Answer:
We will ned (4) four tables.
Explanation:
For the given scenario we will have to build a relational database. The database will have four tables i.e. Competation, Collage, Team and student.
For each table the database fields are mention below. Note that foreign keys are mentioned in itallic.
Competition:
Competition_ID, Competition_Name, Competition_Data, Competition_Name _of_Branch, College_ID
Collage:
Collage_ID, Collage_Name, Collage_Contact, Collage_address
Team:
Name, Color, Team_ID, College_ID, Competition_ID
College:
Student_ID, First_Name, Last_Name, Date_of_Birth, Major, Expected_Gradiuation_Date, Collage_ID, Team_ID, Competation_ID
To organize the required data for your firm's case competitions, you would need to create four tables: Competitions, Colleges, Teams, and Students. These tables would store competition details, college contact information, team names and colors, and student demographics, respectively.
Explanation:Databases organize information into tables that are composed of rows and columns where each row is a record for an entity and each column is an attribute of the entity. For your case competition database, you will need different tables to store related data without redundancy. The goal here is consistency, efficiency, and comprehensibility. Let's break down the tables you would require:
A Competitions table to store each competition's name, date, and branch location.A Colleges table to store participating college's name, contact phone number, and address, with a reference to the competition they are participating in.A Teams table to record each team's name, color, and associated college.A Students table to keep track of all student participants' first name, last name, date of birth, major, and expected graduation date, with a reference to their team.By designing your database with these four tables, you ensure data is grouped logically and related information is linked appropriately to eliminate the need for repeated data entry and promote efficient data management.
int replace_text (Document *doc, const char *target, const char *replacement) The function will replace the text target with the text replacement everywhere it appears in the docu- ment. You can assume the replacement will not generate a line that exceeds the maximum line length; also you can assume the target will not be the empty string The function will return FAILURE if doc, target or replacement are NULL; otherwise the function will return SUCCESS. #de fine DOCUMENT H #de fine MAX PARAGRAPH LINES 20 #de fine MAX PARAGRAPHS 15 #de fine MAX STR SIZE 80 #de fine HIGHLIGHT START STR "[" #define HIGHLIGHT END STR "ן" #de fine SUCCESS #de fine FAILURE - typedef struct { int number of 1ines: char lines[MAX PARAGRAPH LINES 1 [MAX STR SIZE + 11 \ Paragraph; typedef struct { char name [ MAX STR SIZE+ 11 int number of paragraphs; Paragraph paragraphs [MAX PARAGRAPHS]; } Document;
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:
Loop through total paragraphs and total lines.Store the beginning and ending of paragraph in specific variables.Copy the remainging chars .Finally return SUCCESS.The GNU/Linux operating system comes with many built-in utilities for getting real work done. For example, imagine you had to analyze thousands of files as part of a digital forensics investigation. One utility you might use is the wc command, which prints the newline, word, and byte counts for a given file. For example, if a file contained the text "This is the first line.\nThis is the second.", the wc command would print 1 9 43 (i.e., 1 newline character, 9 words, 43 total characters). For this exercise, you will implement a similar utility. Create a new class named WordCount that has a single method named analyze that takes a string parameter named text and returns an array of three integers. This method should count the number of newlines, words, and characters in the given string. (The return value is an array of these three counts.) Note that a "word" is any sequence of characters separated by whitespace. Hint: You can use a Scanner to count the number of words in a string. The Scanner.next() method returns the next word, ignoring whitespace.
Answer:
Detailed solution is given below:
this IDS defeating techniques works by splitting a datagram or packet into multiple fragments and the IDS will not spot the true nature of the fully assembled datagram. the datagram is not reassembled until it reaches its final destination. It would be a processor-intensive task for an IDS to reassemble all fragments itself, and on a busy system the packet will slip through the IDS onto the network. what is this technique called?
A. IP routing or packet dropping
B. IP splicing or packet reassesmbly
C. IDS spoofing or session assembly
D. IP fragmentation or Session splicing
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.
Consider the following code segment: theSum = 0.0 while True: number = input("Enter a number: ") if number == ": break theSum += float(number) How many iterations does this loop perform?
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.
You are to design class called Employee whose members are as given below:
Data members:
char *name
long int ID
A constructor:
// initialize data members to the values passed to the constructor
Methods:
get Person (name, id) //allows user to set information for each person
A function called Print () that prints the data attributes of the class.
Show the working of class in a driver program (one with main).
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.
