The function below tries to create a list of integers by reading them from a file. The file is expected to have a single integer per line and the code works correctly in that case. If any of the lines don't contain just an integer, than this code fails. Fix this code so that it ignores any lines that don't contain integers instead of raising an exception. Do this with try/except blocks. Your code should still return the list of integers from the lines that contain integers.

Answer:

Check the explanation

Explanation:

#Define the class Employee.

class Employee:

   #Declare and initialize the required member variables.

   emp_name = ''

   Id_num = ''

   emp_dept = ''

   emp_job_title = ''

#Create an object of the class Employee.

emp_obj1 = Employee()

#Assign required values to the members of the class for a

#particular object.

emp_obj1.emp_name = 'Susan Meyers'

emp_obj1.Id_num = '47899'

emp_obj1.emp_dept = 'Accounting'

emp_obj1.emp_job_title = 'Vice President'

#Create another object of the class Employee.

emp_obj2 = Employee()

#Assign required values to the members of the class for the

#current object.

emp_obj2.emp_name = 'Marke Jones'

emp_obj2.Id_num = '39119'

emp_obj2.emp_dept = 'IT'

emp_obj2.emp_job_title = 'programming'

#Create another object of the class Employee.

emp_obj3 = Employee()

#Assign required values to the members of the class for the

#current object.

emp_obj3.emp_name = 'Joy Rogers'

emp_obj3.Id_num = '81774'

emp_obj3.emp_dept = 'Manufacturing'

emp_obj3.emp_job_title = 'Engineering'

#Display the details of each employee objects.

print('Employee 1 details:')

print('Employee Name:', emp_obj1.emp_name)

print('Employee ID Number:', emp_obj1.Id_num)

print('Employee Department:', emp_obj1.emp_dept)

print('Employee Job Title:', emp_obj1.emp_job_title)

print()

print('Employee 2 details:')

print('Employee Name:', emp_obj2.emp_name)

print('Employee ID Number:', emp_obj2.Id_num)

print('Employee Department:', emp_obj2.emp_dept)

print('Employee Job Title:', emp_obj2.emp_job_title)

print()

print('Employee 3 details:')

print('Employee Name:', emp_obj3.emp_name)

print('Employee ID Number:', emp_obj3.Id_num)

print('Employee Department:', emp_obj3.emp_dept)

print('Employee Job Title:', emp_obj3.emp_job_title)

Kindly check the attached image below for the code output.

Answer:

The previous andwers needed a few tweaks to be correct.

Explanation:

class Employee:

  #Declare and initialize the required member variables.

  emp_name = ''

  Id_num = ''

  emp_dept = ''

  emp_job_title = ''

#Create an object of the class Employee.

emp_obj1 = Employee()

#Assign required values to the members of the class for a

#particular object.

emp_obj1.emp_name = 'Susan Meyers'

emp_obj1.Id_num = '47899'

emp_obj1.emp_dept = 'Accounting'

emp_obj1.emp_job_title = 'Vice President'

#Create another object of the class Employee.

emp_obj2 = Employee()

#Assign required values to the members of the class for the

#current object.

emp_obj2.emp_name = 'Mark Jones'

emp_obj2.Id_num = '39119'

emp_obj2.emp_dept = 'IT'

emp_obj2.emp_job_title = 'Programmer'

#Create another object of the class Employee.

emp_obj3 = Employee()

#Assign required values to the members of the class for the

#current object.

emp_obj3.emp_name = 'Joy Rogers'

emp_obj3.Id_num = '81774'

emp_obj3.emp_dept = 'Manufacturing'

emp_obj3.emp_job_title = 'Engineer'

#Display the details of each employee objects.

print('Name:', emp_obj1.emp_name)

print('ID Number:', emp_obj1.Id_num)

print('Department:', emp_obj1.emp_dept)

print('Job Title:', emp_obj1.emp_job_title)

print()

print('Name:', emp_obj2.emp_name)

print('ID Number:', emp_obj2.Id_num)

print('Department:', emp_obj2.emp_dept)

print('Job Title:', emp_obj2.emp_job_title)

print()

print('Name:', emp_obj3.emp_name)

print('ID Number:', emp_obj3.Id_num)

print('Department:', emp_obj3.emp_dept)

print('Job Title:', emp_obj3.emp_job_title)

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the solution to the above question.

