Which of the following are true of trademarks?
Check all of the boxes that apply.
They protect logos.
They protect fabric.
They show who the maker of the item is.
They can be registered with the government.

Answer:

total = 0

count = 0

while(True):

   grade = float(input("Enter a grade: "))

   if grade < 0:

       break

   else:

       total += grade

       count += 1

average = total/count

print("The average is: " + str(average))

Explanation:

*The code is in Python.

Initialize the total and count as 0

Create a while loop that iterates until a specific condition is met inside the loop

Inside the loop, ask the user to enter a grade. If the grade is smaller than 0, stop the loop. Otherwise, add the grade to the total and increment the count by 1.

When the loop is done, calculate the average, divide the total by count, and print it

Answer:

In dynamic scoping the current block is searched by the compiler and then all calling functions consecutively e.g. if a function a() calls a separately defined function b() then b() does have access to the local variables of a(). The visible variables with the name of the function in which it was defined are given below.

Explanation:

a. main calls fun1; fun1 calls fun2; fun2 calls fun3

Solution:

b. main calls fun1; fun1 calls fun3

Solution:

c. main calls fun2; fun2 calls fun3; fun3 calls fun1

Solution:

Here variables c, d and e of fun2 are not visible .

d. main calls fun1; fun1 calls fun3; fun3 calls fun2

Solution:

Answer:

a) tie the exception condition to a processing step

Explanation:

Numbering exception conditions, in a fully developed use case description is helpful to tie the exception condition to a processing step

The fully developed use case description is useful for the documentation of the context, purpose, description, conditions, and workflow of each use case. Activity diagrams give a graphical image of the use case workflow and serve the purpose of illustrating the alternative paths through a business process.

Answer:

Given:

Passwords are selected from 4 characters.

Character combinations are 26 lower case alphabetic characters.

Passwords attempts by adversary is at rate of 1 second.

To find:

Expected time to discover the correct password

Explanation:

Solution:

4 character combinations of 26 alphabetic characters at the rate of one attempt at every 1 second = 26 * 4 = 104

So 104 seconds is the time to discover the correct password. However this is the worst case scenario when adversary has to go through every possible combination. An average time or expected time to discover the correct password is:

13 * 4 = 52

You can also write it as 104 / 2 = 52 to discover the correct password. This is when at least half of the password attempts seem to be correct.

Answer:

The csharp program is as follows.

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

using System.Windows.Forms;  

namespace WindowsFormsApplication1

{

   public partial class Form1 : Form

   {

       public Form1()

       {

           InitializeComponent();

       }

private void button1_Click(object sender, EventArgs e)

       {

           string word = textBox1.Text;

               string  ch = word.Substring(0, 1);

               string str = word.Substring(1, word.Length-1);

               string s = str.Insert(str.Length, ch);

               textBox2.Text = s.Insert(s.Length, "ay");            

       }

     private void button3_Click(object sender, EventArgs e)

       {

           Close();

       }

       private void button2_Click(object sender, EventArgs e)

       {

           textBox1.Text = "";

           textBox2.Text = "";

       }

   }

}

Explanation:

1. A string variable to hold the user input is declared and initialized accordingly. The user inputted string is taken from textbox1.

string word = textBox1.Text;

2. A string variable to hold the first character of the user inputted string is declared and initialized.

string  ch = word.Substring(0, 1);

3. A string variable to hold the user inputted string without the first character is declared and initialized accordingly.

string str = word.Substring(1, word.Length-1);

4. A string variable to hold the substring from step 3 along with the inserted characters at the end, is declared and initialized accordingly.

string s = str.Insert(str.Length, ch);

5. The final string is assigned to the textbox 2, which is the PigLatin conversion of the user inputted string.

textBox2.Text = s.Insert(s.Length, "ay");

6. All the above take place when the user clicks Convert to PigLatin button.

7. Two additional buttons, clear and exit are also included in the form.  

8. When the user clicks clear button, both the textboxes are initialized to empty string thus clearing both the textboxes.