Write a program whose input is two integers and whose output is the two integers swapped. Ex: If the input is: 38 then the output is: 83 Your program must define and call a function. SwapValues returns the two values in swapped order. void SwapValues (int* userVali, int* userVal2) LAB ACTIVITY 6.22.1: LAB: Swapping variables 22.1. LAB 0/10 ] main.c Load default template... #include /* Define your function here */ int main(void) { Ovo AWN /* Type your code here. Your code must call the function. */ return 0; 10 }
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.
Firstly, we define required header file and function 'SwapValues()', pass two pointer type integer variables in argument that is 'userVal1' and 'userVal2'.Set integer data type variable 'z' and initialize the value of 'userVal1' in it, then initialize the value of 'userVal2' in 'userVal1' and then initialize the value of 'z' in 'userVal2'.Finally, define the main method in which we set two integer type variables and get input from the user in it then, call and pass those variables and print it.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.
Consider an ERD for a beauty salon in which there is a superclass entity EMPLOYEE with four subclasses:
CASHIER, HAIR_STYLIST, NAIL_TECH, and MANAGER.
A HAIR_STYLIST is a specialization of EMPLOYEE and can also inherit from the MANAGER class. Which of the following terms best describes the relationship between HAIR_STYLIST, EMPLOYEE, and MANAGER?
O multiple inheritance
O tree structure
O single inheritance
O partial and disjoint
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.
ular RSA modulus sizes are 1024, 2048, 3072 and 4092 bits. How many random odd integers do we have to test on average until we expect to find one that is a prime
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.
In this assignment, you will write a complete C program that will act as a simplecommand-line interpreter (i.e., a shell) for the Linux kernel. In writing your shell, you areexpected to use the fork-exec-wait model discussed in class. In particular, you are toimplement the following:• Loop continuously until the user enters quit, which exits your shell.• Inside the loop, you will print your "minor5" prompt and read in the user’scommand, which may consist of a Linux command with 0 or more options andarguments supported by the command. You are expected to read in and processonly 1 command at a time with no pipelining or redirection.• In a child process, you are to execute the command as given, including alloptions and arguments given. If the command is not valid, rather than display an"exec failed" message as shown in class examples, you will simply print outthe command itself with "command not found" as shown in the SAMPLEOUTPUT and then exit the child process. The parent process should wait for thechild process to finish.If you have any questions about this, please contact your instructor, TAs, or IAsassigned to this course to ensure you understand these directions.SAMPLE OUTPUT (user input shown in bold):$ ./a.outminor5> lsa.out grades.txt rec01.txt testdir phone.txt route.txt who.txtdu.txt rec01.c file1 rec01sol.c minor5.cminor5> ls -a -l -ttotal 144-rwx------ 1 cat0299 cat0299 7835 Oct 14 17:39 a.outdrwx------ 4 cat0299 cat0299 4096 Oct 14 17:39 .-rw------- 1 cat0299 cat0299 2665 Oct 14 17:39 minor5.c-rw------- 1 cat0299 cat0299 33 Oct 5 03:30 du.txt-rw------- 1 cat0299 cat0299 33 Oct 5 01:28 file1-rw------- 1 cat0299 cat0299 333 Oct 5 01:02 route.txt
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:
a. Carly’s Catering provides meals for parties and special events. In Chapter 3, you created an Event class for the company. The Event class contains two public final static fields that hold the price per guest ($35) and the cutoff value for a large event (50 guests), and three private fields that hold an event number, number of guests for the event, and the price. It also contains two public set methods and three public get methods.
Now, modify the Event class to contain two overloaded constructors
∎ One constructor accepts an event number and number of guests as parameters. Pass these values to the setEventNumber() and setGuests() methods, respectively. The setGuests() method will automatically calculate the event price.
∎ The other constructor is a default constructor that passes "A000" and 0 to the two-parameter constructor.
Save the file as Event.java
b. In Chapter 3, you also created an EventDemo class to demonstrate using two Event objects. Now, modify that class to instantiate two Event objects, and include the following new methods in the class:
∎ Instantiate one object to retain the constructor default values
∎ Accept user data for the event number and guests fields, and use this data set to instantiate the second object. Display all the details for both objects.
Save the file as EventDemo.java.
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.
You can view the existing Access Control Lists for a set of folders on a Windows system by right-clicking the folder you want to view, selecting Properties, and clicking the:
Answer:
And clicking the security tab option.
