Cost is 7.

4. On adding 17th element cost is (16+1) = 17 and array is resized to size of 32.

5. Then for 18th,19th,20th,.......,32th element total Cost is 15.

So total cost is 8 +9 + 7 +17+15= 56 ..So for adding 32 elements Cost is 56 units.

We are Intrested in finding the Amortized Cost i.e Averga cost over large number of insertion:-

When the array size is doubled:-

N operation of N pushes will have cost of 1+2+4+8+16+.. + 2k where 2k < N, Summing upo the series we have

2k-1 - 2 . But 2k <=n <= 2k-1 So,

The total cost of the sequence of n insertions will be O(n) .

Since the Amortized Cost is the Average of total Cost = Total Cost/n = O(1)

2) resize by increasing the size by 2 elements

For adding 16 elements

1. So Cost for first 8 element its 8.

2. On adding 9th element cost is (8+1) = 9 and array is resized to size of 10

3. Then for 10th element Cost is 1

4. Then for 11th element Cost is 11

5.Then for 12th element Cost is 1

6.Then for 13th element Cost is 13

6.Then for 14th element Cost is 1

6.Then for 15th element Cost is 15

6.Then for 16th element Cost is 1

So total cost is 60 ..So for adding 16 elements Cost is 60 units.

For adding 32 elements

1. So Cost for first 8 element its 8.

2. On adding 9th element cost is (8+1) = 9 and array is resized to size of 10

3. Then for 10th element Cost is 1

4. Then for 11th element Cost is 11

5.Then for 12th element Cost is 1

7.Then for 13th element Cost is 13

8.Then for 14th element Cost is 1

9.Then for 15th element Cost is 15

10.Then for 16th element Cost is 1

11. Then for 17th element Cost is 17

12. Then for 18th element Cost is 1

13.Then for 19th element Cost is 19

14.Then for 20th element Cost is 1

15.Then for 21th element Cost is 21

16.Then for 22th element Cost is 1

17.Then for 23th element Cost is 23 .. . . .. .. Then for 31th element Cost is 31 .. . . . . . . . . .. Then for 32th element Cost is 1 .

So total cost is 260 ..So for adding 32 elements Cost is 260 units.

N operation of N pushes will have cost of 1+2+3+4+5+.. + N-1 = N(N-1)/2

The total cost of the sequence of n insertions will be O(n2) .

Since the Amortized Cost is the Average of total Cost = Total Cost/n = O(n)

Answer:

Explanation:

Find attach the solution

Answer:

The answer is provided in the attached document.

Explanation:

The explanation is provided in the attached document.

Final answer:

The RTT, estimatedRTT, DevRTT, and TimeoutInterval are important TCP parameters used to manage data transmission reliability. Without specific RTT measurements, we cannot calculate exact values for them. The formulas for updating these values are weighted averages and take into account the most recent RTT measurements.

Explanation:

In Transmission Control Protocol (TCP), the round trip time (RTT) is a measure of the time a signal takes to be sent plus the time it takes for an acknowledgment of that signal to be received. This time is used by TCP to adjust the timeout interval for packet retransmission. The variables estimatedRTT and DevRTT are used to calculate the TimeoutInterval, which determine how long TCP waits for an acknowledgment before resending a packet.

To update estimatedRTT and DevRTT after each measured RTT:

However, to provide an accurate answer, we need the specific RTT measurements to calculate the exact values for estimatedRTT, DevRTT, and TimeoutInterval.

Answer:

#include <iostream>

using namespace std;

int main()

{

   double minutes, cost = 0;

   cout<<"Enter the number of minutes: ";

   cin >> minutes;

   if (minutes <= 3)

       cost = 1.99 + (minutes * 2);

   else

       cost = 1.99 + (2 * 3 + (minutes - 3) *0.45);

   cout <<"The cost is: " << cost <<endl;

   return 0;

}

Explanation:

Declare the variables

Ask the user for the minutes

Check the minutes:

If it is less than or equal to 3, calculate the cost by summing multiplication of  the minutes by 2 and 1.99

Otherwise, calculate the cost by summing the multiplication of the first 3 minutes by 2, multiplication of  the remaining minutes by 0.45 and 1.99

Print the the cost

Final answer:

The program calculates the cost of an international call based on the connection fee, the rate for the first three minutes, and the rate for each additional minute, given the total call duration input by the user. It then outputs the total amount due formatted with two decimal places.