 textBox1.Text = "";

           textBox2.Text = "";

9. When the user clicks the exit button, the application closes using the Close() method.

10. The program is done in Visual Studio.

11. The output of the program is attached.

12. The program can be tested for any type of string and any length of the string.

Answer:

weight1 = float(input("Enter the weight of first package: "))

price1 = float(input("Enter the price of first package: "))

weight2 = float(input("Enter the weight of second package: "))

price2 = float(input("Enter the price of second package: "))

if weight1 > 0 and price1 > 0 and weight2 > 0 and price2 > 0:

   unit_cost1 = price1 / weight1

   unit_cost2 = price2 / weight2

   if unit_cost1 < unit_cost2:

       print("Package 1 has a better price.")

   else:

       print("Package 2 has a better price.")

else:

   print("All the entered values must be positive!")

Explanation:

*The code is in Python.

Ask the user to enter the weight and the price of the packages

Check if the all the values are greater than 0. If they are, calculate the unit price of the packages, divide the prices by weights. Then, compare the unit prices. The package with a smaller unit price has a better price.

If all the entered values are not greater than 0, print a warning message

Answer:

Explanation:

The objective here is to determine if the programmer can simply increase the upperbound to 80,000.

Of course Yes, The programmer can simply increase the delay by doubling the upperbound by 80000. The representation can be illustrated as:

( int : i = 0;  i <  40,000; i ++ )

{

  // delay code

}

Which can be modified as:

( int : i = 0;  i <  80,000; i ++ )

{

  // delay code

}

b)  If the code contains a delay loop and we noticed that no delay is being created at run-time. What should we suspect during debugging?

Assuming there is no delay being created at the run-time,

The code is illustrated as:

For ( int : i = 0 ; i < 0 ; i ++ )

{

  // delay code which wont

  //execute since code delay is zero

}

we ought to check whether the loop is being satisfied or not.  At the Initial value of loop variable, is there any break or exit statement is being executed in between loop. Thus, the  aforementioned delay loop wont be executed since the loop wont be executed for any value of i.

Answer:

This program is written using python

It uses less comments (See explanation section for more explanation)

Also, see attachments for proper view of the source code

Program starts here

#Prompt user for price of package 1

price1 = int(input("Enter Price 1: "))

while(price1 <= 0):

     price1 = int(input("Enter Price 1: "))

#Prompt user for weight of package 1

weight1 = int(input("Enter Weight 1: "))

while(weight1 <= 0):

     weight1 = int(input("Enter Weight 1: "))

#Calculate Unit of Package 1

unit1 = float(price1/weight1)

print("Unit cost of Package 1: "+str(unit1))

#Prompt user for price of package 2

price2 = int(input("Enter Price 2: "))

while(price2 <= 0):

     price2 = int(input("Enter Price 2: "))

#Prompt user for weight of package 2

weight2 = int(input("Enter Weight 2: "))

while(weight2 <= 0):

     weight2 = int(input("Enter Weight 2: "))

#Calculate Unit of Package 2

unit2 = float(price2/weight2)

print("Unit cost of Package 2: "+str(unit2))

#Compare units

if unit1 > unit2:

     print("Package 1 has a better price")

elif unit1 == unit2:

     print("Both Packages have the same price")

else:

     print("Package 2 has a better price")

Explanation:

price1 = int(input("Enter Price 1: ")) -> This line prompts the user for price of package 1

The following while statement is executed until user inputs a value greater than 1 for price

while(price1 <= 0):

     price1 = int(input("Enter Price 1: "))

weight1 = int(input("Enter Weight 1: ")) -> This line prompts the user for weight of package 1

The following while statement is executed until user inputs a value greater than 1 for weight

while(weight1 <= 0):

     weight1 = int(input("Enter Weight 1: "))

unit1 = float(price1/weight1) -> This line calculates the unit cost (per weight) of package 1

print("Unit cost of Package 1: "+str(unit1)) -> The unit cost of package 1 is printed using this print statement

price2 = int(input("Enter Price 2: ")) -> This line prompts the user for price of package 2