Explanation:
Lets explain what an object's ACL is. I will use an example to best explain this. Let's suppose that user Bob would want to access a folder in a Windows environment. What supposedly will happen is that Windows will need to determine whether Bob has rights to access the folder or not. In order to do this, an ACE with the security identity of John will be created. These ACEs are the ones that grant John access to the folder and the ACLs of this particular folder that John is trying to access is a list of permissions of everyone who is allowed to access this folder. What this folder will do is the to compare the security identity of John with the folders ACL and determine whether John has Full control of the folder or not.
By right clicking the folder and selecting the security tab, John will be in a position to see a list of the permissions (ACLs) granted to him by the folder.
Choose two prime numbers p and q (these are your inputs). Suppose that you need to send a message to your friend, and you implement the RSA algorithm for secure key generation. You need public and private keys (these are your outputs). You are free to choose other parameters or inputs if you need any. Write a program for the RSA algorithm using any programing language you know to generate your public and private key. The program may also show a message for any wrong inputs such as "you entered a number, which is not a prime".
Answer:
Answer is attached in the doc file
Explanation:
The explanation is given in the file attached.
The screenshot of the output is attached.
Which of the following statement is true for Service Request Floods A. An attacker or group of zombies attempts to exhaust server resources by setting up and tearing down TCP connections B. It attacks the servers with a high rate of connections from a valid source C. It initiates a request for a single connection
Answer:
The answer is "Option A and Option B"
Explanation:
This is a type of attack, which is mainly used to bring down a network or service by flooding large amounts of traffic. It is a high-rate server from legitimate sources, where an attacker or group of zombies is attempting to drain server resources by creating and disconnecting the TCP link, and wrong choices can be described as follows:
In option C, It can't initiate a single request because when the servers were overloaded with legitimate source links, the hacker may then set up and uninstall TCP links.
Use a programmable calculator or computer (or the sum command on a CAS) to estimate 1 + xe−y R dA where [0, 1] × [0, 1]. Use the Midpoint Rule with the following numbers of squares of equal size: 1, 4, 16, 64, 256, and 1024. (Round your answers to six decimal places.)
Answer:
1.141606
1.143191
1.143535
1.143617
1.143637
1.143642
Explanation:
The inverse of a map is a new map where the values of the original map become the keys of the new map and the keys become the values. For example, given the map {1:2, 3:4}, the inverse is the map {2:1, 4:3}. Given the map, d, create the inverse of d. Associate the new map with the variable inverse. You may assume that there are no duplicate values in d (that is no two keys in d map to the same value).
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|
Which of the following statements regarding the current electronic waste scenario is true? Electronic waste increases with the rise of living standards worldwide. The content of gold in a pound of electronic waste is lesser than that in a pound of mined ore. The process of separating densely packed materials inside tech products to effectively harvest the value in e-waste is skill intensive. Sending e-waste abroad can be much more expensive than dealing with it at home. E-waste trade is mostly transparent, and stringent guidelines ensure that all e-waste is accounted for.
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.
A company has a legacy application using a proprietary file system and plans to migrate the application to AWS. Which storage service should the company
use?
A. Amazon DynamoDB
B. Amazon S3
C. Amazon EBS
D. Amazon EFS
Final answer:
For a legacy application using a proprietary file system migrating to AWS, Amazon EFS (Elastic File System) is the best storage service to use. It supports NFS protocol, making it suitable for applications needing a traditional file system interface without requiring alterations to their existing file system usage.
Explanation:
If a company has a legacy application using a proprietary file system and plans to migrate the application to AWS, the best storage service to use would be Amazon EFS (Elastic File System). Amazon EFS offers a scalable file storage solution for use with AWS Cloud services and on-premises resources. It's built to provide easy, scalable, and reliable storage for applications that require a file system interface and file system semantics.
Amazon EFS is highly suitable for legacy applications being migrated to AWS because it supports NFS (Network File System) protocol, allowing those applications to work without requiring any alterations to the file system usage. Unlike options like Amazon DynamoDB, which is a NoSQL database service for applications that need consistent, single-digit millisecond latency at any scale, or Amazon S3, best for object storage with durability and scalability, EFS provides a more traditional file system interface and file system semantics, making it a more direct match for legacy applications. Lastly, while Amazon EBS (Elastic Block Store) provides block-level storage volumes for use with Amazon EC2 instances, it's not designed to be directly accessed by applications as a file system.
Question 3. Using simulation with 10,000 trials, assign chance_of_all_different to an estimate of the chance that if you pick three words from Pride and Prejudice uniformly at random (with replacement), they all have different lengths. Hint: Remember that !