Explanation:

The first 3 lines of code tell the user to input a 5 digit number (ex. 72,910) or a 1 digit number (ex. 3). The next 5 lines of code determine if it is a 5/1 digit number or not. If it is, it prints "Success!". The rest of the code just tells that if it's not a 5/1 digit number, and if it's not, it gives errors and tells the user to try again.

Hope this helps!

Answer:

Explanation:

PiLatinWithMultipleWords.java

import java.util.Scanner;

public class PiLatinWithMultipleWords {

public static void main(String[] args) {

Scanner scan = new Scanner(System.in);

System.out.print("Enter a string to be translated: ");

String s = scan.nextLine();

String words[] = s.split("\\s+");

for(String word: words){

System.out.print(pigLatin(word)+" ");

}

System.out.println();

}

public static String pigLatin (String s){

s = s.toLowerCase();

s = s.substring(1,s.length())+ s.charAt(0) + "ay";

return s;

}

}

Output

Enter a string to be translated: the fox jumps over the lazy dog

hetay oxfay umpsjay veroay hetay azylay ogday

Answer:

Check the explanation

Explanation:

// Clothing.java

public class Clothing {

//Declaring instance variables

private int quantity;

private String color;

//Zero argumented constructor

public Clothing() {

this.quantity = 0;

this.color = "";

}

//Parameterized constructor

public Clothing(int quantity, String color) {

this.quantity = quantity;

this.color = color;

}

// getters and setters

public int getQuantity() {

return quantity;

}

public void setQuantity(int quantity) {

this.quantity = quantity;

}

public String getColor() {

return color;

}

public void setColor(String color) {

this.color = color;

}

public double calculatePrice()

{

return 0.0;

}

}

_____________________________

// Pants.java

public class Pants extends Clothing {

//Declaring instance variables

private int waist;

private int inseam;

//Zero argumented constructor

public Pants() {

this.waist = 0;

this.inseam = 0;

}

//Parameterized constructor

public Pants(int quantity, String color, int waist, int inseam) {

super(quantity, color);

setWaist(waist);

setInseam(inseam);

}

// getters and setters

public int getWaist() {

return waist;

}

public void setWaist(int waist) {

if (waist > 0)

this.waist = waist;

}

public int getInseam() {

return inseam;

}

public void setInseam(int inseam) {

if (inseam > 0)

this.inseam = inseam;

}

public double calculatePrice() {

double tot = 0;

if (waist > 48 || inseam > 36) {

tot = 65.50;

} else {

tot = 50.0;

}

return tot;

}

}

__________________________

// Shirt.java

public class Shirt extends Clothing {

//Declaring instance variables

private String size;

//Zero argumented constructor

public Shirt() {

this.size = "";

}

//Parameterized constructor

public Shirt(int quantity, String color, String size) {

super(quantity, color);

this.size = size;

}

// getters and setters

public String getSize() {

return size;

}

public void setSize(String size) {

this.size = size;

}

public double calculatePrice() {

double tot = 0;

if (size.equalsIgnoreCase("S")) {

tot = getQuantity() * 11.00;

} else if (size.equalsIgnoreCase("M")) {

tot = getQuantity() * 12.50;

} else if (size.equalsIgnoreCase("L")) {

tot = getQuantity() * 15.00;

} else if (size.equalsIgnoreCase("XL")) {

tot = getQuantity() * 16.50;

} else if (size.equalsIgnoreCase("XXL")) {

tot = getQuantity() * 18.50;

}

return tot;

}

}

___________________________

//Test.java

import java.util.ArrayList;

public class Test {

public static void main(String[] args) {

int totShirts=0,totPants=0;

double sprice=0,pprice=0,totWaist=0,totinseam=0,avgWaist=0,avginseam=0;

int cnt=0;

ArrayList<Clothing> clothes=new ArrayList<Clothing>();

Shirt s=new Shirt(8,"Green","XXL");

Pants p1=new Pants(6,"Brown",48,30);

Pants p2=new Pants(4,"Blue",36,34);

clothes.add(s);

clothes.add(p1);

clothes.add(p2);