The following while statement is executed until user inputs a value greater than 1 for price

while(price2 <= 0):

     price2 = int(input("Enter Price 2: "))

weight2 = int(input("Enter Weight 2: ")) -> This line prompts the user for weight of package 2

The following while statement is executed until user inputs a value greater than 1 for weight

while(weight2 <= 0):

     weight2 = int(input("Enter Weight 2: "))

unit2 = float(price2/weight) -> This line calculates the unit cost (per weight) of package 2

print("Unit cost of Package 2: "+str(unit2)) -> The unit cost of package 2 is printed using this print statement

The following if statements compares and prints which package has a better unit cost

if unit1 > unit2:

     print("Package 1 has a better price")

elif unit1 == unit2:

     print("Both Packages have the same price")

else:

     print("Package 2 has a better price")

Answer:

def to_pig_latin(word):

   new_word = word[1:] + word[0] + "ay"

   return new_word

print(to_pig_latin("hello"))

print(to_pig_latin("latin"))

Explanation:

Create a function called to_pig_latin that takes one parameter, word

Inside the function, create a new_word variable and set it to the characters that are between the second character and the last character (both included) of the word (use slicing) + first character of the word + "ay". Then, return the new_word.

Call the to_pig_latin function twice, first pass the "hello" as parameter, and then pass the "latin" as parameter

Answer:

...huh?

Explanation:

Answer:

Replace /* Your solution goes here */  with the following lines of code

for(i = 0;i<SCORES_SIZE-1;i++)

{

bonusScores[i]+=bonusScores[i+1];

}

Explanation:

The above iteration starts from the index element (element at 0) and stops at the second to the last element (last - 1).

Using an iterative variable, i

It adds the current element (element at i) with the next element; element at i + 1.

The full code becomes

#include<iostream>

using namespace std;

int main(void) {

const int SCORES_SIZE = 4;

int bonusScores[SCORES_SIZE];

int i = 0;

bonusScores[0] = 10;

bonusScores[1] = 20;

bonusScores[2] = 30;

bonusScores[3] = 40;

for(i = 0;i<SCORES_SIZE-1;i++)

{

bonusScores[i]+=bonusScores[i+1];

}

/* Your solution goes here */

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

printf("%d ", bonusScores[i]);

}

printf("\n");

return 0;

}

See attachment for .cpp file

Answer:int main() {

const int SCORES_SIZE = 4;

int bonusScores[SCORES_SIZE];

int i;

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

cin >> bonusScores[i];

}

for (i = 0; i < SCORES_SIZE-1; ++i){

bonusScores[i] += bonusScores[i+1];

}

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

cout << bonusScores[i] << " ";

}

cout << endl;

return 0;

}

Explanation: SCORES_SIZE -1 will prevent the for loop from going past the last value in the array. bonusScores[i] += will add the value of bonusScores[i+1] to the original bonusScores[i].

for example, i = 1; 1 < SCORES_SIZE - 1 ; bonusScores[1] += bonusScores[1+1} becomes{ bonusScores[1] + bonusScores{2];

Answer:

Many-to-one (multiple inputs, single output)

Explanation:

Solution

In the case of a recurrent neural network that listens to a audio speech sample, and classifies it according to whose voice it is, the RNN will listen to the audio and will give result by doing classification.

There will be a single output, for the classified or say identified person.

The RNN will take a stream of input as the input is a audio speech sample.

Therefore, there will be multiple inputs to the RNN and a single output, making the best fit Architecture to be of type Many-to-one(multiple inputs, single output).

Answer:

i think html

Explanation:

Answer:

Web browser

Explanation:

Answer:

Explanation:

Relevance of Thrown Items:

Recommendation To Management:

cheers i hope this helped !!

