A Layer 2 firewall is also called a(n) _____. Group of answer choices

packet-filtering router
bastion host
packet-filtering bridge
application-level gateway
circuit-level gateway

Final answer:

The corrected C++ program calculates total ticket sales and the total number of tickets sold, with results written to an output file, properly accumulating sales using += and outputting with fixed precision.

Explanation:

The student's question is about writing a C++ program to calculate total ticket sales after a football game, formatting the output to show the total number of tickets sold and the total sale amount with two decimal places. A critical mistake in the original program is the calculation of total sales, where the total should accumulate all sales rather than being overwritten on each iteration. Below is the corrected version of the program.

Corrected C++ Program:

#include
#include
#include
using namespace std;

int main() {
   double total = 0;
   int nTickets = 0, a, b;
   ifstream infile("tickets.txt");
   while (infile >> a >> b) {
       total += a * b; // Correct accumulation of total sales
       nTickets += b;
   }
   infile.close();
   ofstream outfile("sales_summary.txt");
   outfile << fixed << setprecision(2);
   outfile << "Total Sale amount: " << total << endl;
   outfile << "Number of tickets sold: " << nTickets << endl;
   outfile.close();
   return 0;
}

This program reads ticket data from a file, calculates both the total number of tickets sold and the total sales amount, and writes these results to an output file. Note that the setprecision function is used to format the output as required.

This C++ program reads ticket data from a file, calculates total ticket sales and total number of tickets sold, and outputs the results to a file with proper formatting.

To calculate the total ticket sales for a football stadium, we can write a C++ program that reads ticket data from a file and computes the total number of tickets sold as well as the total sales amount. Below is the corrected version of the program:

This program reads the ticket price and the number of tickets sold from the input file "tickets.txt" and calculates the total sales amount in dollars and the total number of tickets sold. The results are then written to an output file "sales.txt" formatted to two decimal places for the sales amount.

Answer:

Explanation:

Firstly import Scanner and Random libraries which will be used to get input number and to generate random value (Line 1 -2).

Next declare all the necessary variables as given in pseudo-code (Line 7 - 8)

Then use nextInt method to randomly generate integer and assign them to rand1 and rand 2, respectively (Line 10 - 11). Next, sum up the two random values and assign it to variable ans (Line 12).

Display the two random numbers (Line 14) and create a Scanner object and prompt user input for answer (Line 16- 18).

Create an if-else condition to check if the input answer (ans) is equal to the real answer (sum) and display the appropriate message accordingly (Line 20 - 24).

Answer:

#include <iostream>

using namespace std;

int main (){

int num=0;

cin>>num;

do{

cout<<num;

cin>>num;

}while(num>=0 && num<=15)

return 0;

}

Explanation:

In this exercise we have to use the computer language knowledge in C++  to write the code as:

the code is in the attached image.

In a more easy way we have that the code will be:

#include <iostream>

using namespace std;

int main (){

int num=0;

cin>>num;

do{

cout<<num;

cin>>num;

}while(num>=0 && num<=15)

return 0;

}

See more about C++ code at brainly.com/question/25870717

Answer:

The ans will be given in the python script below. A picture of the answer is also attached

Explanation:

print("Welcome To Garage Parking Fee Calculator")

hours = float(input("Type the number of hours parked :  "))

#fee per hour

rate = 2.40

#multiply rate per hour by the number of hours inputted

price = rate * hours

if price < 6:

   price = 6

if price > 20:

   price = 20

print("Parking fee is:  $", +price)      

Answer:

C code explained below

Explanation:

#include <stdio.h>

#include <stdbool.h>

int main(void) {

int userNum;

bool isPositive;

bool isEven;

scanf("%d", &userNum);

isPositive = (userNum > 0);

isEven = ((userNum % 2) == 0);

if(isPositive && isEven){

  printf("Positive even number");

}

else if(isPositive && !isEven){

  printf("Positive number");

}

else{

  printf("Not a positive number");

}

printf("\n");

return 0;