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

{

if(clothes.get(i) instanceof Shirt)

{

Shirt s1=(Shirt)clothes.get(i);

totShirts+=s1.getQuantity();

sprice+=s1.calculatePrice();

}

else if(clothes.get(i) instanceof Pants)

{

Pants pa=(Pants)clothes.get(i);

totPants+=pa.getQuantity();

pprice+=pa.calculatePrice();

totinseam+=pa.getInseam();

totWaist+=pa.getWaist();

cnt++;

}

}

System.out.println("Total number of shirts :"+totShirts);

System.out.println("Total price of Shirts :"+sprice);

System.out.println("Total number of Pants :"+totPants);

System.out.println("Total price of Pants :"+pprice);

System.out.printf("Average waist size is :%.1f\n",totWaist/cnt);

System.out.printf("Average inseam length is :%.1f\n",totinseam/cnt);

}

}

_________________________

Output:

Total number of shirts :8

Total price of Shirts :148.0

Total number of Pants :10

Total price of Pants :100.0

Average waist size is :42.0

Average inseam length is :32.0

Answer:

see explaination

Explanation:

# importing "collections" for namedtuple()

from collections import namedtuple

# function creates student records and return the named tuple

def create_student():

Student = namedtuple('Student',['name','exam1','exam2']) # creating the object Student

# taking input the name and scores of the students

name = input()

exam1 = float(input())

exam2 = float(input())

return Student(name, exam1, exam2) # return the named tuple

# function creates list of student records and return it

def create_class(n):

student_list = []

# creating student data for n students

for i in range(n):

student_list.append(create_student()) # storing data in the student_list

return student_list # return the student data i.e. student_list

# function calculates the final score of each student and returns it

def calculate_score(S):

final_score = 0

if S[1] > S[2]: # if exam1 score is greater than exam2

final_score = S[1] * 0.3 + S[2] * 0.7 # final score is sum of 30% of exam1 and 70% exam2

elif S[1] < S[2]: # if exam2 score is greater than exam1

final_score = S[1] * 0.7 + S[2] * 0.3 # final score is sum of 30% of exam2 and 70% exam1

elif S[1] == S[2]: # if both scores are equal

final_score = S[1] # final score is sum if 30% of exam1 and 70% exam2, but here both scores are equal so final score is equal to any one of the score

else: # if scores are invalid

print('Invalid entries')

return final_score # return final score

# driver function

def main():

n = int(input()) # enter the no. of students in class

if n < 1: # if n is less than 1 return nothing and stop execution

return

student_list = create_class(n) # call create_class to create class of n students

for i in student_list:

print(round(calculate_score(i), 2)) # calculate the final score for each student and print it

if __name__ == "__main__": main()

Answer:

See explaination for the program code

Explanation:

Code below:

package trees;

import java.util.NoSuchElementException;

public class TreeDriver<T extends Comparable<? super T>> {

private Entry<T> root;

publicTreeDriver() {

root = null;

}

public static<T extends Comparable<? super T>>TreeDriver<T> createTestTree() {

return newTreeDriver<T>();

}

public void insert(T value) {

root = insert(value, root);

}

public void remove(T value) {

root = remove(value, root);

}

public boolean contains(T value) {

return valueOf(find(value, root)) != null;

}

private T valueOf(Entry<T> entry) {

return entry == null ? null : entry.element;

}

private Entry<T> insert(T value, Entry<T> entry) {

if (entry == null)

entry = new Entry<T>(value);

else if (value.compareTo(entry.element) < 0)

entry.left = insert(value, entry.left);

else if (value.compareTo(entry.element) > 0)

entry.right = insert(value, entry.right);

else

throw new RuntimeException("Duplicate Entry : " + value.toString());

return entry;

}

private Entry<T> remove(T value, Entry<T> entry) {

if (entry == null)

throw new NoSuchElementException("Entry not found : " + value.toString());

if (value.compareTo(entry.element) < 0)

entry.left = remove(value, entry.left);

else if (value.compareTo(entry.element) > 0)

entry.right = remove(value, entry.right);

else {

// Entry found.

