Write an application that prompts a user for a month, day, and year. Display a message that specifies whether the entered date is (1) not this year, (2) in an earlier month this year, (3) in a later month this year, or (4) this month. Save the file as PastPresentFuture.jav

Final answer:

The student's question is about creating a Python program to map names to phone numbers using a dictionary, where after inputting contacts, the program can search for and retrieve phone numbers by name.

Explanation:

The student's question involves writing a program to manage a list of contacts, specifically to map names to phone numbers. To solve this task, one would typically use a data structure known as a dictionary in programming, where each name (key) is associated with a corresponding phone number (value). Below is a simple example of how one could write such a program in Python:

# Prompt for the number of contacts
N = int(input('Enter the number of contacts: '))
# Initialize an empty dictionary to store contacts
contacts = {}
# Loop to enter the contacts into the dictionary
for _ in range(N):
   name, phone_number = input('Enter contact name and phone number: ').split()
   contacts[name] = phone_number
# Prompt for the name to search
search_name = input('Enter a name to search for their phone number: ')
# Output the phone number for the given name
if search_name in contacts:
   print(f'The phone number for {search_name} is {contacts[search_name]}')
else:
   print('Name not found in contact list.')
This script takes the number of contacts as input, then iterates and adds each name-phone pair to the contacts dictionary. When the user enters a name to search, the program looks up the name in the dictionary and prints out the associated phone number if found.

Answer:

Following is given the code in JAVA as required. It analyzes an object falling for 10 seconds. All the necessary descriptions are given inside the code as comments  The output of the program is also attached at the end.

I hope it will help you!

Explanation:

Final answer:

The question involves creating a program in Computers and Technology to calculate the distance of free-fall in meters and feet over 10 seconds using physics formulas and programming methods for calculations and conversions.

Explanation:

The student's question involves writing a program that calculates the distance an object falls in meters over 10 seconds, as well as converting that distance to feet. This problem applies the physics concept of free-fall, using the equation of motion h = (1/2) * g * t^2, where h is the height, g is the acceleration due to gravity (9.81 m/s2), and t is the time in seconds. The student should also write a method to convert the distance from meters to feet (1 meter = 3.28084 feet).

The main method should include a loop that calls these methods to generate a table of distances fallen and the corresponding times. To convert meters to feet, the following formula can be used:

feet = meters * 3.28084

Answer:

#include <iostream>

#include <stack>

#include <queue>

using namespace std;

bool checkfor(stack<int> &stk, int x){

   queue<int> q;

   bool flag = false;

   while(!stk.empty()){

       int p = stk.top();

       stk.pop();

       if(p == x){

           flag = true;

       }

       q.push(p);

   }

   while(!q.empty()){

       stk.push(q.front());

       q.pop();

   }

   while(!stk.empty()){

       q.push(stk.top());

       stk.pop();

   }

   // from queue to stack

   while(!q.empty()){

       stk.push(q.front());

       q.pop();

   }

   return flag;

}

int main(){

   stack<int> stk;

   // pushing data to stack

   stk.push(1);

   stk.push(3);

   stk.push(43);

   stk.push(2);

   stk.push(5);

   cout<<(checkfor(stk, 5)?"Found": "Not Found")<<'\n';

   cout<<(checkfor(stk, 10)?"Found": "Not Found")<<'\n';

   return 0;

}

Explanation:

Answer:

string_class.h:

include <string>

using namespace std;

class string_class

{

private:

  string current_string;

public:

  string_class();

  string_class(string temp);

  bool palindrome();

  string replace_all(string Old, string New);

  friend ostream& operator<<(ostream& os,string_class& temp);

};

string_class.cpp:

#include "string_class.h"

#include<algorithm>

using namespace std;

string_class::string_class()

{

  this->current_string = "";

}

string_class::string_class(string temp)

{

  this->current_string = temp;

}

bool string_class::palindrome()

{

  string temp = this->current_string;

  reverse(temp.begin(), temp.end());

  if (temp == this->current_string)

      return true;

  return false;