}

Answer:

int replace(Document *doc, const char *target, const char *replacement){

int i, j, k;

int beginning;

char temp[MAX_STR_SIZE + 1] ;

char *beginning, *end, *occ, *line;

if(doc == NULL || target == NULL || replacement == NULL)

return FAILURE;

for(i = 0; i < doc->total_paragraphs; i++){

for(j = 0; j < doc->paragraphs[i]->total_lines; j++){

line = doc->paragraphs[i]->lines[j];

beginning = line;

end = beginning + strlen(line);

strcpy(temp, "");

while(beginning < end && (occ = strstr(beginning, target))!= NULL){

len = occ - beginning;

strncat(temp, beginning, len);

strcat(temp, replacement);

beginning = occ + strlen(target);

}

strcat(temp, beginning);

strcpy(doc->paragraphs[i]->lines[j], temp);

}

}

return SUCCESS;

}

Explanation:

Answer:

Bill is building a project network that involves testing a prototype. he must design the prototype (activity 1), build the prototype (activity 2), and test the prototype (activity 3). activity 1 is the predecessor for activity 2 and activity 2 is the predecessor for activity 3. if the prototype fails testing, bill must redesign the prototype; therefore, activity 3 is a predecessor for activity 1. this is an example of

b. looping

Explanation:

Here's the modified Event class with two overloaded constructors and the EventDemo class with the required methods:

`java

// Event.java

package event;

public class Event {

public static final int CUTOFF_VALUE = 50;

public static final int PRICE_PER_GUEST = 35;

private final int eventNumber;

private final int guests;

private final int price;

public Event(int eventNumber, int guests) {

   this.eventNumber = eventNumber;

   this.guests = guests;

   this.price = calculatePrice();

}

public Event() {

   this(0, 0);

}

private int calculatePrice() {

   return PRICE_PER_GUEST * guests;

}

public int getPrice() {

   return price;

}

public int getEventNumber() {

   return eventNumber;

}

public int getGuests() {

   return guests;

}

// EventDemo.java

package event;

import java.util.Scanner;

public class EventDemo {

public static void main(String[] args) {

Scanner input = new Scanner(System.in);

   // Create an Event object with default values

   Event defaultEvent = new Event();

   System.out.println("Default event details:");

   System.out.println("Event number: " + defaultEvent.getEventNumber());

   System.out.println("Number of guests: " + defaultEvent.getGuests());

   System.out.println("Price per guest: $" + defaultEvent.getPrice());

   // Create an Event object with user input

   System.out.println("Enter event number:");

   int eventNumber = input.nextInt();

   System.out.println("Enter number of guests:");

   int guests = input.nextInt();

   Event userEvent = new Event(eventNumber, guests);

   System.out.println("Event details:");

   System.out.println("Event number: " + userEvent.getEventNumber());

   System.out.println("Number of guests: " + userEvent.getGuests());

   System.out.println("Price per guest: $" + userEvent.getPrice());

}

The `Event` class now contains two overloaded constructors, one that accepts an event number and number of guests as parameters and another default constructor that passes "A000" and 0 to the two-parameter constructor.

The `EventDemo` class demonstrates how to create two `Event` objects, one with default values and another with user input. It then displays the details of both objects.

Answer:

Partial and disjoint

Explanation:

Since there is overlapping in relationship of HAIR_STYLIST and MANAGER it can't be tree structure.

A MANAGER can or can't be HAIR_STYLIST. In order for the relationship to be multiple inhertiance am entity in sub-class has to be union of all subclasses

In single inheritance, a sub-class has to be a union of a single super class.

In partial and disjoint, some entity in super class may or may not be related to a sub-class.

Final answer:

Creating a function to print the kth top movie year involves programming skills, understanding of data structures, and error handling. In the example provided, Python is used to access a list of movie years sorted by rank and print the year at the provided rank, illustrating both functional logic and basic error management.