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
Heather wants a transition effect applied to the links in the gameLinks list in which a gradient-colored bar gradually expands under each link during the hover event. To create this effect, you will use the after pseudo-element and the content property to insert the bar. Create a style rule for the nav#gameLinks a::after selector that: a) places an empty text string as the value of the content property, b) places the content with absolute positioning with a top value of 100% and a left value of 0 pixels, c) sets the width to 0% and the height to 8 pixels, d) changes the background to a linear gradient that moves to right from the color value rgb(237, 243, 71) to rgb(188, 74, 0), e) sets the border radius to 4 pixels, and f) hides the bar by setting the opacity to 0.
Answer:
Hi there Zelenky! This is a good question to practice style sheets and effects. Please find my answer below.
Explanation:
Below CSS contains the code to answer all parts of the question.
nav#gameLinks a::after {
content: ‘’;
top: 100%;
left: 0px;
width: 0%;
height: 8px;
position: absolute;
background: -webkit-gradient(linear, right, left, from(rgb(237, 243, 71)), to(rgb(188, 74, 0));
border-radius: 4px;
opacity: 0;
}
UNIX treats file directories in the same fashion as files; that is, both are defined by the same type of data structure, called an inode. As with files, directories include a nine-bit protection string. If care is not taken, this can create access control problems. For example, consider a file with protection mode 644 (octal) contained in a directory with protection mode 730. How might the file be compromised in this case? What are the limitations? Hint: This is a relatively famous exploit. Give specific commands that will accomplish the exploit.
Answer:
The explanation is listed in the Explanation section below.
Explanation:
All the UNIX directories or files have their summary contained in such an' inode ' format. The inode includes data regarding the current location and size of a file, final release duration, last change time, authorization, etc. Directories are always shown as files and also have an inode connected to them.Unless the folder is big, a list of references to existing data points is implicitly pointed to by inode.
It is composed of the relevant areas:
Type of file.Connect number.Position of data throughout the file.Admin privileges to access files.The system or device command is mkdir and rm (mode) to build directory.
Given dictionaries, d1 and d2, create a new dictionary with the following property: for each entry (a, b) in d1, if a is not a key of d2 (i.e., not a in d2) then add (a,b) to the new dictionary for each entry (a, b) in d2, if a is not a key of d1 (i.e., not a in d1) then add (a,b) to the new dictionary For example, if d1 is {2:3, 8:19, 6:4, 5:12} and d2 is {2:5, 4:3, 3:9}, then the new dictionary should be {8:19, 6:4, 5:12, 4:3, 3:9} Associate the new dictionary with the variable d3
Done in Python please!
Here's what I have:
d3 = {}
for i,v in d1.items():
if v in d2.keys():
d3[i] = d2[v]
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
The manager of a football stadium wants you to write a program that calculates the total ticket sales after each game. There are four types of tickets—box, sideline, premium, and general admission. After each game, data is stored in a file in the following form:
ticketPrice numberOfTicketsSold
...
Sample data are shown below:
250 5750
100 28000
50 35750
25 18750
The first line indicates that the ticket price is $250 and that 5750 tickets were sold at that price. Output the total number of tickets sold and the total sale amount into an output file. Format your output with two decimal places. (You are required to generate an output file that has the results.)
So far my answer is
#include
#include
#include
using namespace std;
int main() {
double total = 0;
int nTickets = 0;
std::ifstream infile("tickets.txt");
int a, b;
while (infile >> a >> b)
{
total = a*b;
nTickets = nTickets + b;
}
cout << "Total Sale amount: " << total << endl;
cout << "Number of tickets sold: " << setprecision(2) << nTickets << endl;
system("pause");
return 0;
}
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:
#include <iostream>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.
Find the number of times a value appears in a list, and create a new list that contains the index positions where the value occurs in the list argument.
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:
If the user input 5 for the size and 1,2,3,4,5 for the list and 5 for the element, then it will print 5.Explanation:
The above code is in python language which is used to take the size for the list, then take the value to add on the list.Then the element is entered from the user to search on the list.Then the element is searched on the list with the help of for loop.It prints the location when the element is matched.What is ISP? What is peering? Understand that carriers usually don’t charge one another for peering. Instead, "the money is made" in the ISP business by charging the end-points in a network—the customer organizations and end users that an ISP connects to the Internet.
Answer:
ISP: internet service provider the company that is able to provide you with access to the Internet services. Example: AT & T.