}

string string_class::replace_all(string Old, string New)

{

  for (int i = 0; i < int(this->current_string.length()); i++)

  {

      string replace = "";

      for (int j = 0; j < int(Old.length()); j++)

          replace += this->current_string[i + j];

      if (replace == Old)

      {

          string final = "";

          for (int j = i + Old.length(); j < int(this->current_string.length()); j++)

              final += this->current_string[j];

          this->current_string = final;

      }

  }

  return this->current_string;

}

ostream& operator<<(ostream& co,string_class& temp)

{

  co << temp.current_string;

  return co;

}

test.cpp:

#include "string_class.h"

#include <iostream>

#include <string>

using namespace std;

int main()

{

  string_class s;

  cout << "**************" << endl

      << "Test#1: tesing default constructor and overloaded operator<< with chaining\n"

      << s << "1st blank line" << endl << s << "2nd blank line" << endl

      << "Test#1 Ended" << endl

      << "**************" << endl;

  string_class r("hello");

  cout << "**************" << endl

      << "Test#2: tesing explicit-value constructor and overloaded operator<< with chaining\n"

      << r << "1st blank line" << endl << r << "2nd blank line" << endl

      << "Test#2 Ended" << endl

      << "**************" << endl;

  cout << "**************" << endl

      << "Test#3: tesing palindrome\n"

      << "**************" << endl;

  string response = "Y";

  string ss;

  while (response == "Y" || response == "y")

  {

      cout << "Enter String: ";

      cin >> ss;

      string_class main_string(ss);

      if (main_string.palindrome())

      {

          cout << ss << " is a palindrome\n";

      }

      else

      {

          cout << ss << " is not a palindrome\n";

      }

      cout << "Would you like to try another string? (Y or N): ";

      cin >> response;

  }

  cout << "Test#3 Ended" << endl

      << "**************" << endl;

  cout << "**************" << endl

      << "Test#4: tesing replace_all\n"

      << "**************" << endl;

  response = "y";

  string current, old_substring, new_substring;

  while (response == "Y" || response == "y")

  {

      cout << "Enter value for current_string: ";

      cin >> current;

      string_class current_string(current);

      cout << "Enter old_substring: ";

      cin >> old_substring;

      cout << "Enter new_substring: ";

      cin >> new_substring;

      cout << "Current string = " << current << endl

          << "New string = " << current_string.replace_all(old_substring, new_substring)

          << endl;

      cout << "Would you like to try another string? (Y or N): ";

      cin >> response;

  }

  cout << "**************" << endl

      << "Test#4: tesing replace\n"

      << "**************" << endl;

  return 0;

}

Explanation:

Answer:

Detailed answer is provided in the explanation section

Explanation:

Conversion of ARMv8 machine code into ARMv8 Assembly Language instructions:

ARMv8 machine code ARMv8 Assembly Language

0x0000000080001294 : EB01001F                    CMP X0,X1

0x0000000080001298 : 5400006A

0x000000008000129C : 91000400

0x00000000800012A0 : 17FFFFFD

0x00000000800012A4 : D61F03C0

Answer:

The answer is "True".

Explanation:

A single quote is a punching mark also known as an apostrophe located on the US Standard keyboard near the Enter button.

Answer:

weightPound = float(input("Enter your weight in pounds "))

heightInches = float(input("Enter your height in inches "))

weightKg = weightPound*0.453592

heightMeter = heightInches*0.0254

BodyMassIndex = weightKg/(heightMeter*heightMeter)

print("")

print("Your Weight in kilograms is: ")

print(weightKg)

print("Your Height in meters is: ")

print(heightMeter)

print("Your Body Mass Index is: ")

print(BodyMassIndex)

Explanation:

Answer:

datatable

Explanation:

Answer:

the answer is

data table

Explanation:

i think so

Answer:

Explanation: One of the noise difficult to remove is "Flicker noise". This noise is a 1/f power spectral density that occurs in almost all electronic devices, which increases the overall noise level. It also shows up with variety of other effects, such impurities in conductive channels of electronic components.