Answer:

The csharp program is given below.

using System;

using System.Collections.Generic;

using System.ComponentModel;

using System.Data;

using System.Drawing;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

using System.Windows.Forms;

namespace WindowsFormsApplication1

{

   public partial class Form1 : Form

   {

       public Form1()

       {

           InitializeComponent();

      }

       private void button6_Click(object sender, EventArgs e)

       {

           int total = (Convert.ToInt32(textBox1.Text) +  Convert.ToInt32(textBox2.Text) + Convert.ToInt32(textBox3.Text) + Convert.ToInt32(textBox4.Text) + Convert.ToInt32(textBox5.Text));

           int dozen = total / 12;

           int eggs = total % 12;

           textBox6.Text = dozen.ToString();

           textBox7.Text = eggs.ToString();

       }

       private void button7_Click(object sender, EventArgs e)

       {

           textBox1.Text = "";

           textBox2.Text = "";

           textBox3.Text = "";

           textBox4.Text = "";

           textBox5.Text = "";

           textBox6.Text = "";

           textBox7.Text = "";

       }

        private void button8_Click(object sender, EventArgs e)

       {

           Close();

       }

    }

}

Explanation:

1. The integer variables are declared for total eggs, number of dozens and number of eggs.

2. The input of each text box is converted into integer by using the Convert.ToInt32() method on the value of that particular text box.

3. All the inputs are added and the sum is assigned to variable, total.

4. The number of dozens are obtained by dividing total by 12 and assigning the value to the variable, dozen.

5. The number of extra eggs are obtained by taking the modulus of total and 12 and the value is assigned to the variable, eggs.

6. The integer values in the variables, dozen and eggs, are converted into string using the ToString() function with that particular value.

dozen.ToString();

eggs.ToString();

7. The text boxes are assigned the respective values of dozens and number of eggs.

textBox6.Text = dozen.ToString();

textBox7.Text = eggs.ToString();

8. Two additional buttons, clear and exit, are also added.

9. The clear button erases the contents of all the text boxes. The Text property of each textbox is set to “” thereby clearing all the text fields.

10. The exit button closes the application using Close() function.

11. The program is made in visual studio and the output is attached.

Answer:

Explanation:

public class Team {

   private String teamName;

   private int teamWins;

   private int teamLosses;

   public String getTeamName() {

       return teamName;

   }

   public void setTeamName(String teamName) {

       this.teamName = teamName;

   }

   public int getTeamWins() {

       return teamWins;

   }

   public void setTeamWins(int teamWins) {

       this.teamWins = teamWins;

   }

   public int getTeamLosses() {

       return teamLosses;

   }

   public void setTeamLosses(int teamLosses) {

       this.teamLosses = teamLosses;

   }

   public double getWinPercentage() {

       return teamWins / (double) (teamWins + teamLosses);

   }

}

Following are the Java program to define the Team class and calculate its  value:

class Team //defining the class Team

{

   private String teamName;//defining String variable

   private int teamWins, teamLosses;//defining integer variable

   //defining the set method to set value the input value

   public void setTeamName(String teamName)//defining setTeamName method that takes one String parameter

   {

       this.teamName = teamName;//using this keyword that sets value in teamName

   }

   public void setTeamWins(int teamWins) //defining setTeamWins method that takes one integer parameter

   {

       this.teamWins = teamWins;//using this keyword that sets value in teamWins

   }

   public void setTeamLosses(int teamLosses)//defining setTeamLosses method that takes one integer parameter

   {

       this.teamLosses = teamLosses;//using this keyword that sets value in teamLosses

   }

   //defining the get method that returns the input value

   public String getTeamName() //defining getTeamName method

   {

       return teamName;//return teamName value

   }

   public int getTeamWins()  //defining getTeamWins method

   {

       return teamWins;//return teamWins value

   }

   public int getTeamLosses() //defining getTeamLosses method

   {

       return teamLosses;//return teamLosses value

   }

   public double getWinPercentage()//defining getWinPercentage method

   {              

       return ((teamWins * 1.0) / (teamWins + teamLosses));//using the return keyword that returns percentage value

   }      

}

Please find the complete code in the attached file and its output file in the attached file.

Class definition:

Find out more about the Class here:

brainly.com/question/17001900

Philosophy, I beleive. He tought Sikander liturature and eloquence, but his most famous teachings were of philosophy.