Explanation:

Creating a function in a programming language to print the kth top movie year involves selecting a programming language (like Python), then utilizing data about movie years and their ranks. The function, print_kth_top_movie_year, would likely access a pre-sorted or dynamically sorted list or array of movie years based on some criteria, such as box office success, critical acclaim, or other metrics. Given the specificity of the task, it assumes access to structured data about movies.

For example, in Python, one might start with creating a list of movie years and their rankings. Then, the function would use the rank (an integer) as an index to find and print the respective year from the list. It's important to handle cases where the rank provided is out of range, ensuring the function gracefully handles such errors without crashing.

An illustration of this in code (assuming Python and a predefined list of movie years named movie_years_ranked):

def print_kth_top_movie_year(k):
   try:
       print(movie_years_ranked[k-1]) # Adjust for zero-based indexing
   except IndexError:
       print("Rank out of range.")

Answer:

//The Employee Class

public class Employee {

   char name;

   long  ID;

//The constructor  

public Employee(char name, long ID) {

       this.name = name;

       this.ID = ID;

   }

//Method Get Person

   public void getPerson (char newName, long newId){

       this.ID = newName;

       this.ID = newId;

   }

//Method Print

   public void print(){

       System.out.println("The class attributes are: EmpName "+name+" EmpId "+ID);

   }

}

The working of the class is shown below in another class EmployeeTest

Explanation:

public class EmployeeTest {

   public static void main(String[] args) {

       Employee employee1 = new Employee('a', 121);

       Employee employee2 = new Employee('b', 122);

       Employee employee3 = new Employee('c', 123);

       employee1.print();

       employee2.print();

       employee3.print();

   }

}

In the EmployeeTest class, Three objects of the Employee class are created.

The method print() is then called on each instance of the class.

The relation R in question is a partial order relation characterized as reflexive, antisymmetric, and transitive, based on the properties of birth dates among a set of people.

The relation R described in the question operates under two conditions within the domain of people's birth dates: first, a person x is related to person y if they have the same birth date; second, x is related to y if his birth date is earlier than y's birth date. Given that there are at least two people with the same birth date and at least two people with different birth dates, the correct characterization for this relation is that it is reflexive, antisymmetric, and transitive. These characteristics define a partial order relation. Reflexive because every person is related to themselves (has the same birth date), antisymmetric because if person x is related to person y due to an earlier birth date, then y cannot be related to x in the same manner, and transitive because if person x is born earlier than person y, who in turn is born earlier than person z, then x is also born earlier than z.

Answer:

In RSA, the bit size n of the public modulus N is often of the form n=c⋅2k with c a small odd integer. c=1 (n=512, 1024, 2048, 4096.. bit) is most common, but c=3 (n=768, 1536, 3072.. bit) and c=5 (n=1280..) are common. One reason for this is simply to limit the number of possibilities, and similar progressions are found everywhere in cryptography, and often in computers where binary rules (e.g. size of RAM).

The difficulty of factoring (thus, as far as we know, the security of RSA in the absence of side-channel and padding attacks) grows smoothly with n. But the difficulty of computing the RSA public and private functions grows largely stepwise as n increases (more on why in the next paragraph). The values of n just below a step is thus more attractive than the value just above a step: they are about as secure, but the later is more difficult/slow in actual use. And, not coincidentally, the common RSA modulus sizes are just below such steps.

One major factor creating a step is when one more word/storage unit becomes required to store a number. When the storage unit is b-bit, there is such step every b bits for the RSA public function x↦xemodN; and a step every r⋅b bits for the RSA private function x↦xdmodN, with r=1 for the naïve implementation, and r equal to the number of factors of N when using the CRT with factors of equal bit size (most usually r=2, but I have heard of plans up to r=8). On any modern general-purpose CPU suitable for RSA, b is a power of two and at the very least 25, creating a strong incentive that n is at least a multiple of 26 (r=2 is common, and the only reasonable choice for n below about a thousand).

Note: n=1984=31⋅26 is not unseen in the field of Smart Cards, because the next multiple of 26 would break the equivalent of a sound barrier in a common transport protocol, ISO/IEC 7816-3 T=0. For a list of common RSA modulus size in this field, search LENGTH_RSA_1984.