Error detection and correction in wireless signals work by modulating the carrier signals of a waveform through radio transmission.

Data is not compressed when transmitted, rather data is broken into fragments known as packets and transmitted by method of packet switched networking and applying a protocol called "Transmission control protocol".

Final answer:

The answer discusses the challenges of removing noise from analog and digital signals, error detection and correction with wireless signals, and the data compression process.

Explanation:

Analog vs. Digital Signal Noise

Most Difficult Noise to Remove: The most challenging noise to remove from an analog signal is cumulative noise that builds up with each copy due to continuous variability. In digital signals, noise is less problematic as it can be regenerated and removed.

Error Detection and Correction with Wireless Signals: Error detection and correction in wireless signals involve adding redundant data to check for errors and correct them if detected.

Data Compression Process: Data is compressed before transmission by reducing redundant information, utilizing encoding schemes like lossless or lossy compression.

Answer:

Before the project

Explanation:

We must o administrate and elaborate a prioritization about this redundant data, if we started after, we're going to lose a general view about the data, and when we're talking about redundant data always must be prevention and not monitoring, because any network must have redundant data, if we try to fix a network or database with this problem, we could delete delicate data.

Answer:

The answer to your question is true

Answer:

Following is the program in Java language:

import java.util.*;//import package

public class Main // main class

{

public static void main(String[] args) // main method

{

    String  firstname,lastname; // Declare the two String variables

    Scanner ob=new Scanner(System.in); // create a object of scanner //class

    System.out.println("Enter the first name:");  // Prompt the user for // enter the first name

   firstname=ob.nextLine();// Read in the first name by user

    System.out.println("Enter the last name:"); // Prompt the user for last //name

   lastname=ob.nextLine();// Read in the last name by user

   System.out.println("Hello " + firstname +' ' +lastname); // print the //firstname,lastname

}

}

Output:

Enter the first name:

San

Enter the last name:

ert

Hello San ert

Explanation:

Following are the description of program

Answer:

The answer is "False".

Explanation:

CBIS is part of IT, it is a structured combination of programs, technologies and human resources for the distribution of timely, standardized, accurate and useful decision-making knowledge.

Answer:

Option C.

Explanation:

Search engine use of pre-stored search that results is not mostly accurate.

Search engines shouldn't search the whole Internet each time when user requests for a search because It might be taken more time than the user might have expected and Many users were not involved in exploring the whole Internet.

The question given is incomplete. From the internet source I got it as follows:

Assume a five layer TCP/IP model. There are A bytes of application data. A B byte header is added at the transport layer, a C byte header is added at the internet layer and a D byte header is added at the network interface layer. The network interface layer also adds an E byte trailer. Express the answers to the questions below in terms of A, B, C, D, and E a. b. c. d.

How many total bytes are sent to the physical layer?

What percentage of the bytes sent to the physical layer is overhead?

How many total bytes are in the transport layer packet?

What percentage of the bytes in the transport layer packet is overhead?

Following is given the answer of above question part by part. The diagrams will help you to understand easily. I hope it will help you!

Answer:

The code is as below. The solution is given in python and can be converted  to other programming language.

Explanation:

As the language is not identified, it is assumed that the requirement of the language is python

As the complete question is missing, here the link to complete question is given in the comments above.

From the complete question, it is given that the approximated solution is for x86 instruction. thus as the x86 instructions are for integers only thus the complete code is as follows: Here the code is given as

#Defining the function

def relerror(x):

   x=float(x);

   y=int(x)

   f_h=y**(0.72);

   f=x**(0.72);

   absol=abs(f_h-f);

   rel=absol/f;

   relp=(absol/f)*100;

   print('The relative error for',x,' as value of x when approximated as x86 routine is ',round(relp,2),'%')

   #Main program calling given function

x=input('Please enter a value in float')

relerror(x)

Answer:

Following is attached the solution of each part step-by-step. I hope it will help you!