if (entry.left != null && entry.right != null) {

// Replace with in-order successor (the left-most child of the right subtree)

entry.element = findMin(entry.right).element;

entry.right = removeInorderSuccessor(entry.right);

// Replace with in-order predecessor (the right-most child of the left subtree)

// entry.element = findMax(entry.left).element;

// entry.left = removeInorderPredecessor(entry.left);

} else

entry = (entry.left != null) ? entry.left : entry.right;

}

return entry;

}

private Entry<T> removeInorderSuccessor(Entry<T> entry) {

if (entry == null)

throw new NoSuchElementException();

else if (entry.left != null) {

entry.left = removeInorderSuccessor(entry.left);

return entry;

} else

return entry.right;

}

private Entry<T> removeInorderPredecessor(Entry<T> entry) {

if (entry == null)

throw new NoSuchElementException();

else if (entry.right != null) {

entry.right = removeInorderPredecessor(entry.right);

return entry;

} else

return entry.left;

}

private Entry<T> findMin(Entry<T> entry) {

if (entry != null)

while (entry.left != null)

entry = entry.left;

return entry;

}

private Entry<T> findMax(Entry<T> entry) {

if (entry != null)

while (entry.right != null)

entry = entry.right;

return entry;

}

private Entry<T> find(T value, Entry<T> entry) {

while (entry != null) {

if (value.compareTo(entry.element) < 0)

entry = entry.left;

else if (value.compareTo(entry.element) > 0)

entry = entry.right;

else

return entry;

}

return null;

}

private void printInOrder(Entry<T> entry) {

if (entry != null) {

printInOrder(entry.left);

System.out.println(entry.element);

printInOrder(entry.right);

}

}

public void printInOrder() {

printInOrder(root);

}

private static class Entry<T extends Comparable<? super T>> {

T element;

Entry<T> left;

Entry<T> right;

Entry(T theElement) {

element = theElement;

left = right = null;

}

}

private static class Test implements Comparable<Test> {

private String value;

public Test(String value) {

this.value = value;

}

public String toString() {

return value;

}

atOverride // Replace the at with at symbol

public int compareTo(Test o) {

return this.value.compareTo(o.toString());

}

}

private static class Test1 extends Test {

public Test1(String value) {

super(value);

}

}

public static vo id main(String[] args) {

TreeDriver<Test> tree =TreeDriver.createTestTree();

int size = 20;

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

tree.insert(new Test1(String.valueOf(i)));

}

tree.insert(new Test1("100"));

tree.remove(new Test1("10"));

tree.remove(new Test1(String.valueOf(15)));

tree.remove(new Test1(String.valueOf(20)));

tree.printInOrder();

System.out.println("Contains (10) : " + tree.contains(new Test1("10")));

System.out.println("Contains (11) : " + tree.contains(new Test1(String.valueOf(11))));

}

}

Final answer:

To create a balanced binary search tree from a sorted array, you can use a recursive approach. Here is an example Java code that can be used as a starting point. The code creates a balanced binary search tree by selecting the middle element as the current node and calling the createBalancedBST method on the left and right subarrays.

Explanation:

In order to create a balanced binary search tree from a sorted array, you can use a recursive approach. Here is an example Java code that can be used as a starting point:

import java.util.ArrayList;

class Node<T> {
   Node<T> left;
   Node<T> right;
   T value;

   Node(T value) {
       this.value = value;
       this.left = null;
       this.right = null;
   }
}

public class TreeDriver<T extends Comparable<T>> {
   Node<T> root;

   TreeDriver() {
       this.root = null;
   }
   
   private Node<T> createBalancedBST(T[] arr, int start, int end) {
       if (start > end)
           return null;
       
       int mid = (start + end) / 2;
       Node<T> node = new Node<>(arr[mid]);
       
       node.left = createBalancedBST(arr, start, mid - 1);
       node.right = createBalancedBST(arr, mid + 1, end);
       
       return node;
   }
   
   private void inOrderTraversal(Node<T> node) {
       if (node == null)
           return;
       
       inOrderTraversal(node.left);
       System.out.print(node.value + " ");
       inOrderTraversal(node.right);
   }
   
   public void createAndPrintBalancedBST(T[] arr) {
       this.root = createBalancedBST(arr, 0, arr.length - 1);
       inOrderTraversal(this.root);
   }
   
   public static void main(String[] args) {
       TreeDriver<String> tree = new TreeDriver<>();
       String[] arr = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M","N"};
       tree.createAndPrintBalancedBST(arr);
   }
}

The createBalancedBST method takes an array arr, a start index, and an end index. It recursively creates a balanced binary search tree by selecting the middle element as the current node and calling the createBalancedBST method on the left and right subarrays.

