Write the routines with the following declarations: void permute( const string & str ); void permute( const string & str, int low, int high ); The first routine is a driver that calls the second and prints all the permutations of the characters in string str. If str is "abc", then the strings that are output are abc, acb, bac, bca, cab, and cba. Use recursion for the second routine.

Answer:

Answer explained

Explanation:

as PI 3 B+ is quadcore processor(i.e. 4 cores)

omp_get_thread_num() is used for getting thread number.(i.e. 0,1,2,3)

omp_get_num_threads() is used for getting number of threads.(i.e. 4)

1. Output should be,

Hello from thread 0 of 4

Hello from thread 1 of 4

Hello from thread 2 of 4

Hello from thread 3 of 4

2. #pragma omp parallel is used to fork(create child) additional threads to carry out the work enclosed in the construct in parallel.

3. We can Compile this code using,

  gcc -o omp_helloc -fopenmp hello.c

4. We can Run this code using,

   ./hello

5.omp_get_num_threads() is used for getting number of threads.(i.e. 4)

Answer:

Clock 2.5GHz

L1 I cache 32KB, 8way, 64B line size, 4 cycle access latency

L1 Dcache write-back, write-allocate; MSHR with 0 (lockup

cache), 1, 2, and 64 (unconstrained non-blocking

cache) entries, write-back buffer with 16 entries

L2 cache 256KB, 8way, 64B line size, 10 cycle access latency

L3 cache 2MB per core, 64B line size, 36 cycle access latency

Memory DDR3-1600, 90 cycle access latency

Issue width 4

Instruction window size 36

ROB Size 128

Load Buffer Size 48

Store Buffer Size 32

b)

parallelism took this a step further by providing more parallelism and hence more

latency-hiding opportunities. It is likely that the use of instruction- and threadlevel

parallelism will be the primary tool to combat whatever memory delays are

encountered in modern multilevel cache systems.

that of the lockup cache setup (hit-under-0-miss). For the integer programs: the average performance

(measured as CPI) improvement is 7.08% for hit-under-1-miss, 8.36% for hit-under-2-misses, and 9.02%

for hit-under-64-misses (essentially the unconstraint non-blocking cache), compared to lockup cache. For

the floating point programs, the three numbers are 12.69%, 16.22%, and 17.76%, respectively

c)

Non-blocking caches are an effective technique for tolerating cache-miss latency. They can reduce

miss-induced processor stalls by buffering the misses and continuing to serve other independent access

requests. Previous research on the complexity and performance of non-blocking caches supporting

non-blocking loads showed they could achieve significant performance gains in comparison to blocking

caches. However, those experiments were performed with benchmarks that are now over a decade old.

Furthermore the processor that was simulated was a single-issue processor with unlimited run-ahead

capability, a perfect branch predictor, fixed 16-cycle memory latency, single-cycle latency for floating

point operations, and write-through and write-no-allocate caches. These assumptions are very different

from today's high performance out-of-order processors such as the Intel Nehalem. Thus, it is time to

re-evaluate the performance impact of non-blocking caches on practical out-of-order processors using

up-to-date benchmarks. In this study, we evaluate the impacts of non-blocking data caches using the latest

SPECCPU2006 benchmark suite on practical high performance out-of-order (OOO) processors.

Simulations show that a data cache that supports hit-under-2-misses can provide a 17.76% performance

gain for a typical high performance OOO processor running the SPECCPU 2006 benchmarks in

comparison to a similar machine with a blocking cache.

Explanation:

Answer:

#include <iostream>

#include<iomanip>

using namespace std;

int main()

{

double mass,e,m,v;

cout<<"Enter the mass : ";

cin>>mass;

cout<<"The mass is "<<mass<<" kg\n";

if(mass<=0){

cout<<" mass must be greater than zero";

return 0;

}

e= mass * 9.81;

m= mass * 1.62;

v = mass * 8.87;

cout.setf(ios::fixed);

cout<<"Location"<<right<<setw(10)<<"Weight\n";

cout<<"Earth"<<right<<setw(15)<<setprecision(4)<<e<<endl;

cout<<"Moon"<<right<<setw(15)<<setprecision(4)<<m<<endl;

cout<<"Venus"<<right<<setw(15)<<setprecision(4)<<v<<endl;

cout<<endl<<endl;

if(m>=1500)

cout<<"Object is light weight";

else

cout<<"Object is heavy weight";

}

Explanation:

Final answer:

An object's weight is calculated by multiplying its mass by the acceleration due to gravity, which varies by celestial body. On Earth, the weight is the mass times 9.81 m/s², while on the Moon it is mass times 1.62 m/s², and on Venus, the weight is calculated with an acceleration due to gravity of 8.87 m/s².

Explanation:

The mass of an object is a measure of the amount of matter in that object and does not change regardless of the object's location. On the other hand, weight is the force exerted by gravity on an object and is calculated by multiplying the mass (in kilograms) by the acceleration due to gravity (in meters per second squared, m/s²). This acceleration due to gravity varies depending on the celestial body.

On Earth, the acceleration due to gravity is approximately 9.81 m/s², which means that an object's weight can be calculated as its mass multiplied by 9.81. For example, a mass of 100 kg on Earth would have a weight of 981 Newtons (N).

The Moon's gravitational acceleration is much lower, around 1.62 m/s². Thus, the same 100 kg object would weigh 162 N on the Moon. It's clear that, although mass remains consistent, weight varies significantly depending on the acceleration due to gravity at the location.