Explanation:

Answer:

Software and Hardware virtualization.

Explanation:

Virtualization is a process or the ability of a system to run multiple operating systems and virtually using all the physical resources, as though they are individual systems.

Hardware virtualization allows for hardware components like the storage, CPU, memory etc to be virtually used by multiple virtual machines. Software virtualization uses software managers to run individual guest operating systems on virtual machines.

Answer:

The program to this question as follows:

Program:

quote="You can always edit a bad page. You can’t edit a blank page."

name="Jodi Picoult"

print("Quote:\n",quote)

print ('\t\t\t\t\t\t\t',"Author name-", name)

Output:

Quote:

You can always edit a bad page. You can’t edit a blank page.

       Author name- Jodi Picoult

Explanation:

In the above python code, two variable "quote and name" is defined, in which variable both variable holds some string value.

Eucalyptus is the correct choice for creating a privately hosted computing infrastructure with easy interaction with AWS products, offering compatibility through AWS APIs.

Craig is looking for a solution to create a computing infrastructure that allows for internal hosting but also provides ease of interaction with AWS services. The correct answer to this question is c. Eucalyptus. Eucalyptus is an open-source software platform that enables organizations to implement on-premises Infrastructure as a Service (IaaS) clouds with strong compatibility with AWS products.

It allows businesses to create private clouds that are interoperable with AWS, using AWS APIs to manage resources within the Eucalyptus private cloud. This means developers can use the same tools and skills they would for AWS, providing a seamless transition between the private cloud and AWS services when needed for scalability or additional services.

A program that reads a person's first and last names, separated by a space is shown below.

Given that;

Write a program that reads a person's first and last names, separated by a space.

Here's a solution in C++ that should do the trick:

#include <iostream>

#include <string>

using namespace std;

int main() {

 string firstName;

 string lastName;

 // Prompt user for input

 cout << "Please enter your first and last names, separated by a space: ";

 cin >> firstName >> lastName;

 // Output the last name, comma, first name

 cout << lastName << ", " << firstName << endl;

 return 0;

}

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Final answer:

To solve the programming exercise, include cin to read two strings representing the first and last name, then output them in 'Last name, First name' format using cout, followed by a newline.

Explanation:

To complete the coding exercise, you need to read in a person's first and last name and then display the last name followed by a comma, and then the first name. You can achieve this by using the cin to read the names and cout to output the result in the required format.

The program will look like this:

#include
#include
using namespace std;
int main() {
   string firstName;
   string lastName;
   // Reading the first and last name
   cin >> firstName >> lastName;
   // Output in 'Last name, First name' format
   cout << lastName << ", " << firstName << endl;
   return 0;
}

Simply include the code segment within the "/* Your solution goes here */" section in your program. When you run this code, it will prompt you to input a first and last name, and after you provide it, the program will print out the last name, a comma, and then the first name followed by a newline as specified in the exercise.

Explanation:

Virtualization can be defined as a virtual or software-based representation of virtual applications, storage, network and servers. Through this process it is possible to run a variety of different operating systems on the same physical server at the same time, which reduces information technology expenses, increases the speed and agility of the systems.

Some of the benefits of virtualization are:

Explanation:

The internet is a tool that has revolutionized the way communication is carried out. As it is a dynamic tool with easy access for any user, it is possible to use different platforms or social media to realize disruptive or non-traditional forms of earnings.

Some of them could be the dissemination of links that would generate advertising for other companies, or the free dissemination in social media of services, and several other innovations that use the internet as the main means to manufacture an extra income, such as online teaching, advertising for third parties etc.