As an aside, it is significantly simpler to code quotient estimation in Euclidian division (something much used in RSA) when the number of bits of the divisor is known in advance. This creates an incentive to reduce the number of possible bit sizes for the modulus. The two simplest cases are when the number of bits is a multiple of b, and one more than a multiple of b; the former won.

Answer:

Following are the program in the C++ Programming Language.

#include <iostream>

using namespace std;

//define function for swapping

void SwapValues(int* userVal1,int* userVal2){  

//set integer variable to store the value

int z = *userVal1;

//interchange their value

*userVal1 = *userVal2;  

//interchange their value

*userVal2 = z;

}

//define main method

int main()

{    

//declare variables

int x,y;

//get input from the user

cin>>x>>y;

//Call the method to swap the values

SwapValues(&x,&y);

//print their values

cout<<x<<" "<<y;

return 0;

}

Output:

3 8

8 3

Explanation:

Following are the description of the program.

To swap two integers in C, you define a function using pointers. The program reads two integers, swaps them using this function, and then prints the swapped values. Implementing SwapValues ensures correct swapping.

To swap two integers in C, you need to use a function that takes pointers as arguments. A particularly clever application of pointer manipulation allows us to swap the values of two variables.

Firstly, let's define the function SwapValues:

void SwapValues(int* userVal1, int* userVal2) { int temp = *userVal1; *userVal1 = *userVal2; *userVal2 = temp; }

Inside main(), we will call this function:

#include <stdio.h>
void SwapValues(int* userVal1, int* userVal2);
int main(void) { int num1, num2;
scanf("%d %d", &num1, &num2);
SwapValues(&num1, &num2);
printf("%d %d\n", num1, num2);
return 0; }

This code reads two integers, swaps them using the SwapValues function, and prints the result.

n where n is the number of chances user takes to enter a blank number and n>=1.

Explanation:

The loop starts with a universal condition where it is initialized using a true value. Hence the iteration count goes to 1. The user is asked to enter a number after 1st iteration. If number is a blank number, the loop is terminated, else the loop goes on until the users enters a blank number. Hence the iterations depend on the number of chances taken by the user to enter a blank number. Since the user is going to enter a number at least once, the minimum value of n will be 1.

Final answer:

The provided code segment includes an indeterminate while loop that performs iterations based on user input. The loop stops only when the user inputs a colon, with the number of iterations being entirely dependent on the user.

Explanation:

The question concerns a while loop in a code segment that runs indefinitely until the user enters a specific character (please note there is a typo in the original question, it should be "if number == ":"). We are asked how many iterations this loop performs. In this case, the number of iterations is indeterminate, as it depends entirely on the user's actions. Each time the user inputs a number, the loop completes one iteration. The loop only stops (breaks) when the user inputs a colon (":").

Since the user can potentially input an infinite number of numbers before entering the colon, there's no fixed number of iterations that we can provide. The loop is designed to accumulate a sum of the numbers entered by the user, which is a common programming task in various programming languages.

Answer:

No, the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the manner in which the checksum for the packet is calculated. If the corresponding bits (that would be added together) of two 16-bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same. Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error

Explanation:

No, the receiver cannot be confident that there haven't been any bit mistakes.

UDP stands for user data protocol. UDP is defined as a messaging protocol that makes it easier for computers to communicate with one another via a network. A communications protocol called User Datagram Protocol (UDP) is largely used to provide low-latency, loss-tolerant connections between internet-based applications. UDP allows data to be transferred before the receiving party provides an agreement, which speeds up transfers.

A checksum field on UDP allows for the detection of IP header errors; if a checksum error is found, the IP header checksum is recalculated before the IP packet is delivered. Therefore, it cannot ensure that your package will reach the receiver. However, UDP does guarantee that if it is delivered, it will be error-free.

Thus, no, the receiver cannot be confident that there haven't been any bit mistakes.