The inOrderTraversal method is used to print the values of the created BST in order.

Answer:

A. Java Script

Explanation:

java gives languages

Answer:

A. Java Script

Explanation:

That is the only one they teach and is easier than others!

Answer:

see explaination

Explanation:

SIG{INT} = sub {

`mv unprocessed.txt processed.txt`;

print "\n";

exit;

};

`ps aux > unprocessed.txt`;

print "Press 'q' to exit or 'ctrl-c' to process the file and exit:\n";

$inp = <>;

if ($inp == 'q')

{

exit;

}

Answer:

See explaination

Explanation:

import java.io.File;

import java.io.IOException;

import java.util.Scanner;

import java.util.StringTokenizer;

public class SecretMessage {

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

{

File file = new File("secret.txt");

StringBuilder stringBuilder = new StringBuilder();

String str; char ch; int numberOfTokens = 1; // Changed the count to 1 as we already consider first workd as 1

if(file.exists())

{

Scanner inFile = new Scanner(file);

StringTokenizer line = new StringTokenizer(inFile.nextLine()); // Since the secret.txt file has only one line we dont need to loop through the file

ch = line.nextToken().toUpperCase().charAt(0); // Storing the first character of first word to string builder as mentioned in problem

stringBuilder = stringBuilder.append(ch);

while(line.hasMoreTokens()) { // Looping through each token of line read using Scanner.

str= line.nextToken();

numberOfTokens += 1; // Incrementing the numberOfTokens by one.

if(numberOfTokens == 5) { // Checking if it is the fifth word

ch = str.toUpperCase().charAt(0);

stringBuilder = stringBuilder.append(ch);

numberOfTokens =0;

}

}

System.out.println("----Secret Message----"+ stringBuilder);

}

}

}

Answer:

The correct answer to the following question will be "Wattage ".

Explanation:

The quantity of power needed for the operation of such an electrical system seems to be a Wattage.

Answer: Wait, you're asking strangers to write a report for you? Just use common sense--tell her to forward the emails to the police department, or to provide the login details or the IPv4 logs (depending on the email service) to work off of. Considering that she most likely has the email linked to her phone, you should get into contact with her phone provider to locate the daughter. I don't know what else to say considering that you're asking strangers for an essay to be written for you. Hopefully what I said helps, but I doubt that anyone's going to write a full two page essay.

APA offers authors a dependable structure they can use each time they write. Authors' arguments or research are more effectively organized when they are consistent.

The APA format helps papers on usually complex issues to be more understandable. It makes reading and understanding papers easier.

Utilize common sense and instruct her to transmit the emails to the police department, or, depending on the email service, to supply the login information or the IPv4 logs for investigators to use.

You should get in touch with her phone provider to find the daughter because she almost certainly has the email connected to her phone.

Therefore, I don't know what else to say, considering that you're asking strangers for an essay to be written for you.

Learn more about APA format here:

https://brainly.com/question/12548905

#SPJ2

Answer:

C. spectrum used by most mobile phone firms that does not travel well through solid objects.

Explanation:

Obstructions on the path of radio waves can result to poor network coverage. This is also called Phone reception or coverage black spot. Just as an obstruction on the path of light waves causes the emergence of shadows, so also an obstruction from waves coming from the cell tower. Common culprits are windows, doors, roofing materials, and walls. These solid objects cause an interruption of network for network providers inside a building.

The use of antennas connected from outside the building to inside can be helpful in reducing the effect of this obstruction. Problems resulting could be low signal, difficulty hearing a caller at the other end, slow internet connection, etc.