Explanation:

The question relates to creating a C++ program that calculates the total cost of an international call from New York to New Delhi. The cost includes a connection fee, a fixed cost for the first three minutes, and a variable cost for each additional minute. The C++ program must prompt the user for the duration of the call in minutes and calculate the total amount due based on the given rates. It should then display the final amount formatted to two decimal places.

Here is an outline of the program that performs this calculation:

#include
#include
using namespace std;
int main() {
   const double connectionFee = 1.99;
   const double firstThreeMinuteCost = 2.00;
   const double additionalMinuteCost = 0.45;
   int minutes;
   double amountDue;
   cout << "Enter the number of minutes the call lasted: ";
   cin >> minutes;
   if (minutes <= 3) {
       amountDue = connectionFee + firstThreeMinuteCost;
   } else {
       amountDue = connectionFee + firstThreeMinuteCost + (minutes - 3) * additionalMinuteCost;
   }
   cout << fixed << setprecision(2);
   cout << "The total amount due for the call is $" << amountDue << endl;
   return 0;
}
The user is prompted for the total minutes of the call. The program calculates the cost considering all the constraints and displays the result with two decimal places using fixed and setprecision manipulators.

Answer:

Explanation:

Find attach the solution

The expressions of relational algebra for the given queries and the query trees for each with the SQL statement are explained.

Given are expressions for the given queries in relational algebra, draw the query trees, and provide the corresponding SQL statements:

Let's assume we have the following tables:

PC (model, speed, ram, hd, price)

Printer (model, color, type, price)

Laptop (model, speed, ram, hd, screen, price)

Product (model, type, price)

Manufacturer (name, country)

a. Relational algebra expression: π model (σ speed ≥ 3.00 (PC))

Query tree:

    π model

      |

     σ speed >= 3.00

      |

    PC

SQL statement:

SELECT model

FROM PC

WHERE speed >= 3.00;

b. Relational algebra expression: π model (σ color = 'color' (Printer))

Query tree:

    π model

      |

     σ color = 'color'

      |

   Printer

SQL statement:

SELECT model

FROM Printer

WHERE color = 'color';

(Note: Replace 'color' with the actual value representing color printers.)

c. Relational algebra expression: π name (σ hd ≥ 100 (Laptop ⋈ Manufacturer))

Query tree:

     π name

       |

      σ hd >= 100

       |

  Laptop ⋈ Manufacturer

SQL statement:

SELECT DISTINCT m.name

FROM Laptop l

INNER JOIN Manufacturer m ON l.model = m.model

WHERE l.hd >= 100;

d. Relational algebra expression: π model, price (σ name = 'B' (Product ⋈ Manufacturer))

Query tree:

     π model, price

        |

       σ name = 'B'

        |

  Product ⋈ Manufacturer

SQL statement:

SELECT p.model, p.price

FROM Product p

INNER JOIN Manufacturer m ON p.model = m.model

WHERE m.name = 'B';

e. Relational algebra expression: π name (Manufacturer) - π name (σ type = 'PC' (Product ⋈ Manufacturer)) ⋈ π name (σ type = 'Laptop' (Product ⋈ Manufacturer))

Query tree:

      π name (Manufacturer)

          |

          -

         / \

        /   \

  π name    π name (σ type = 'Laptop' (Product ⋈ Manufacturer))

              |

             σ type = 'PC'

              |

        Product ⋈ Manufacturer

SQL statement:

SELECT DISTINCT m.name

FROM Manufacturer m

WHERE m.name NOT IN (

   SELECT DISTINCT m2.name

   FROM Product p

   INNER JOIN Manufacturer m2 ON p.model = m2.model

   WHERE p.type = 'PC'

)

AND m.name IN (

   SELECT DISTINCT m3.name

   FROM Product p

   INNER JOIN Manufacturer m3 ON p.model = m3.model

   WHERE p.type = 'Laptop'

);

Please note that the SQL statements assume the actual column names in the tables may vary, and you need to adjust them accordingly.