Similarly, on Venus, the acceleration due to gravity is 8.87 m/s². To find the weight of an object on Venus, you multiply its mass by this acceleration. So, a 100 kg object would weigh approximately 887 N on Venus.

The distinction between mass and weight is important as it highlights that weight can change locationally due to variations in gravitational pull, while mass is an intrinsic property of an object that does not change with location.

Answer:

Logic:

profit = retailPrice - wholeSalePrice

salePrice = retailPrice - (0.25 * retailPrice)

saleProfit = salePrice - wholeSalePrice

Java code:

double retailPrice, wholeSalePrice;

double profit = retailPrice - wholeSalePrice;

double salePrice = retailPrice - (0.25 * retailPrice);

double saleProfit = salePrice - wholeSalePrice;

Explanation:

The logic just represented the words in equation format. That is, profit is retailPrice minus wholesaleprice.

SalePrice is 25% deducted from retail price.

SaleProfit is saleprice minus wholesaleprice.

The whole logic is then represented in Java assignment statement using type double.

Answer:

C++.

#include <iostream>

using namespace std;

////////////////////////////////////////////////////////////////

int main() {

   int weekly_hours = 0;

   int hourly_rate;

   float gross_pay = 0;

   cout<<"Enter weekly hours worked: ";

   cin>>weekly_hours;

   cout<<"Enter hourly rate: ";

   cin>>hourly_rate;

   cout<<endl;

   ////////////////////////////////////////////////

   if (weekly_hours > 40) {

       gross_pay = (weekly_hours*hourly_rate) + ((weekly_hours*hourly_rate)*0.5);

   }

   else

       gross_pay = weekly_hours*hourly_rate;

   cout<<"Weekly gross pay: $"<<gross_pay;

   ////////////////////////////////////////////////

   return 0;

}

Answer:

B

Explanation:

Craig could of left the coffee shop but instead he could just shoulder surfing so no one could see all the other options are just plain Wrong shoulder surfing can do no harm

Answer:

Option B i.e., a production plan.

Explanation:

While sales are estimated, the manufacturing schedule for estimating the necessary raw materials should be produced.

So when the revenue is already estimated then at that time the manufacturing schedule should be produced approximately that raw material which is needed, so the following option is correct according to the scenario.

Answer:

#include<bits/stdc++.h>

using namespace std;

int main(){

  // Defining Variables

  int no_of_weeks;

  int total_cases = 0;

  //Declaring Vector of Pair of Integer and string

  std::vector<pair<int,string>> data;

  // Taking Input for the Number of Weeks

  cout<<"Enter No. of Weeks\n";

  cin >> no_of_weeks;

  // Running the Loop for no_of_weeks times

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

      int A,B,C;

      // Taking Input for different types of flus

      cout<<"Enter No. of Cases of Flu A, B, C for week" << i + 1 << " seperated by space : \n";

      cin >> A >> B >>C;

      // Adding all the cases in a week

      int cases_in_a_week = A + B + C;

      // Updating total cases

      total_cases += cases_in_a_week;

      // Declaring the level variable

      string level;

      // Updating the level of the week corresponding to each case

      if(cases_in_a_week < 500) level = "Low";

      else if(cases_in_a_week >= 500 && cases_in_a_week < 2000) level = "Moderate";

      else level = "Widespread";

      // Storing the Week's information by using a vector of pairs

      // in which pair's first is the number of cases which is of type int

      // while the second is the level of the flu which is of the type string

      data.push_back(make_pair(cases_in_a_week,level));

  }

  // Linking the stdoutput to the flu_report.txt file

  // this also creates the file with the same name if it doesn't exists

  freopen("flu_report.txt", "w", stdout);

  // Printing the respective output data with Bar Chart of stars for each level

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

      //printing the week no. and number of cases

      cout<<i+1<<" "<<data[i].first<<" "<<data[i].second<<" |";

      //calculating the number of stars

      int stars = data[i].first/250;

      //printing the stars of the bar chart

      for(int j = 0; j < stars ; j++) cout<<"*";

      cout<<endl;

  }

  //printing the total number of cases

  cout<<total_cases;

}

Explanation:

Answer:

C code explained below

Explanation:

I have provided the proper commented code below.

I hope that you find the answer helpful.

CODE:

-------------------------------------------------------------------------------------------------------------

#include<bits/stdc++.h>

using namespace std;

int main(){

  // Defining Variables

  int no_of_weeks;

  int total_cases = 0;

  //Declaring Vector of Pair of Integer and string

  std::vector<pair<int,string>> data;

  // Taking Input for the Number of Weeks

  cout<<"Enter No. of Weeks\n";

  cin >> no_of_weeks;

  // Running the Loop for no_of_weeks times

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

      int A,B,C;

      // Taking Input for different types of flus

      cout<<"Enter No. of Cases of Flu A, B, C for week" << i + 1 << " seperated by space : \n";

      cin >> A >> B >>C;

      // Adding all the cases in a week

      int cases_in_a_week = A + B + C;

      // Updating total cases

      total_cases += cases_in_a_week;

      // Declaring the level variable

      string level;

      // Updating the level of the week corresponding to each case

      if(cases_in_a_week < 500) level = "Low";

      else if(cases_in_a_week >= 500 && cases_in_a_week < 2000) level = "Moderate";

      else level = "Widespread";