Options:

a) The client wants to develop internal resources and capabilities

b) The client is looking for the lowest possible costs

c) The client has unique business requirement that must be satisfied by this software application

d) The client has some existing technology in place whose requirements must be met by the new software

Answer:

d) The client has some existing technology in place whose requirements must be met by the new software

Explanation:

For a client to purchase a software, the client has an already existing technology that requires that software for adequate functioning. The analyst will recommend the new software to the client based on the technology he/she previously has and the requirements he wants to meet.

The other options are not impossible, but they are not the primary reason why the software will be purchased, the overall aim for all the company's activities may be to develop internal resources and capabilities or to meet some unique business requirements, but these are secondary. The primary purpose is that the client has a technology that cannot function properly without the recommended software.

Answer:

lst = []

n = int(input("Input an array size for you words array: "))

print("Now please enter " + str(n) + " words")

max = 0

min = 1000

index_min = 0

index_max = 0

for i in range(n):

   s = input("Input a word: ")

   lst.append(s)

   if len(s) >= max:

       max = len(s)

       index_max = i

   if len(s) <= min:

       min = len(s)

       index_min = i

print("The longest word is :" + lst[index_max])

print("The shortest word is :" + lst[index_min])

Explanation:

Create an empty list, lst

Get the size from the user

Create a for loop that iterates "size" times

Inside the loop, get the strings from the user and put them in the lst. Find the longest and shortest strings and their indices using if structure.

When the loop is done, print the longest and shortest

Answer:

Check the explanation

Explanation:

As per requirement submitted above kindly find below solution.

This demonstration is using SQL Server.

Table Names used as

department

project

works_on

SQL query :

select proj_name,dept_name,count(emp_no) as 'most employees ' from

project,department,works_on

where

project.proj_no=works_on. proj_no and

department. dept_no=works_on. dept_no

group by proj_name,dept_name

having count(emp_no)=(

select max(empCount) from (

select proj_name,dept_name,count(emp_no) empCount from project,department,works_on where project. proj_no=works_on. proj_no and department. dept_no=works_on. dept_no

group by proj_name,dept_name) as emp);

Answer:

import java.util.Scanner;

public class nu3 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter number of cycles");

       int numCycles = in.nextInt();

       //Call the method

       printShampooInstructions(numCycles);

   }

   public static void printShampooInstructions(int numCycles){

                if (numCycles<1){

              System.out.println("Too few");

          }

          if (numCycles>4){

              System.out.println("Too Many");

          }

          else {

              for (int i = 1;i <= numCycles; i++) {

                  System.out.println( numCycles+ ": Lather and rinse");

                  numCycles--;

              }

              System.out.println("     Done");

          }

   }

}

Explanation:

This is solved using Java programming language

The method printShampooInstructions() is in bold in the answer section

I have also provided a complete program that request user to enter value for number of cycles, calls the method and passes that value to it

The logic here is using if .... else statements to handle the different possible values of Number of cycles.

In the Else Section a loop is used to decrementally print the number of cycles

Answer:

def print_shampoo_instructions(num_cycles):

   if num_cycles < 1:

       print("Too few.")

   elif num_cycles >4:

       print("Too many.")

   else:

       N = 1

       for N in range (N,num_cycles+1):

           print(N,": Lather and rinse.")

       print("Done.")

Explanation:

Final answer:

The function shampoo_instructions() controls repetitions based on input, preventing infinite loops, akin to the shampoo instructions' issue. It emphasizes the importance of iteration control in programming loops to avoid endless execution. Understanding how to manage loop cycles prevents program malfunction due to infinite loops.

Explanation:

shampoo_instructions() is a function that takes a parameter num_cycles and prints instructions based on the input. If num_cycles is less than 1, it prints 'Too few.'; if more than 4, it prints 'Too many.'; otherwise, it prints the cycle number followed by 'Lather and rinse.' num_cycles times.The directions on shampoo create an infinite loop since they lack an iteration variable to specify the number of repetitions. In programming, developers must include a mechanism to control loops' execution, preventing infinite iterations.

In a loop, each iteration represents a cycle, akin to the instructions in shampoo; therefore, the concept of controlling the number of repetitions is fundamental to avoid infinite loops and ensure proper program execution.