Answer:

import java.util.Scanner;

public class TestClock {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter Value in grams");

       double grams = in.nextDouble();

       double pounds = 453.592*grams;

       System.out.println(grams+" grams is equal to "+pounds+" pounds");

   }

}

Explanation:

The program that converts grams to pounds is as follows;

def grams_to_pounds(x):

  weight_in_pounds = x / 453.592

  return weight_in_pounds

w1 = grams_to_pounds(500)

w2 = grams_to_pounds(1000)

w3 = grams_to_pounds(2000)

print(w1)

print(w2)

print(w3)

learn more on python code here: https://brainly.com/question/1620423?referrer=searchResults

In the twentieth century, Xerox’s PARC had outstanding research personnel—at the time almost half of the world’s top 100 computer scientists

Explanation:

Answer:

Option A and Option D are the correct options.

Explanation:

They decided to allow their users with a different perspective when viewing a Salesforce1 transaction records rather than the specific web edition. So, they accomplish the following scenario by adding and allocating a Mobile Design to a profile and by adding the following page to the mobile card section.

Answer:

Program for the above question in python:

mass = float(input("Enter the mass in grams")) # It is used to take the mass as input.  

density = float(input("Enter the density grams per cm cube")) # It is used to take the input for the grams.

print("Volume ={:.2f}".format(mass/density)) #It is used to print the Volume.

Output:

Explanation:

To write a program that calculates and outputs the volume of an object using mass and density, you need to collect the mass and density from the user as input and use the formula volume = mass / density. Format the output to two decimal places using a print or format function that specifies the decimal precision.

To write a program that calculates and outputs the volume of an object using mass and density, you would need to collect the mass and density from the user as input. Then, you can use the formula volume = mass / density to calculate the volume. To format the output to two decimal places in most programming languages, you can use a print or format function that specifies the decimal precision. Here's an example in Python:

This program prompts the user to enter the mass and density, calculates the volume using the formula, and then formats the output to two decimal places using the '{:.2f}' format specifier.

https://brainly.com/question/20474319

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Explanation:

What happens is that the security of a company's information and data system can be affected by its people because it is the organization's own employees who can leak sensitive information, for their own benefit or for the benefit of others. That is why it is extremely relevant that organizations should limit the level of access that each user has to their information system.

Adequate training in the use of these systems is essential for employees to use it properly, to correct system failures, and for top management to have adequate knowledge in the preparation of system security plans.

It is the security policies that will prevent unauthorized access and greater control over the security of important data and information, so it is important that any company with a data information system keeps itself up to date with the software and implements necessary security resources, in a way maintain security and always update and analyze the operation of technological equipment.

Answer:

Routing table and routing table population.

Explanation:

A router is an intermediate network device used to determine a good path for routing packets. It is a layer 3 device. Each interface of the router is a broadcast domain, that is, it represents a network.

It uses its routing table to find paths to a network. The router's routing table can populated (to add more routes) statically and dynamically.

To populate a routing table statically, use the format, "IP rout 'network address' 'subnet mask' 'next hop address'".

To populate dynamically, the protocols like RIP, OSPF, BGP, EIGRP etc, are used. The static routing is good for small networks because routing table population is manual while dynamic routing is good for large networks because it's routing table population is automatic.

Answer:

a) with Array

class array{

   int arrA[MAX],location,item,nA;

   int arrB[MAX],nB;

   int arr_merge[MAX+MAX],nM;

   public:

   array(){

      location=0;item=0;nA=0;

      nB=0;nM=0;

   }

   void init(); //initial data assignment  

void traverse(); //process is display (assumed)void insert();

   void del();

   void search();

   void sort();

   void merge();

};

void array :: del(){

    clrscr();

    int i;

    cout<<"\n\n*****Deleting Element*****\n";

    if(nA < 0){

   cout<<"\nArray is Empty\nDeletion Not Possible\n";

   goto end;

    }

    cout<<"\nEnter Location of deletion : ";

    cin>>location;

    location--;

    if(location<0 || location>=nA)

    {

   cout<<"\n\nInvalid Position\n";

   goto end;

    }

    cout<<"\nItem deleted is : "<<arrA[location];

    for(i=location;i<nA;i++){

    arrA[i] = arrA[i+1];