To learn more about UDP, refer to the link below:

https://brainly.com/question/16984740

#SPJ6

Answer:

Complete code is given below:

Explanation:

addi $t0, $zero, 12 # x = 12

addi $t1, $zero, 21    # y = 21

addi $t2, $zero, 32    # z = 32

addi $t3, $zero, 3

mul $t3, $t3, $t0

mul $t3, $t3, $t0  # $t3 = 3x^2

addi $t4, $zero, 10

mul $t4, $t4, $t1  # $t4 = 10y

addi $t5, $zero, 5

mul $t5, $t5, $t2  # $t5 = 5z

add $a0, $t3, $t4

add $a0, $a0, $t5  # $a0 = 3x^2 + 10y + 5z

addi $v0, $zero, 1

syscall

Answer:

The sample output is been attached below

Explanation:

//main.cpp

#include <iostream>

using namespace std;

#include "IntList.h"

int main() {

//tests constructor, destructor, push_front, pop_front, display

cout << "\nlist1 constructor called";

IntList list1;

cout << "\npushfront 10";

list1.push_front(10);

cout << "\npushfront 20";

list1.push_front(20);

cout << "\npushfront 30";

list1.push_front(30);

cout << "\nlist1: ";

list1.display();

cout << "\npop";

list1.pop_front();

cout << "\nlist1: ";

list1.display();

cout << "\npop";

list1.pop_front();

cout << "\nlist1: ";

list1.display();

cout << "\npop";

list1.pop_front();

cout << "\nlist1: ";

list1.display();

cout << endl;

cout << "list1 destructor called" << endl;

return 0;

}

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

//IntList.cpp

#include <iostream>

#include "IntList.h"

using namespace std;

IntList::IntList() : head(NULL), tail(NULL) { }

IntList::~IntList() {

while(!empty()){

pop_front();

}

}

void IntList::push_front(int value) {

/*Creates a temporary pointer of type IntNode, assigning a memory address (block of memory)

to temp_ptr of type IntNode and then initializing with a constructor. */

IntNode* temp_ptr = new IntNode(value);

/*Then we assign the current address of head (if just empty list, this is NULL) to the

dereferenced temporary pointer's next memory address */

(*temp_ptr).next = this->head;

/*Assign the memory address of temporary pointer (which points to the newly created block of memory) to head. */

this->head = temp_ptr;

return;

}

void IntList::display() const {

/* If empty list, exit out of display function */

if(empty()) {

return;

}

//Since the list is not empty, print the data which the dereferenced head points to

cout << (*head).data;

//Initialize a new pointer, which is assigned to the next pointer in the linked list

IntNode* plc_ptr = (*head).next;

// Create a while loop that checks whether the pointer points to anything.

// This checks to see whether it is the end of the linked list

while(plc_ptr != NULL) {

// Prints the dereferenced pointers data

cout << " " << (*plc_ptr).data;

/*Updates the memory address which the temporary pointer points to,

going through the list each time it goes through the while loop */

plc_ptr = (*plc_ptr).next;

}

return;

}

bool IntList::empty() const {

//The linked list is empty if the head doesn't point to any memory addres/is NULL

if(head == NULL) {

return true;

}

return false;

}

void IntList::pop_front() {

//First check to see if the linked list is empty/if there is a first item that needs to be deleted

if(empty()) {

return;

}

//Set a temporary pointer to the next of the first item, which is the memory address of the second item

IntNode* temp_ptr = (*head).next;

//Delete the memory address for head

delete head;

//Reassign to head the memory address of what was the memory address of the second item

head = temp_ptr;

return;

}

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

//IntList.h

#ifndef _INTLIST_H

#define _INTLIST_H

struct IntNode {

int data;

IntNode* next;

IntNode(int data) : data(data), next(NULL) {}

};

class IntList {

public:

IntList();

~IntList();

void display() const;

void push_front(int);

void pop_front();

bool empty() const;

private:

IntNode* head;

IntNode* tail;

};

#endif

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

Answer:

Python code explained below

Explanation:

# python code

import sys

import readline

from sys import stdin

import random

d1 = {2:3, 8:19, 6:4, 5:12}

d2 = {2:5, 4:3, 3:9}

d3 = {}

#for each entry (a, b) in d1

for i,v in d1.items():

  # if a is not a key of d2

  if i not in d2.keys():

      # add (a,b) to the new dictionary

      d3[i] = v

# for each entry (a, b) in d2

for i,v in d2.items():

  # if a is not a key of d1

  if i not in d1.keys():

      #add (a,b) to the new dictionary

      d3[i] = v

print "d3: ", d3

#output: d3: {8: 19, 3: 9, 4: 3, 5: 12, 6: 4}

Answer:

d1 = {2:3, 8:19, 6:4, 5:12}

d2 = {2:5, 4:3, 3:9}

d3 = {}

for key,value in d1.items():  

       if key not in d2.keys():

                   d3[key] = value

for key,value in d2.items():

        if key not in d1.keys():

                   d3[key] = value

print(d3)

Explanation:

The code is written in python as instructed from the question.

d1 = {2:3, 8:19, 6:4, 5:12}  This is the d1 dictionary entry with the key-value pair.

d2 = {2:5, 4:3, 3:9}   This is the d2 entry with the key-value pair.

d3 = {}  This is an empty dictionary to unpack values

for key,value in d1.items():  This code loops through dictionary d1 and get the key-value pair

if key not in d2.keys():  If any of key in d1 is not in d2 keys.

d3[key] = value  Then add the key-value pair to the empty d3

for key,value in d2.items():  This code loops through dictionary d2 and get the key-value pair

if key not in d1.keys():   If any of key in d2 is not in d1 keys

d3[key] = value  Then add the key-value pair to the empty d3

print(d3)  Display the final key-value pairs of d3

Final answer:

The chance of picking three words with different lengths from Pride and Prejudice can be estimated using a simulation of 10,000 independent trials, checking for unique word lengths in each trial.

Explanation:

To estimate the chance that if you pick three words from Pride and Prejudice uniformly at random (with replacement), they all have different lengths, you need to perform a simulation with 10,000 trials. This simulation involves generating lengths for three different words and checking if all lengths are unique, then repeating this trial 10,000 times to estimate the probability of this event.

Each word length can be thought of as a random variable, and since words are picked with replacement, the trials are independent. Use the principle of statistical independence as well as discrete distribution and the concept of distinguishability of random variables to structure the simulation. For example, you could use a list of words from the book and their lengths. In each trial, three word lengths would be randomly selected with replacement, and you would check if all three are different. The series of trials would give you a proportion of trials where all the word lengths are different. However, this problem is not a binomial distribution, as each trial can have more than two outcomes.

In this Python code, a simulation is performed to determine the number of times, out of 10,000 trials, that two words chosen randomly with replacement from the text "Pride and Prejudice" have different lengths.

Import random module: This module is used for random number generation.

Define the list of words from "Pride and Prejudice": Replace the ellipsis (...) with the actual list of words from the text.

Set the number of trials and initialize num_different to 0:

trials is the number of times you want to repeat the experiment.

num_different will keep track of the number of times two words have different lengths.

Run the simulation using a for loop:

The loop runs trials times.

Inside the loop, two words are randomly chosen from the list using random.choice().

Check if the lengths of the two words are different:

If the lengths are different, increment num_different.

Print the result:

After the loop, print the final count of num_different.

import random

# Assume you have a list of words from "Pride and Prejudice"

pride_and_prejudice_words = ["word1", "word2", "word3", ...]  # Replace ellipsis with the actual words

trials = 10000

num_different = 0

# Run the simulation

for _ in range(trials):

   # Randomly pick two words with replacement

   word1 = random.choice(pride_and_prejudice_words)

   word2 = random.choice(pride_and_prejudice_words)

   # Check if the lengths of the two words are different

   if len(word1) != len(word2):

       num_different += 1

# Print the result

print("Number of times two words have different lengths:", num_different)

This code will give you an estimate of the number of times two words with different lengths are chosen randomly from "Pride and Prejudice" in 10,000 trials.