Explanation: Complete question and answer is attached

Answer:

See Explaination

Explanation:

import java.util.Arrays;

/**

*

* atauthor xxxx //replace at with the at symbol

*/

public class Test {

public static void main(String[] args) {

java.util.Scanner input = new java.util.Scanner(System.in);

// Enter values for list1

System.out.print("Enter list1: ");

int size1 = input.nextInt();

int[] list1 = new int[size1];

for (int i = 0; i < list1.length; i++)

list1[i] = input.nextInt();

// Enter values for list2

System.out.print("Enter list2: ");

int size2 = input.nextInt();

int[] list2 = new int[size2];

for (int i = 0; i < list2.length; i++)

list2[i] = input.nextInt();

if (equal(list1, list2)) {

System.out.println("Two lists are identical");

}

else {

System.out.println("Two lists are not identical");

}

}

public static boolean equal(int[] list1, int[] list2) {

if(list1.length == list2.length)

Arrays.sort(list1);

return true;

else

Arrays.sort(list2);

return false;

// Hint: (1) first check if the two have the same size.

// (2) Sort list1 and list2 using the sort method.

// (3) Compare the corresponding elements from list1 and list2.

// return false, if not match. Return true if all matches.

}

}

Answer:

See Explaination

Explanation:

def process_string(s):

global char#global varibles

char=[]

global count

count=[]

for i in s:

#checking whether i is alphabet or not

if(i.isalpha()):

#converting to lower

i=i.lower()

if i in char:

#if char already present increment count

count[char.index(i)]+=1

else:

#if char not present append it to list

char.append(i)

count.append(1)

#menu

print("1. Enter a string")

print("2. Display the string")

print("3. Reverse the string")

print("4. Append a string to the existing one")

print("5. Slice the string")

print("6. Display the number of occurences of each letter")

print("7. Quit")

op=input("Enter option:")

#as I don't know what you have covered

#I am trying to make it simple

#checking whether input is numeric or not

#you can use try except if you want

while(not op.isnumeric()):

op=input("Enter valid option:")

op=int(op)

global x

while(op!=7):

if(op==1):

x=input("Enter a string: ")

elif(op==2):

print("The string is:",x)

elif(op==3):

x=x[::-1]#reversing the string

print("The string is:",x)

elif(op==4):

y=input("Enter a string: ")

x=x+y #string concatnation

elif(op==5):

a=input("Enter first integer: ")

while(not a.isnumeric()):

a=input("Enter valid first integer: ")

a=int(a)

b=input("Enter second integer: ")

while(not b.isnumeric()):

b=input("Enter valid second integer: ")

b=int(b)

x=x[a:b]#string slicing

#you can also use x.slice(a,b)

elif(op==6):

process_string(x)

for i in range(len(char)):

print(char[i],",",count[i])

else:

#incase of invalid input

print("Invalid option")

print("1. Enter a string")

print("2. Display the string")

print("3. Reverse the string")

print("4. Append a string to the existing one")

print("5. Slice the string")

print("6. Display the number of occurences of each letter")

print("7. Quit")

op=input("Enter option:")

while(not op.isnumeric()):

op=input("Enter valid option:")

op=int(op)

Answer:

See attachment please

Answer:

The answer is "loss of 1,260,000"

Explanation:

Following are the important points of the question:

The cost of research and development is = $3840000

net income of the company is=  $2580000

Find: loss or profit .

According to the rate of the cost, the company cost is higher than the company rate. So, the company goes in loss stage, then  

Loss = total cost - net income

Loss = $3840000 - $2580000

Loss= $ 1,260,000

The computer company would face a net loss of $1,260,000.

The question is asking about the calculation of net income or loss for a computer company after accounting for research and development costs. To determine the net income or loss, we subtract the research and development (R&D) costs from the company's income before these costs are accounted for.

So, if the net income before R&D costs is $2,580,000 and the R&D costs are $3,840,000, we perform the following calculation:

Net Income after R&D = Net Income before R&D - R&D costs

Net Income after R&D = $2,580,000 - $3,840,000

Net Income after R&D = -$1,260,000

This result is a net loss of $1,260,000 after accounting for R&D costs.

Answer:

The flowchart to this question can be given in attachment.

Explanation:

In the given statement, a flowchart is defined, in which we use an oval shape to start the, in the next step square shape is used, that initializes the value 0 in z variable, and an input box is used for user input in x variable.