Aristotle taught the world science. He was considered the best scientists of his time.

Answer:

one-sided looking

Explanation:

Larry purposely looks away

Answer:

Explanation:

PROGRAM

main.c

#include <fcntl.h>

#include <stdio.h>

#include <stdlib.h>

#include <sys/stat.h>

#include <sys/types.h>

#include <unistd.h>

#include "myio.h"

int checkInt(char *arg);

int main(int argc, char *argv[]) {

  int doubles, i, ints, lines;

  char newline;

  FILE *out;

  int x, y, z;

  newline = '\n';

  if (argc != 5) {

     printf("Usage is x y z output_filename\n");

     return 0;

  }

  if (checkInt(argv[1]) != 0)

     return 0;

  ints = atoi(argv[1]);

  if (checkInt(argv[2]) != 0)

     return 0;

  doubles = atoi(argv[2]);

  if (checkInt(argv[3]) != 0)

     return 0;

  lines = atoi(argv[3]);

  out = fopen(argv[4], "a");

  if (out == NULL) {

     perror("File could not be opened");

     return 0;

  }

  for (x = 0; x < ints; x++) {

     int n = ReadInteger();

     printf("%d\n", n);

     fprintf(out, "%d\n", n);

  }

  for (y = 0; y < doubles; y++) {

     double d = ReadDouble();

     printf("%lf\n", d);

     fprintf(out, "%lf\n", d);

  }

  for (z = 0; z < lines; z++) {

     char *l = ReadLine();

     printf("%s\n", l);

     fprintf(out, "%s\n", l);

     free(l);

  }

  fclose(out);

  return 0;

}

int checkInt(char *arg) {

  int x;

  x = 0;

  while (arg[x] != '\0') {

     if (arg[x] > '9' || arg[x] < '0') {

        printf("Improper input. x, y, and z must be ints.\n");

        return -1;

     }

     x++;

  }

  return 0;

}

myio.c

#include <limits.h>

#include <stdio.h>

#include <stdlib.h>

#include <unistd.h>

char *ReadInput(int fd) {

  char buf[BUFSIZ];

  int i;

  char *input;

  int r, ret, x;

  i = 1;

  r = 0;

  ret = 1;

  input = calloc(BUFSIZ, sizeof(char));

  while (ret > 0) {

     ret = read(fd, &buf, BUFSIZ);

     for (x = 0; x < BUFSIZ; x++) {

        if (buf[x] == '\n' || buf[x] == EOF) {

           ret = -1;

           break;

        }

        input[x*i] = buf[x];

        r++;

     }

     i++;

     if (ret != -1)

        input = realloc(input, BUFSIZ*i);

  }

  if (r == 0)

     return NULL;

  input[r] = '\0';

  input = realloc(input, r+1);

  return(input);

}

int ReadInteger() {

  char *input;

  int go, num, x;

  go = 0;

  do {

     go = 0;

     printf("Input an integer\n");

     input = ReadInput(STDIN_FILENO);

     for (x = 0; x < INT_MAX; x++) {

        if (x == 0&& input[x] == '-')

           continue;

        if (input[x] == 0)

           break;

        else if (input[x]> '9' || input[x] < '0') {

           go = 1;

           printf("Improper input\n");

           break;

        }

    }

  } while (go == 1);

  num = atoi(input);

  free(input);

  return num;

}

double ReadDouble(void) {

  int dec, exp;

  char *input;

  int go;

  double num;

  int x;

  do {

     go = 0;

     dec = 0;

     exp = 0;

     printf("Input a double\n");

     input = ReadInput(STDIN_FILENO);

     for (x = 0; x < INT_MAX; x++) {

        if (x == 0&& input[x] == '-')

           continue;

        if (input[x] == 0)

           break;

        else if (input[x] == '.' && dec == 0)

           dec = 1;

        else if (x != 0&& (input[x] == 'e' || input[x] == 'E') && exp == 0) {

           dec = 1;

           exp = 1;