      // Storing the Week's information by using a vector of pairs

      // in which pair's first is the number of cases which is of type int

      // while the second is the level of the flu which is of the type string

      data.push_back(make_pair(cases_in_a_week,level));

  }

  // Linking the stdoutput to the flu_report.txt file

  // this also creates the file with the same name if it doesn't exists

  freopen("flu_report.txt", "w", stdout);

  // Printing the respective output data with Bar Chart of stars for each level

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

      //printing the week no. and number of cases

      cout<<i+1<<" "<<data[i].first<<" "<<data[i].second<<" |";

      //calculating the number of stars

      int stars = data[i].first/250;

      //printing the stars of the bar chart

      for(int j = 0; j < stars ; j++) cout<<"*";

      cout<<endl;

  }

  //printing the total number of cases

  cout<<total_cases;

}

Answer:

Data encapsulation

Explanation:

When we send data to someone else across another network, this data travels down a stack of layers of the TCP/IP model. Each layer of the TCP/IP layer encapsulates data or hides the data by adding headers and sometimes trailers. This packaging is what is known as data encapsulation. The data starts at the application layer and trickles down beneath. As it moves down the layers, it is encoded or compressed to a standard format and sometimes encrypted for security purposes.

Answer:

Here is the program in C.

#include<stdio.h>   // header file used to input output operations

int permutation(int n, int r);  // function to computer permutation

int combination(int n, int r);  //function to compute combination

int factorial(int number);  // function to computer factorial

void par_input(int n, int r);    // function take take input value of n and r

int main()  //start of main() function body

{    

   int n,r;  // declare n and r variables

   par_input(n,r);  //calls par_input() function to get values of n and r

}

int permutation(int n, int r)   // method to calculate permutation

{

   return factorial(n) / factorial(n-r);  //formula for permutation

}

int combination(int n, int r)  // method to calculate combination

{

   return permutation(n, r) / factorial(r); //formula for combination

}

int factorial(int number)  // to find factorial of a number

{

   int fact = 1;      

   while(number > 0)  //loop continues until value of number gets <=0

   {

       fact = fact *number;  

       number--;

   }      

   return fact;

}

void par_input(int n, int r)  //function to take input values

{

   printf("Enter n: ");  //prompts user to enter the value of n

   scanf("%d", &n);      //reads the value of n from user

   printf("Enter r: "); ////prompts user to enter the value of r

   scanf("%d", &r); //reads the value of r from user

  //calls the combination and permutation methods to display results

   printf("The permutation is = %d\n", permutation(n, r));    

   printf("The Combination is = %d", combination(n, r));      

}

Explanation:

The output of the program is attached in the screen shot.

Here's the implementation for the requested tasks:

### Job Class

```csharp

using System;

public class Job

{

   private static int nextJobNumber = 1;

   public int JobNumber { get; }

   public string CustomerName { get; set; }

   public string JobDescription { get; set; }

   public double EstimatedHours { get; set; }

   public double Price => EstimatedHours * 45.00;

   public Job(string customerName, string jobDescription, double estimatedHours)

   {

       JobNumber = nextJobNumber++;

       CustomerName = customerName;

       JobDescription = jobDescription;

       EstimatedHours = estimatedHours;

   }

   public override bool Equals(object obj)

   {

       if (obj == null || GetType() != obj.GetType())

       {

           return false;

       }

       Job otherJob = (Job)obj;

       return JobNumber == otherJob.JobNumber;

   }

   public override int GetHashCode()

   {

       return JobNumber.GetHashCode();

   }

   public override string ToString()

   {

       return $"Job Number: {JobNumber}, Customer Name: {CustomerName}, Job Description: {JobDescription}, Estimated Hours: {EstimatedHours}, Price: {Price:C}";

   }

}

```

### JobDemo Form

```csharp

using System;

using System.Collections.Generic;

using System.Windows.Forms;

namespace JobDemo

{

   public partial class JobDemoForm : Form

   {

       private List<Job> jobs = new List<Job>();

       public JobDemoForm()

       {

           InitializeComponent();

       }

       private void addButton_Click(object sender, EventArgs e)

       {

           string customerName = customerNameTextBox.Text;

           string jobDescription = jobDescriptionTextBox.Text;

           double estimatedHours;

           if (double.TryParse(estimatedHoursTextBox.Text, out estimatedHours))

           {

               Job job = new Job(customerName, jobDescription, estimatedHours);

               if (!jobs.Contains(job))

               {

                   jobs.Add(job);

                   outputListBox.Items.Add(job.ToString());

                   totalLabel.Text = $"Total: {CalculateTotal():C}";

               }

               else

               {

                   MessageBox.Show("Duplicate job number. Please enter a unique job number.");

               }

           }

           else

           {

               MessageBox.Show("Invalid estimated hours. Please enter a valid number.");

           }

       }

       private double CalculateTotal()

       {

           double total = 0;

           foreach (var job in jobs)

           {

               total += job.Price;

           }

           return total;

       }

   }

}

```

### RushJob Class

```csharp

public class RushJob : Job, IComparable<RushJob>

{

   private const double RushJobPremium = 150.00;

   public RushJob(string customerName, string jobDescription, double estimatedHours)

       : base(customerName, jobDescription, estimatedHours)

   {

   }

   public override double Price => base.Price + RushJobPremium;

   public int CompareTo(RushJob other)

   {

       return JobNumber.CompareTo(other.JobNumber);

   }

}