Answer:

Code segment II produces correct output for all values of str, but code segment I produces correct output only for values of str that contain "pea" but not "pear".

Explanation:

The main issue with the first code segment is the way how the if else if condition are arranged. The "pea" is checked at the earliest time in the first code segment and therefore so long as there is the "pea" exist in the string (regardless there is pear or pearl exist in the string as well), the if condition will become true and display "pea" to terminal. This is the reason why the code segment 1 only work for the values of str that contain "pea".

Answer:

A

To ensure that Susan no longer has access without erasing her files I'll take the following steps:

B

The benefits that accrue to retaining her information will depend on the value of information that her account possesses. As the former executive IT personnel, this is most likely the case.

In the long run, the safest thing to do might be to just back up all information she created and had access to to a remote storage or data warehouse.

Cheers!

Answer:

def mph_and_minutes_to_miles():

   miles_per_hour = float(input())

   minutes_traveled = float(input())

   hours_traveled = minutes_traveled / 60.0

   miles_traveled = hours_traveled * miles_per_hour

   print('Miles: {:f}'.format(miles_traveled))

mph_and_minutes_to_miles()

Explanation:

Create a function called mph_and_minutes_to_miles that takes no parameter

Get the code from the main part and put them inside the function

Call the function as requested

Answer:

Check the explanation

Explanation:

Solution(Simplified main code):

# Function written to simplify main code

def mph_and_minutes_to_miles(miles_per_hour, minutes_traveled):

   hours_traveled = minutes_traveled / 60.0

   miles_traveled = hours_traveled * miles_per_hour

   return miles_traveled

miles_per_hour = float(input())

minutes_traveled = float(input())

print('Miles: %f' % mph_and_minutes_to_miles(miles_per_hour, minutes_traveled))

Kindly check the attached image below for the code screenshot and output.

Answer:

Check the explanation

Explanation:

IPSec-based VPN is configured in two different modes namely

IPSec Tunnel mode and IPSec Transport mode so here we are using IPsec transport mode which is used for end to end communications between a client and a server in this original IP header is remain intact except the IP protocol field is changed and the original protocol field value is saved in the IPSec trailer to be restored when the packet is decrypted on the other side due to this arrangement you need to use application based firewall because there are certain specific rules that can address the particular field (explained above-change in IP protocol field) which at the other end need to be same so at the type of decryption, original IP protocol field can be matched.

Answer:

The program in cpp for the given scenario is shown below.

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <queue>

using namespace std;

int main()

{

   //object created of file stream

   ofstream out, out1;

   //file opened for writing

   out.open("words.txt");

   out1.open("word.txt");

   //queues declared for two files

   queue<string> first, second;

   //string variables declared to read from file

   string s1, s2;

   //object created of file stream

   ifstream one("words.txt");

   ifstream two("word.txt");

   //first file read into the que

   while (getline(one, s1)) {

       first.push(s1);

   }

   //second file read into the queue

   while (getline(two, s2)) {

       second.push(s2);

}

   //both files compared

   if(first!=second)

        cout<<"files are not the same"<<endl;

   else

        cout<<"files are identical"<<endl;

   //file closed

   out.close();

   return 0;

}

OUTPUT:

files are identical

Explanation:

1. The object of the file output stream are created for both the files.

2. The two files are opened using the objects created in step 1.

3. The objects of the file input stream are created for both the files.

4. The queues are declared for both the files.

5. Two string variables are declared.

6. Inside a while loop, the text from the first is read into the string variable. The value of this string variable is then inserted into the queue.

7. The loop continues till the string is read and end of file is not reached.

8. Inside another while loop, the text from the second file is read into the second string variable and this string is inserted into another queue.

9. The loop continues till the end of file is not reached.

10. Using if-else statement, both the queues are compared.

11. If any character in the first queue does not matches the corresponding character of the second queue, the message is displayed accordingly.

12. Alternatively, if the contents of both the queues match, the message is displayed accordingly.

13. In the given example program, the message is displayed as "files are identical." This is because both the files are empty and the respective queues are considered identical since both the queues are empty.

14. Since queues are used, the queue header file is included in the program.

15. An integer value 0 is returned to indicate successful execution of the program.