Learn more about SQL statements click;

https://brainly.com/question/34389274

#SPJ3

Answer:

c. ​Group use cases by actor.

Explanation:

A use case is used to describe interaction between systems and users to achieve a goal. Mostly, a list of possible interaction between the system and the user is identified. The user can be a single person or a group. By using use case, the requirement of a systems is identified from a users point of view.

Answer:

Using an array list is preferable

Explanation:

A Linked List is an arranged collection of information of the same type in which each element is connected to the next using pointers.

An array list contains set of similar data objects stored in a well arranged memory locations under a common heading or a variable name.

Array elements can be accessed randomly using the array index and can be looked up anytime which makes it correct.Random accessing is not possible in linked list.

Answer:

I would use an array list over a linked list to store the given information from the question.

Explanation:

Let us define what an array list and linked list are

Array list: Array List and Linked List both maintains insertion order and  implements List interface. Array List basically uses a dynamic array to keep or store  the elements.

Linked list: Linked List internally uses a  list that is doubly linked to keep or store the elements

Now,

Using an array list will be of advantage  than a linked list as far as look ups are concerned. also we do have some idea about the size of information,

while a Linked list  will be useful when we don't know the size of data. also retrieving the information from the array list is much more simpler than linked list as  random access is possible with arrays against sequential traversal for lists.

Answer:

Check the explanation

Explanation:

BaseballGame:

public class BaseballGame {

  protected String[] names = new String[2];

  protected int[][] scores;

  protected int innings;

  public BaseballGame() {

      innings = 9;

      scores = new int[2][9];

      for(int i = 0; i < 9; i++)

          scores[1][i] = scores[0][i] = -1;

  }

  public String getName(int team) {

      return names[team];

  }

  public void setNames(int team, String name) {

      names[team] = name;

  }

  public int getScore(int team, int inning) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

      return scores[team][inning];

  }

  public void setScores(int team, int inning, int score) throws Exception {

      if(team < 0 || team >= 2)

          throw new Exception("Team is ut of bounds.");

      if(inning < 0 || inning >= innings)

          throw new Exception("Inning is ut of bounds.");

      if(score < 0)

          throw new Exception("Score is ut of bounds.");

      for(int i = 0; i < inning; i++)

          if(scores[team][i] == -1)

              throw new Exception("Previous scores are not set.");

      scores[team][inning] = score;

  }

}

HighSchoolBaseballGame:

public class HighSchoolBaseballGame extends BaseballGame {

  public HighSchoolBaseballGame() {

      innings = 7;

      scores = new int[2][7];

      for(int i = 0; i < 7; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

LittleLeagueBaseballGame:

public class LittleLeagueBaseballGame extends BaseballGame {

  public LittleLeagueBaseballGame() {

      innings = 6;

      scores = new int[2][6];

      for(int i = 0; i < 6; i++)

          scores[1][i] = scores[0][i] = -1;

  }

}

Answer:

1. n

2. we can slect that permutation without any comparison and the array wil be sorted in to one time.

3. we cannot sort the list without any comparision test.

Explanation:

1. The total number of permutations available are n! because there are ' n ' distinct elements in an array.

2. Here we are given the total ordering permutations, among those available permutations there will be one permutation in which all the elements of the array are sorted. so, we can slect that permutation without any comparison and the array wil be sorted in to one time.

3. If there is no total ordering permutation then we cannot sort the list without any comparision test. In this case we should know the the number of inversion required to see the array. Based on the permutations we can select the permutation which requires minimum inversions. Here inversion sort works well.

See attachment for the details of 4 and 5.