```

### JobDemo3 Form

```csharp

using System;

using System.Collections.Generic;

using System.Linq;

using System.Windows.Forms;

namespace JobDemo

{

   public partial class JobDemo3Form : Form

   {

       private List<RushJob> rushJobs = new List<RushJob>();

       public JobDemo3Form()

       {

           InitializeComponent();

       }

       private void addButton_Click(object sender, EventArgs e)

       {

           string customerName = customerNameTextBox.Text;

           string jobDescription = jobDescriptionTextBox.Text;

           double estimatedHours;

           if (double.TryParse(estimatedHoursTextBox.Text, out estimatedHours))

           {

               RushJob rushJob = new RushJob(customerName, jobDescription, estimatedHours);

               if (!rushJobs.Any(rj => rj.JobNumber == rushJob.JobNumber))

               {

                   rushJobs.Add(rushJob);

                   outputListBox.Items.Add(rushJob.ToString());

                   totalLabel.Text = $"Total: {CalculateTotal():C}";

               }

               else

               {

                   MessageBox.Show("Duplicate job number. Please enter a unique job number.");

               }

           }

           else

           {

               MessageBox.Show("Invalid estimated hours. Please enter a valid number.");

           }

       }

       private double CalculateTotal()

       {

           double total = 0;

           foreach (var rushJob in rushJobs)

           {

               total += rushJob.Price;

           }

           return total;

       }

   }

}

This code should fulfill the requirements you've specified. Let me know if you need further assistance!

Answer:

Answer explained

Explanation:

The main difference between parse_text_into_words( v1 and v2) is I have used splitlines also in v1 so the escape sequence '\n' will not be present in the list of words. while in v2 only split function is used so it will have '\n' present in words list.

Code:

def read_file(fileName = None):

  try:

      File = open(fileName,"r") # opening file in read mode

      for row in File:          # for each line in File

          print(row,end="")

  except IOError:      # handling exception of file opening

      print("Could not read file:",fileName)

def parse_text_into_words_v1(text):

  list_words = []

  list_lines = text.splitlines()

  for i in list_lines:

      list_words.extend(i.split(' '))

  return list_words

def parse_text_into_words_v2(text):

  list_words = []

  list_words.extend(text.split(' '))

  return list_words

def determine_difference(a_list,b_list):

  print("first list:",a_list)

  print("second list:",b_list)

  a_set = set(a_list)

  b_set = set(b_list)

  diff = a_set - b_set

  diff = list(diff)

  print("\nDifference between 2 sets: ",diff)

if __name__ == '__main__':

  print("Reading file using function read_file() to read \"data.txt\" :")

  read_file("data.txt")

  print("\n\n")

  t = '''vhfhyh ghgggj ghchvjhvj'''

  print("\nDemonstrating implementation of parse_text_into_words_v1:")

  l = parse_text_into_words_v1(t) # calling function with text parameter

  print("list of words:")

  print(l)

  print("\nDemonstrating implementation of parse_text_into_words_v2:")

  b = parse_text_into_words_v2(t) # calling function with text parameter

  print("list of words:")

  print(b)

  print("\nDemonstrating difference between two lists")

  a_list = [1,2,3,4,5,6,7,8,9,10]

  b_list = [2,4,6,8,10]

  determine_difference(a_list,b_list) # passing two list to take difference

the function and loop will be

def factorial(x):

   total = 1

   if x != 1 and x != 0:

       for i in range(x,1,-1):

           total *= i

   return total

The program is an illustration of loops

Loops are program statements that are used to perform repeated operations

The program in python, where comments are used to explain each line is as follows:

#This gets input for N

N = int(input())

#This initializes factorial to 1

factorial = 1

#This opens the while loop

while N >1:

   #This calculates the factorial

   factorial*=N

   N-=1

#This prints the factorial

print(factorial)

Read more about loops at:

https://brainly.com/question/14284157

Answer:

Java code explained below

Explanation:

CourseGradePrinter.java

import java.util.Scanner;

public class CourseGradePrinter {

public static void main (String [] args) {

final int NUM_VALS = 4;

int[] courseGrades = new int[NUM_VALS];

int i = 0;

courseGrades[0] = 7;

courseGrades[1] = 9;

courseGrades[2] = 11;

courseGrades[3] = 10;

for(i=0; i<NUM_VALS; i++){

  System.out.print(courseGrades[i]+" ");

}

System.out.println();

for(i=NUM_VALS-1; i>=0; i--){

  System.out.print(courseGrades[i]+" ");

}

return;

}

}

Output:

7 9 11 10

10 11 9 7

Answer:

var count = 0;

var counterElement = document.getElementById("counter");

counterElement.innerHTML = count;

var interval = setInterval(function () {

   count++;

   counterElement.innerHTML = count;

   if (count === 4) {

       clearTimeout(interval);

   }

}, 100);

Explanation:

Ootions: TRUE OR FALSE

Answer: FALSE

Explanation: According to (Steer,2015) there are many platforms for learning and teaching purposes in corporate Organisations, they include PINTEREST,WIKI,GOOGLE+, LINKEDIN,TWITTER etc, The social media platforms are available for effective and efficient Communication and has been the main driving force for Businesses the world over.

These social media platforms have aided the growth and expansion of product marketing strategies and enhanced the over all business efficiency.

Answer:

Hi there Tonyhh! Please find the implementation below.

Explanation:

You can save the below code in a file called "price_discount.py". The code is implemented in python 2.0+. To make it work in Python 3.0+, simply update the code to use "input" rather than "raw_input".

price_discount.py

def calculate_discount(price, quantity):

 total = 0.95 * price * quantity;

 return total;

price = raw_input("Enter a price: ");

try:

 price = int(price);

 while price <= 0:

   print("Input a valid number: ")

   price = raw_input("Enter a price: ");

 quantity = raw_input("Enter a quantity: ");

 try:

   quantity = int(quantity);

   if quantity >= 10:

     print(calculate_discount(price, quantity));

 except ValueError:

   print("Invalid quantity!")

except ValueError:

 print("Invalid input!");

 price = -1;

Answer:

def name_facts(firstname):

   print("Your name is",len(firstname),"letters long",end=", ")