Peering: a telecommunication method that allows two networks to connect and exchange traffic directly without having to pay a third party to carry traffic across.
Explanation:
An Internet Service Provider (ISP) facilitates access to the Internet for its customers, varying in services and size. Peering between ISPs allows direct data interchange without costs, contrasting with ISP revenue generated from charging end-users and organizations for Internet access. The concept of net neutrality emphasizes equal treatment of all Internet data to prevent service disparities.
Explanation:An Internet Service Provider (ISP) is a company that provides customers with access to the Internet. ISPs can vary in size from small community providers to large multinational companies, and they offer different types of Internet connections such as broadband, DSL, and fiber optics. A critical aspect of the ISP business is the practice of peering, which is a direct interconnection between the networks of two ISPs, allowing them to exchange traffic without the involvement of a third party. Peering is typically done without the exchange of money between ISPs; instead, they benefit mutually from the direct flow of data between their networks. This practice contrasts with the way ISPs earn revenue, which is primarily through charging end-users and customer organizations for Internet access. The fees for Internet access can vary depending on the speed and reliability of the connection provided. As the Internet's infrastructure continues to evolve, debates such as those surrounding net neutrality have emerged, stressing the importance of treating all Internet traffic equally to avoid creating disparities between users based on their ability to pay for faster services.
Create a program to compute the fee for parking in a garage for a number of hours. The program should: 1. Prompt the user for how many hours parked 2. Calculate the fee based on the following: a. $2.50/hour b. minimum fee is $6.00 c. maximum fee is $20.00 3. Print the result python
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)
Write a loop that sets each array element to the sum of itself and the next element, except for the last element which stays the same. Be careful not to index beyond the last element. Ex: Initial scores: 10, 20, 30, 40 Scores after the loop: 30, 50, 70, 40 The first element is 30 or 10 20, the second element is 50 or 20 30, and the third element is 70 or 30 40. The last element remains the same.
Answer:
The solution code is written in Python
numList = [10, 20, 30, 40] for i in range(0, len(numList) - 1): numList[i] = numList[i] + numList[i + 1] print(numList)Explanation:
Firstly, create a sample number list, numList (Line 1)
Create a for-loop that will traverse through the array element from 0 till the second last of the element (len(numList) - 1) (Line 3)
Set the current element, numList[i], to the sum of the current element, numList[i] and the next element, numList[i+1]
Print the modified numList (Line 6) and we can see the output as follows:
[30, 50, 70, 40]
Suppose that a program performs an intermixed sequence of (stack) push and pop operations. The push operations put the integers 0 through 9 in order onto the stack; the pop operations print out the return values. Which of the following sequence(s) could not occur?
a. 4321098765
b. 2143658790
c. 0465381729
d. 4687532901
Answer:
c and d.
c. 0465381729
d. 4687532901
Explanation:
Once an item has been stacked on top of another item, there
is no way to pop them in a different order.
The variable planet_distances is associated with a dictionary that maps planet names to planetary distances from the sun. Write a statement that deletes the entry for the planet name "Pluto".
Answer:
Check the attached image
Explanation:
Check the attached image
Write an if-else statement to describe an integer. Print "Positive even number" if isEven and is Positive are both true. Print "Positive number" if isEven is false and is Positive is true. Print "Not a positive number" otherwise. End with newline.
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;
}
Create an array w with values 0, 0.1, 0.2, . . . , 3. Write out w[:], w[:-2], w[::5], w[2:-2:6]. Convince yourself in each case that you understand which elements of the array that are printed.
Answer:
w = [i/10.0 for i in range(0,31,1)]
w[:] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]
w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]
w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]
w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]
Explanation:
List slicing (as it is called in python programming language ) is the creation of list by defining start, stop, and step parameters.
w = [i/10.0 for i in range(0,31,1)]
The line of code above create the list w with values 0, 0.1, 0.2, . . . , 3.
w[:] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8, 2.9, 3.0]
since start, stop, and step parameters are not defined, the output returns all the list elements.
w[:-2] = [0.0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 1.0, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0, 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 2.8]
w[:-2] since stop is -2, the output returns all the list elements from the beginning to just before the second to the last element.
w[::5]= [0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]
w[::5] since step is -2, the output returns the list elements at index 0, 5, 10, 15, 20, 25,30
w[2:-2:6] = [0.2, 0.8, 1.4, 2.0, 2.6]
the output returns the list elements from the element at index 2 to just before the second to the last element, using step size of 6.