        }

        else if (input[x]> '9' || input[x] < '0') {

           go = 1;

           printf("Improper input\n");

           break;

        }

     }

  } while (go == 1);

  num = strtod(input, NULL);

  free(input);

  return num;

}

char *ReadLine(void) {

  printf("Input a line\n");

  return(ReadInput(STDIN_FILENO));

}

char *ReadLineFile(FILE *infile) {

  int fd;

  fd = fileno(infile);

  return(ReadInput(fd));

}

myio.h

#ifndef _myio_h

#define _myio_h

/*

* Function: ReadInteger

* Usage: i = ReadInteger();

* ------------------------

* ReadInteger reads a line of text from standard input and scans

* it as an integer. The integer value is returned. If an

* integer cannot be scanned or if more characters follow the

* number, the user is given a chance to retry.

*/

int ReadInteger(void);

/*

* Function: ReadDouble

* Usage: x = ReadDouble();

* ---------------------

* ReadDouble reads a line of text from standard input and scans

* it as a double. If the number cannot be scanned or if extra

* characters follow after the number ends, the user is given

* a chance to reenter the value.

*/

double ReadDouble(void);

/*

* Function: ReadLine

* Usage: s = ReadLine();

* ---------------------

* ReadLine reads a line of text from standard input and returns

* the line as a string. The newline character that terminates

* the input is not stored as part of the string.

*/

char *ReadLine(void);

/*

* Function: ReadLine

* Usage: s = ReadLine(infile);

* ----------------------------

* ReadLineFile reads a line of text from the input file and

* returns the line as a string. The newline character

* that terminates the input is not stored as part of the

* string. The ReadLine function returns NULL if infile

* is at the end-of-file position. Actually, above ReadLine();

* can simply be implemented as return(ReadLineFile(stdin)); */

char *ReadLineFile(FILE *infile);

Answer:

change bullets to numbers

Explanation:

100%

Answer:

Explanation:

The objective of this question is to  compute a program  that involve using  a  C++ code for  designing and implementing  a Bank Account Management Simulator with integrated file storage for saving details to secondary storage.

When writing this code I notice the words are more than 5000 maximum number of characters the text editor can contain, so i created a word document for it. The attached file  to the word document can be found below.

Answer:

The programming language is not stater; However, I'll answer this question using C++.

This program does not use comments (See explanation)

See Attachment for program file

Program starts here

#include <iostream>

using namespace std;

int main()

{

string response;

cout<<"What kind of swallow?\n";

cin>>response;

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

{

 response[i]=toupper(response[i]);

}

   if(response == "AFRICAN")

   {

    cout<<"Yes, it could grip it by the husk.";

}

else if(response == "EUROPEAN")

   {

    cout<<"A five-ounce bird could not carry a one-pound coconut.";

}

else

{

 cout<<"You really are not fit to be a king.";

}

   return 0;

}

Explanation:

string response; -> A string variable to hold user input is declared

cout<<"What kind of swallow?\n"; -> prompts user for input

cin>>response; -> user input is stored here

The following iteration converts user input to uppercase; so that the program will work for inputs like African, AFRICAN, AFriCAN, etc.

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

{

 response[i]=toupper(response[i]);

}

The following if statement prints "Yes, it could grip it by the husk." if user input is AFRICAN

   if(response == "AFRICAN")

   {

    cout<<"Yes, it could grip it by the husk.";

}

Otherwise, if user input is EUROPEAN, it prints; "A five-ounce bird could not carry a one-pound coconut."

else if(response == "EUROPEAN")

   {

    cout<<"A five-ounce bird could not carry a one-pound coconut.";

}

Lastly; for every inputs different from AFRICAN and EUROPEAN, the program displays "You really are not fit to be a king."

else

{

 cout<<"You really are not fit to be a king.";

}

Camera and Mic

These 2 things are the most likely things to get you in trouble. Unless you have a 100% protected device, and honestly even if you do, cover up the camera and close your mic. This will definitely save you later.