   if(firstname[0] == 'A'):

       print("does start with the letter A",end=", ")

   else:

       print("does not start with the letter A", end=", ")

   if(firstname.count('Z')>0 or firstname.count('X')>0):

       print("and does not contain a Z or X")

   else:

       print("and does not contain a Z or X")

#Testing

name_facts("Allegra")

name_facts("Xander")

Explanation:

Answer:

// Comments are used for explanatory purpose

// Program starts here

#include<iostream>

using namespace std;

int main()

{

// Declare firstname

string firstname;

//Prompt to enter firstname

cout<<"What's your first name: ";

// Accept input

cin>>firstname;

// A. Get string's length

int length = firstname.length();

// Print length

cout<<"Length = "<<length<<endl;

// Copy string to chat array

char char_array[length + 1];

// copying the contents of the string to char array

strcpy(char_array, firstname.c_str());

// B. Check if it starts with letter A

if(char_array[0] == 'A')

{

cout<<"Your Firstname begins with letter A"<<endl;

}

else

{

cout<<"Your Firstname doesn't begin with letter A"<<endl;

}

// Check if it contains X or Z

int k = 0;

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

if(char_array[i] == 'Z' || char_array[i] == 'X')

{

k++;

}

if(k == 0)

{cout<<"Your Firstname doesn't contain Z or X";}

else

{cout<<"Your Firstname contains Z or X";}

}

return 0;

}

Answer:

The bandwidth is 3200 Kbps

Explanation:

Given:

The length of the point to point link = 50 km = 50×10³ = 50000 m

Speed of transmission = 2 × 10⁸ m/s

Therefore the propagation delay is the ratio of the length of the link to speed of transmission

Therefore,  [tex]t_{prop} = \frac{distance}{speed}[/tex]

[tex]t_{prop} = \frac{50000}{2*10^{8} }=2.5*10^{-4}[/tex]

Therefore the propagation delay in 25 ms

100 byte = (100 × 8) bits

Transmission delay = [tex]\frac{(100*8)bits }{(x)bits/sec}[/tex]

If propagation delay is equal to transmission delay;

[tex]\frac{(100*8)bits }{(x)bits/sec}= \frac{50000}{2*10^{8} }[/tex]

[tex]x=\frac{100*8*2*10^{8} }{50000}[/tex]

[tex]x=\frac{100*8}{2.5*10 ^-4}=3200000[/tex]

x = 3200 Kbps

Answer:

The complete question is here:

Consider a point-to-point link 2 km in length. At what bandwidth would propagation delay (at a speed of 2 x 108 m/s) equal transmission delay for

(a) 100-byte packets?

(b) 512-byte packets?

Explanation:

Given:

Length of the link = 50 km

Propagation delay = distance/speed = d/s

To find:

At what speed would propagation delay (at a speed 2 x 10^8 meters/second) equal transmit delay for 100 byte packets?

Solution:

Propagation Delay = t prop

                                 = 50 * 10^3 / 2 * 10^8

                                 = 50000 / 200000000

                                 = 0.00025

                                 = 25 ms

(a) When propagation delay is equal to transmission delay for 100 packets:

Transmission Delay = packet length / data rate = L / R

So when,

Transmission Delay = Propagation Delay

100 * 8  / bits/sec = 50 * 10^3  / 2 * 10^8

Let y denotes the data rate in bit / sec

speed = bandwidth = y bits/sec

         100 * 8  / y bits/sec = 50 * 10^3  / 2 * 10^8

                        y bit/sec   =  100 * 8  / 25 * 10^ -5

                                         = 800 / 0.00025

                                     y  = 3200000 bit/sec

                                      y = 3200 Kbps

(b) 512-byte packets

512 * 8  / y bits/sec = 50 * 10^3  / 2 * 10^8

                        y bit/sec   =  512 * 8  / 25 * 10^ -5

                                         = 4096 / 0.00025

                                     y  = 16384000 bit/sec

                                      y = 16384 Kbps

Answer:

Here is the Python program:

import random  #to generate random numbers

low = 1  #lowest range of a guess

high = 100  #highest range of a guess

attempts=0   # no of attempts

def guess_number(low, high, attempts):  #function for guessing a number

   print("Guess a number between 1 and 100")      

#prompts user to enter an integer between 1 to 100 as a guess

   number = random.randint(low, high)  

#return randoms digits from low to high

   for _ in range(attempts):  #loop to make guesses

       guess = input("Enter a number: ")   #prompts user to enter a guess

       num = int(guess)  #reads the input guess

       if num == number:  # if num is equal to the guess number

           print('Congratulations! You guessed the right number!')    

           command= input("Do you want to play another game? ")

#asks user if he wants to play another game

           if command=='Y':  #if user enters Y  

               guess_number(low,high,attempts)  #starts game again

           elif command=='N':  #if user types N

               exit() #exits the program

       elif num < number:   #if user entered a no less than random no

           print('Too low, Try Higher')  # program prompts user to enter higher no

       elif num > number: #if user entered a no greater than random no

           print('Too high, Try Lower') # program prompts user to enter lower no

       else:  #if user enters anything else

           print("You must enter a valid integer.")  

#if user guessed wrong number till the given attempts

   print("You failed to guess the number correctly in {attempts} attempts.".format(attempts=attempts))  

   exit() # program exits

Explanation:

The program has a function guess_number which takes three parameters, high and low are the range set for the user to enter the number in this range and attempts is the number of tries given to the user to guess the number.

random.randint() generates integers in a given range, here the range is specified i.e from 1 to 100. So it generates random number from this range.

For loop keeps asking the user to make attempts in guessing a number and the number of tries the user is given is specified in the attempts variable.

If the number entered by the user is equal to the random number then this message is displayed: Congratulations! You guessed the right number! Then the user is asked if he wants to play one more time and if he types Y the game starts again but if he types N then the exit() function exits the game.

If the number entered as guess is higher than the random number then this message is displayed Too high, Try Lower and if the number entered as guess is lower than the random number then this message is displayed Too low, Try Higher.