Complete question:

Using a simulation with 10,000 trials, assign num_different to the number of times, in 10,000 trials, that two words picked uniformly at random (with replacement) from Pride and Prejudice have different lengths. Hint 1: What function did we use in section 1 to sample at random with replacement from an array? Hint 2: Remember that != checks for non-equality between two items. trials = 10000 num_different = ... for ... in ...: num_different =... num_different

Answer:

Electronic waste increases with the rise of living standards worldwide.

Explanation:

Electronic waste are electronic product nearing the end of their usefulness or discard ones, it include computers, phones ,oven, scanners, keyboard,circuit board,clock, lamp etc

Their amount increases every year with the rise of living standards worldwide which constitute to million of tonnes in circulation or these waste end up in landfill which at time contaminate the soil and water by it lead and cadmium content .The amount of electronic waste is expected to increase to about 52.2 million metric tonnes by 2021

Final answer:

True statements regarding the e-waste scenario include the increase of e-waste with higher living standards, the skill-intensive process of recycling, and the high content of gold in e-waste compared to mined ore. However, the e-waste trade is not transparent and is often improperly managed.

Explanation:

Among the current statements about the electronic waste (e-waste) scenario, several are true. The statement that e-waste increases with the rise of living standards worldwide is true. As technological advancements continue and consumerism grows, the amount of e-waste generated increases.

Also true is the fact that the process of separating densely packed materials inside tech products to effectively harvest the value in e-waste is skill-intensive, often leading to unsafe and unregulated labor practices in developing countries. It is indeed accurate that the content of gold in a pound of electronic waste is often greater than that in a pound of mined ore, making it a highly valuable resource.

However, contrary to one of the statements, the e-waste trade is not mostly transparent, and stringent guidelines do not ensure that all e-waste is accounted for; in fact, there is a significant amount of e-waste that is illegally traded or improperly handled.

Answer:

Detailed solution is given below:

Answer:

/ to access the input scanf()

// and output printf()

#include <stdio.h>  

// to access the functions like

// pipe(), fork(), execvp(), dup2()

#include <unistd.h>  

// to access the string functions like

// strtok()

#include <string.h>

// to access function wait()

#include <sys/wait.h>

#include <stdlib.h>

int main()

{

  // declare a variable to hold the process id

  pid_t p_id;

  // declare a variable to hold the index value

  int array_index;

  // declare the string to hold the user input

  // as command

  char userIn_Command[128];

  // use a continuous loop

  while (1)

  {

      // display the prompt for the user

      printf("minor5> ");

      // read the input from the user

      scanf("%[^\n]", userIn_Command);

      // check the condition that whether the user

      // inputs a command called "quit"

      // If the user inputs quit command then exit from

      // the script

      if (strcmp(userIn_Command, "quit") == 0)

      {

          printf("\n");

          break;

      }

      // if there are any usage of pipelining or redirection

      // display the error message and exit from the script

      if (strchr(userIn_Command, '|') != NULL || strchr(userIn_Command, '>') != NULL ||

          strchr(userIn_Command, '<') != NULL)

      {

          printf("Error: Cannot perform the multiple command operations or directions\n");

          break;

      }

      // declare the variables to hold the process

      int p_pids[2];

      // create the system call to pipe() from kernal

      // and display the error message

      if (pipe(p_pids) < 0)

      {

          printf("Error: Pipe creation failed!\n");

      }

      // create the child process

      p_id = fork();

      // condition to check whether the fork() is

      // created. If so, display an error message

      if (p_pids < 0)

      {

          printf("Error: fork() failed!\n");

          break;

      }

      // if the child process is created

      if (p_id == 0)

      {

          // close pipe in child process

          close(p_pids[0]);

          // create duplicate of the standard output

          dup2(p_pids[1], STDOUT_FILENO);

          // close the pipe in child process

          close(p_pids[1]);

          // declare a variable to store path

          // to execute the command

          char *command_path;

          // declare an array of string to hold the options

          char * args[32];