    }

    arrA[nA-1]=0;

    nA--;

end:

     getch();

}

void array :: search(){

    clrscr();

    int i,found=-1;

    cout<<"\n\n*****Searching Element*****\n";

    cout<<"\nEnter Item value to be search : ";

    cin>>item;

    for(i=0;i<nA;i++){

    if(arrA[i] == item){

       found=i+1;

       break;

    }

    }

    if(found==-1)

    cout<<"\nSearch NOT FOUND\n";

    else

    cout<<"\nSearch is FOUND at "<<found<<" location\n";

    getch();

}

void array :: sort(){

    clrscr();

    int i,j,temp;

    cout<<"\n\n*****Sorting Element*****\n";

    for(i=0;i<nA;i++){

   for(j=i;j<nA;j++){

     if(arrA[i] > arrA[j]){

        temp    = arrA[i];

        arrA[i] = arrA[j];

        arrA[j] = temp;

     }

   }

    }

   cout<<"\nData are Sorted\n";

   getch();

}

// b) with pointer

#include <stdio.h>

#include <conio.h>

#include <math.h>

#include <alloc.h>

void main()

{

char *a[10],dum[10],s;

int i,k,j,n;

clrscr();

printf("enter the no of std....");

scanf("%d",&n);

printf("enter the name of students ");

for(k=0;k<n;k++)

scanf("%s",a[k]);

for(i=1;i<n;i++)

{

for(j=1;j<n-i;j++)

{if(strcmp(a[j-1],a[j])>0)

 {strcpy(*dum,*a[j-1]);

  strcpy(*a[j-1],*a[j]);

  strcpy(*a[j],*dum);

}

}  }

for(i=0;i<n;i++)

printf("%s",a[i]);

getch();

}

c) with cursor implementations of lists.

#include<stdio.h>

#include<stdlib.h>

typedef struct Node  

{

       int data;

       struct Node *next;

}node;

void insert(node *pointer, int data)

{

       /* Iterate through the list till we encounter the last node.*/

       while(pointer->next!=NULL)

       {

               pointer = pointer -> next;

       }

       /* Allocate memory for the new node and put data in it.*/

       pointer->next = (node *)malloc(sizeof(node));

       pointer = pointer->next;

       pointer->data = data;

       pointer->next = NULL;

}

int find(node *pointer, int key)

{

       pointer =  pointer -> next; //First node is dummy node.

       /* Iterate through the entire linked list and search for the key. */

       while(pointer!=NULL)

       {

               if(pointer->data == key) //key is found.

               {

                       return 1;

               }

               pointer = pointer -> next;//Search in the next node.

       }

       /*Key is not found */

       return 0;

}

void delete(node *pointer, int data)

{

       /* Go to the node for which the node next to it has to be deleted */

       while(pointer->next!=NULL && (pointer->next)->data != data)

       {

               pointer = pointer -> next;

       }

       if(pointer->next==NULL)

       {

               printf("Element %d is not present in the list\n",data);

               return;

       }

       /* Now pointer points to a node and the node next to it has to be removed */

       node *temp;

       temp = pointer -> next;

       /*temp points to the node which has to be removed*/

       pointer->next = temp->next;

       /*We removed the node which is next to the pointer (which is also temp) */

       free(temp);

       /* Beacuse we deleted the node, we no longer require the memory used for it .  

          free() will deallocate the memory.

        */

       return;

}

void print(node *pointer)

{

       if(pointer==NULL)

       {

               return;

       }

       printf("%d ",pointer->data);

       print(pointer->next);

       }

}

Running Time:

a) 2 minute

b) 1 minutes  

c) 1 & half minutes

Explanation:

a) Initialize the array and traverse it by using for loop. Sort the array with the help of a temporary variable.  

b) Declare a pointer to the array and take the input from user using for loop.

c) Declare a Structure containing an integer and a pointer. Iterate through the list, allocate memory for the new node and insert the data in it.