If the user fails to guess the number in given attempts then the following message is displayed and program ends. You failed to guess the number correctly in {attempts} attempts.

You can call this function in main by passing value to this function to see the output on the screen.

guess_number(1, 100, 5)

The output is shown in screenshot attached.

Answer:

b. the IP address can be spoofed, so if you want to read response from the deceived party, you can use IP spoofing to hide yourself.

Explanation:

Answer:

The ARPANET network became functional in 1969, linking scientific and academic researchers across the United States.

Explanation:

ARPA Net is the network that has become functional in 1969 in united States. The basic purpose of this network is to link all the researchers and scientists across united states. The full form of ARPANet is Advanced Research Project Agency Network.

The importance of this network is increase because small chunks of data that are called packets has been proposed for data transmission in the network. The data is divided into small packets and send it over the network, at destination point these packets are combined together and become orignal information.

Answer:

#include <iostream>

#include<math.h>

using namespace std;

int findGcd(int x, int y) {

int t;

while(1) {

t= x%y;

if(t==0)

return y;

x = y;

y= t;

}

}

int main() {

int v1,v2,p,q,flag = 0;

cout<<"Please enter 1st prime number: ";

cin>>p;

v1 = p/2;

cout<<"Please enter 2nd prime number: ";

cin>>q;

v2 = q/2;

for(int i = 2; i <= v1; i++)

{

if(p%i == 0)

{

cout<<"The number is not prime";

flag = 1;

return 0 ;

}

}

for(int i = 2; i <= v1; i++)

{

if(q%i == 0)

{

cout<<"You entered a number which is not prime";

flag = 1;

return 0;

}

}

double n=p*q;

double tr;

double phi= (p-1)*(q-1);

double e=7;

if( flag == 0)

{

while(e<phi) {

tr = findGcd(e,phi);

if(tr==1)

break;

else

e++;

}

cout<<"public key: "<<e<<endl;

double d1=1/e;

double d=fmod(d1,phi);

cout<<"private key: "<<d<<endl;

double message;

cout<<"please enter a message: ";

cin>>message;

double cipher = pow(message,e);

double m = pow(cipher,d);

cipher = fmod(cipher,n);

m = fmod(m,n);

cout<<"Encrypted message: "<<cipher;

}

}

Explanation:

Answer:

The cpp program for conversion is given below.

#include <iostream>

using namespace std;

int main() {

// declaring and initializing constant variables with standard values

   const  double INCHES_IN_MILE=63360;

   const  double FEET_IN_MILE=5280;

   const  double YARDS_IN_MILE=1760;

   // declaring and initializing miles with a random value

   double miles=4.0;

// declaring and computing inches in miles

double inch = INCHES_IN_MILE*miles;

// declaring and computing feet in miles

double feet = FEET_IN_MILE*miles;

// declaring and computing yards in miles

double yard = YARDS_IN_MILE*miles;

// displaying all the calculated values

std::cout<<miles<<" miles is "<<inch<<" inches"<<std::endl;

std::cout<<miles<<" miles is "<<feet<<" feet"<<std::endl;

       std::cout<<miles<<" miles is "<<yard<<" yards"<<std::endl;

       return 0;      

}

OUTPUT

4 miles is 253440 inches

4 miles is 21120 feet

4 miles is 7040 yards

Explanation:

The steps in the program are as described.

1. All the three constant variables, mentioned in the question, are declared as double and initialized. These variables hold the fixed conversion values for inches, feet and yards in one mile.

2. The variables are declared constant using the keyword, const.

3. The variable to hold value for mile, miles, is declared as double and initialized with a random value.

4. Next, three more variables are declared as double. These variables will hold the calculated values for inches, feet and yards in the given value of mile.

5. The values are calculated for inches, feet and yards for the given value of variable, miles and stored in the variables mentioned in step 3.

6. These values are displayed to the console along with explanations.

7. The program ends with the return statement.

8. The value of variable, miles, is hard coded in the program and not taken from the user.

9. There is no interaction with the user in this program as this is not mentioned in the question.

10. The program works for all numeric value of the variable, miles, since double accepts integers also.

11. The program can be tested by changing the value of variable, miles.

Answer:

See attached file for detailed code.

Explanation:

See attached file.

Answer:

public class Rectangle {

   private double width;

   private double heigth;

   public Rectangle(double width, double heigth) {

       this.width = width;

       this.heigth = heigth;

   }

   public double getWidth() {

       return width;

   }

   public void setWidth(double width) {

       this.width = width;

   }

   public double getHeigth() {

       return heigth;

   }

   public void setHeigth(double heigth) {

       this.heigth = heigth;

   }

   public double perimeter(double width, double heigth){

       double peri = 2*(width+heigth);

       return peri;

   }

   public double area(double width, double heigth){

       double area = width*heigth;

       return  area;

   }

}

class RectangleTest{

   public static void main(String[] args) {

       //Creating two Rectangle objects

       Rectangle rectangle1 = new Rectangle(4,40);

       Rectangle rectangle2 = new Rectangle(3.5, 35.7);

       //Calling methods on the first Rectangel objects

       System.out.println("The Height of Rectangle 1 is: "+rectangle1.getHeigth());

       System.out.println("The Width of Rectangle 1 is: "+rectangle1.getWidth());

       System.out.println("The Perimeter of Rectangle 1 is: "+rectangle1.perimeter(4,40));

       System.out.println("The Area of Rectangle 1 is: "+rectangle1.area(4,40));

       // Second Rectangle object

       System.out.println("The Height of Rectangle 2 is: "+rectangle2.getHeigth());

       System.out.println("The Width of Rectangle 2 is: "+rectangle2.getWidth());

       System.out.println("The Perimeter of Rectangle 2 is: "+rectangle2.perimeter(4,40));

       System.out.println("The Area of Rectangle 2 is: "+rectangle2.area(4,40));

   }

}

Explanation:

The answer & explanation for this question is given in the attachment below.

The SQL query selects vendor contacts' first names followed by last initials and phone numbers without area codes from the Vendors table. It filters vendors with the 559 area code and sorts results alphabetically by first name and last name.

Below is the SQL SELECT statement that fulfills your requirements:

```sql

SELECT

   CONCAT(SUBSTRING_INDEX(ContactName, ' ', 1), ' ', LEFT(SUBSTRING_INDEX(ContactName, ' ', -1), 1), '.') AS Contact,

   SUBSTRING(VendorPhone, 5) AS Phone

FROM

   Vendors

WHERE

   VendorPhone LIKE '559%'

ORDER BY

   SUBSTRING_INDEX(ContactName, ' ', 1),

   SUBSTRING_INDEX(ContactName, ' ', -1);

```

This query selects the ContactName column from the Vendors table, concatenates the first name and the first letter of the last name, followed by a period, to create the Contact column.

It then extracts the phone number without the area code from the VendorPhone column and stores it in the Phone column.

Finally, it filters the results to include only vendors with the 559 area code and sorts the result set by first name and last name.

Answer:

Attached is the solution. Hope it helps:

a) The class of context-free languages is not closed under intersection, as shown by the example of[tex]A = [ a^m b^n c^n | m, n \geq 0 ][/tex] and [tex]B = [a^n b^n c^m | m, n \geq 0 ][/tex]

Given that;

We have seen in class that the sets of both regular and context-free languages are closed under the union, concatenation, and star operations.

(a) We're given two languages A and B:

[tex]A = [ a^m b^n c^n | m, n \geq 0 ][/tex]

[tex]B = [a^n b^n c^m | m, n \geq 0 ][/tex]

We want to show that the intersection of A and B is not context-free.

To do this, we can use the fact that the language [tex]C = [a^n b^n c^n | n \geq 0 ][/tex] is not context-free.

Assume for contradiction that the intersection of A and B, which we'll call D, is context-free.

Since context-free languages are closed under intersection, we'd have D = A ∩ B is also a context-free language.

Now, notice that D is a subset of C (if a string belongs to D, it must belong to C as well).

However, C is not context-free, as given.

This contradicts our assumption that D is context-free.

Therefore, we can conclude that the class of context-free languages is not closed under intersection, as shown by the example of[tex]A = [ a^m b^n c^n | m, n \geq 0 ][/tex] and [tex]B = [a^n b^n c^m | m, n \geq 0 ][/tex]

(b) Using the result from part (a), we can now show that the set of context-free languages is not closed under complementation.

Let's consider the complement of a context-free language.

If the complement of a context-free language were always context-free, we could take the complement of C and obtain a context-free language.

However, as established earlier, C is not context-free.

Hence, we can conclude that the set of context-free languages is not closed under complementation.

Learn more about Programs here:

brainly.com/question/14368396

#SPJ3

Answer:

Explanation:

hello we will follow a step by step process for this code, i hope you find it easy.

Mips Equivalent code:

  sw      $0,0($fp)

.L7:

       lw      $2,12($fp)

       addiu   $2,$2,-1

       lw      $3,0($fp)

       slt     $2,$3,$2

       beq     $2,$0,.L2

       nop

       lw      $2,0($fp)

       sw      $2,8($fp)

       lw      $2,0($fp)

       addiu   $2,$2,1

       sw      $2,4($fp)

.L5:

       lw      $3,4($fp)

       lw      $2,12($fp)

       slt     $2,$3,$2

       beq     $2,$0,.L3

       nop

       lw      $2,8($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $3,24($2)

       lw      $2,4($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $2,24($2)

       slt     $2,$2,$3

       beq     $2,$0,.L4

       nop

       lw      $2,4($fp)

       sw      $2,8($fp)

.L4:

       lw      $2,4($fp)

       addiu   $2,$2,1

       sw      $2,4($fp)

       b       .L5

       nop

.L3:

       lw      $3,8($fp)

       lw      $2,0($fp)

       beq     $3,$2,.L6

       nop

       lw      $2,0($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $2,24($2)

       sw      $2,16($fp)

       lw      $2,8($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $3,24($2)

       lw      $2,0($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       sw      $3,24($2)

       lw      $2,8($fp)

       dsll    $2,$2,2

       daddu   $2,$fp,$2

       lw      $3,16($fp)

       sw      $3,24($2)

.L6:

    lw      $2,0($fp)

       addiu   $2,$2,1

       sw      $2,0($fp)

       b       .L7

        nop

cheers  i hope this helps