          // tokenize the command by using the delimiter at " "(single space)

          char *cmd_token = strtok(userIn_Command, " ");

          // store the token value

          command_path = cmd_token;

          args[0] = cmd_token;

          array_index = 1;

          // loop until all the options in the command are

          // tokenized

          while (1)

          {

              // get the next token

              cmd_token = strtok(NULL, " ");

              // condition to check whether the token is null

              // or not

              if (cmd_token == NULL)

              {

                  break;

              }

              // store the token if it is not null

              args[array_index] = cmd_token;

              // increment the index

              array_index++;

          }

          // last parameter to the command should be NULL */

          args[array_index] = NULL;

          /* calling exec function with command path and parameters */

          if (strcmp(args[0], "cd") == 0 || strcmp(args[0], "history") == 0 ||

              strcmp(args[0], "exit") == 0)

          {

              printf("%s: Command not found\n", args[0]);

              break;

          }

          if (execvp(command_path, args) < 0 )

          {

              printf("%s: Command not found\n", args[0]);

              break;

          }

      }

      else

      {          

          /* closing writing end of pipe in parent process */

          close(p_pids[1]);

          /* reading ouput written to pipe in child process and

          * writing to console */

          while (1)

          {

              char output[1024];

              int n = read(p_pids[0], output, 1024);

              if (n <= 0)

              {

                  break;

              }

              output[n] = '\0';

              printf("%s", output);

          }

          /* closing read end of pipe1 */

          close(p_pids[0]);

          /* waiting until child process complete its execution */

          wait(NULL);

      }

      /* skipping newline character read while scanf() */

      getchar();

  }

  exit(0);

}

Explanation:

Answer:

1.141606

1.143191

1.143535

1.143617

1.143637

1.143642

Explanation:

Answer:

Check the attached image

Explanation:

Check the attached image

Answer:

D. IP Fragmentation or Session Splicing

Explanation:

The basic premise behind session splicing, or IP Fragmentation, is to deliver the payload over multiple packets thus defeating simple pattern matching without session reconstruction. This payload can be delivered in many different manners and even spread out over a long period of time. Currently, Whisker and Nessus have session splicing capabilities, and other tools exist in the wild.

Answer:

Program :

list_1=[]#take the empty list.

size=int(input("Enter the size of the list: "))#take the size of the list from the user

for x in range(size): #for loop which insert the elemnt on the list.

   list_1.append(int(input("Enter the "+str(x+1)+" element of the list: ")))#take the user input and insert the element.

element=int(input("Enter the element to be searched: "))#it take the elemnt to search.

loc=1#intialize the location value.

count=0

for x in list_1:#for loop to check the element.

   if(x==element): #check the element.

       print(loc,end=", ")#print the location of the element.

       count=count+1

   loc=loc+1

if(count==0):

   print("The element is not present on the list")#print when elemnt are not present.

Output:

Explanation:

Answer:

The answer to these questions can be defined as below:

Explanation:

Question 1:

In the best case big-oh the bubble and selection sort is used, where bubble style simply swaps the items, whereas the sort of sorting takes place by choosing the element.

Question 2:

In the worst case, the merge and selection is used, where Merge-sort splits the list into two sub-lists equally and incessantly names itself in the sublists to still be sorted, whereas Heap-sort is an anti-recursive type.  

Question 3:

In the worst case of swapping the bubble sort algorithm is used.

Question 4:

In sorts of swap selection, bubble, and quick sort is used.

Question 5:

Insertion sorts provide the shift sorts.

Question 6:

The best big-oh is o(n), in the worst case, best case complexity of different sorts are different.

Answer:

The inverse of d would be:

inverse = { }

for key, value in d.items ( ) :

     inverse [value] =key|        

Answer:

The inverse of d would be:

inverse = { }

for key, value in d.items ( ) :

    inverse [value] =key|                                                                                                                                                                                                

Answer:

Answer is attached in the doc file

Explanation:

The explanation is given in the file attached.

The screenshot of the output is attached.