Convert C code to MIPS assembly using registers for array operations and loops, implementing selection sort for efficient sorting of array elements.

To convert the given C code snippet to MIPS assembly, we need to translate each part of the algorithm into MIPS instructions while considering the provided register assignments.

1. Initialization:

  -Registers: `$s0` holds the base address of the array, `$s1` holds `n`, `$t0` holds `position`, `$t1` holds `c`, `$t2` holds `d`, and `$t3` holds `swap`.

2. Outer Loop (`for c`):

  - Use a label (`outer_loop`) to mark the start of the outer loop.

  - Initialize `c` to `0` and check if `c < n - 1`.

  - Increment `c` after each iteration.

3. Inner Loop (`for d`):

  - Inside the outer loop, set `position` to `c`.

  - Use another label (`inner_loop`) to mark the start of the inner loop.

  - Initialize `d` to `c + 1` and check if `d < n`.

  - Compare `array[position]` with `array[d]` and update `position` if `array[d]` is smaller.

4.Swap Condition:

  - After the inner loop, check if `position` is not equal to `c`.

  - If true, swap `array[c]` and `array[position]` using `$t3` (`swap`) as a temporary register.

5. Assembly Code:

 assembly

  # Register assignments:

  # $s0 - base address of array

  # $s1 - n

  # $t0 - position

  # $t1 - c

  # $t2 - d

  # $t3 - swap  

  # Outer loop (for c)

  li $t1, 0          # c = 0

  outer_loop:

      blt $t1, $s1, end_outer_loop  # if c >= n, exit outer loop      

      # Inner loop (for d)

      move $t0, $t1    # position = c

      addi $t2, $t1, 1 # d = c + 1

      inner_loop:

          blt $t2, $s1, end_inner_loop  # if d >= n, exit inner loop          

          # Load array[position] and array[d]

          lw $t4, 0($s0)           # $t4 = array[position]

          lw $t5, 0($s0)($t2)      # $t5 = array[d]          

          # Compare array[position] > array[d]

          bgt $t4, $t5, update_position

          j next_iteration_inner_loop

          update_position:

              move $t0, $t2  # position = d        

          next_iteration_inner_loop:

              addi $t2, $t2, 1  # d++

              j inner_loop      

      end_inner_loop:      

      # Swap if position != c

      bne $t0, $t1, swap_elements

      j next_iteration_outer_loop      

      swap_elements:

          lw $t3, 0($s0)($t1)     # $t3 = array[c]

          lw $t4, 0($s0)($t0)     # $t4 = array[position]        

          # Swap array[c] and array[position]

          sw $t4, 0($s0)($t1)     # array[c] = array[position]

          sw $t3, 0($s0)($t0)     # array[position] = swap      

      next_iteration_outer_loop:

          addi $t1, $t1, 1  # c++

          j outer_loop      

  end_outer_loop:

  - Explanation: The MIPS assembly code mirrors the logic of the C code by using load (`lw`) and store (`sw`) instructions to access array elements, branches (`bgt`, `bne`, `blt`) for conditional logic, and loops (`j`) to control flow. It effectively implements the selection sort algorithm to sort the array in ascending order.  

  - Efficiency: This implementation efficiently sorts the array in-place using O(n^2) time complexity, suitable for moderately sized arrays given the constraints of MIPS architecture.

Complete Question;

convert the following c code to mips. assume the address of base array is associated with $s0, n is associated with $s1, position is associated with $t0, c is associated with $t1, d is associated with $t2, and swap is associated with $t3 for

for ( c = 0 c < (n - 1) c++)

{

position = c

for ( d = c + 1 d < n d ++)

{ if (array[position] > array[d])

position = d

}

if (position != c)

{

swap array[c];

array[c] = array[position];

array[position] = swap;

}

}

Answer: isolation, predominant gender.

Explanation:

Isolation some students felt they were not as smart as their peers and as such were afraid to make contributions because they didn't want to be rediculed by Their peers or tutors.

Predominant gender most calsses are dominated by a certain gender which makes the other gender harbour fears and insecurity

I had difficulty in maths in high school, was always afraid when in class to make contributions because i was insecure, if i had spoken up or met my tutors maybe i would have being better at it