Answer:

Engineering and Science

Explanation:

Answer:

import java.util.Scanner;

public class ShippingCharge

{

   static double wt_2=1.50;

   static double wt_6=3;

   static double wt_10=4;

   static double wt_more=4.75;

   static double charge;

   static double weight;

public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);

   do

   {

 System.out.print("Enter the weight of the package: ");

 weight = sc.nextDouble();

 if(weight<=0)

     System.out.println("Invalid input value!");

}while(weight<=0);

 if(weight<=2)

    charge=wt_2;

else if(weight<=6 && weight>2)

    charge=wt_6;

else if(weight<=10 && weight>6)

    charge=wt_10;

else

    charge=wt_more;

System.out.println("Shipping charges for the entered weight are $"+charge);

}

}

OUTPUT

Enter the weight of the package: 0

Invalid input value!

Enter the weight of the package: 4

Shipping charges for the entered weight are $3.0

Explanation:

1. The variables to hold all the shipping charges are declared as double and initialized.

   static double wt_2=1.50;

   static double wt_6=3;

   static double wt_10=4;

   static double wt_more=4.75;

2. The variable to hold the user input is declared as double.

   static double charge;

   static double weight;

3. The variable to hold the final shipping charge is also declared as double.  

4. Inside main(), an object of Scanner class is created. This is not declared static since declared inside a static method, main().

Scanner sc = new Scanner(System.in);

5. Inside do-while loop, user input is taken until a valid value is entered.

6. Outside the loop, the final shipping charge is computed using multiple if-else statements.

7. The final shipping charge is then displayed to the user.

8. All the code is written inside class since java is a purely object-oriented language.

9. The object of the class is not created since only a single class is involved.

10. The class having the main() method is declared public.  

11. The program is saved with the same name as that of the class having the main() method.

12. The program will be saved as ShippingCharge.java.

13. All the variables are declared as static since they are declared outside main(), at the class level.

Answer: Provided in the explanation section

Explanation:

C++ Code

#include<iostream>

#include<fstream>

#include<string>

using namespace std;

// read plain text from file

void readPlaneText(char *file,char *txt,int &size){

  ifstream inp;

  inp.open(file);

  // index initialize to 0 for first character

  int index=0;

  char ch;

  if(!inp.fail()){

      // read each character from file  

      while(!inp.eof()){

          inp.get(ch);

          txt[index++]=ch;

      }

  }

  // size of message

  size=index-1;

}

// read key

int **readKey(char *file,int **key,int &size){

  ifstream ink;

  //

  ink.open(file);

      if(!ink.fail()){

          // read first line as size

      ink>>size;

      // create 2 d arry

      key=new int*[size];

      for(int i=0;i<size;i++){

          key[i]=new int[size];

      }

      // read data in 2d matrix

      for(int i=0;i<size;i++){

          for(int j=0;j<size;j++){

              ink>>key[i][j];

          }  

      }

  }

  return key;

}

// print message

void printText(string txt,char *msg,int size){

  cout<<txt<<":\n\n";

  for(int i=0;i<size;i++){

      cout<<msg[i];

  }

}

// print key

void printKey(int **key,int size){

  cout<<"\n\nKey matrix:\n\n";

  for(int i=0;i<size;i++){

      for(int j=0;j<size;j++){

          cout<<key[i][j]<<" ";

      }  

      cout<<endl;

  }

}

void encrypt(char *txt,int size,int **key,int kSize,char *ctxt){

  int *data=new int[kSize];

  for(int i=0;i<size;i=i+kSize){  

  // read key size concecutive data

      for(int a=0;a<kSize;a++){

          data[a]=txt[i+a]-'a';

      }

      // cipher operation

      for(int a=0;a<kSize;a++){

          int total=0;

          for(int b=0;b<kSize;b++){

              total+=key[a][b]*data[b];

          }  

          total=total%26;

          ctxt[i+a]=(char)('a'+total);

      }      

  }

}

int main(int argc,char **argv){

  char text[10000];

  char ctext[10000];

  int **key;

  int keySize;

  int size;

  // input

  key=readKey(argv[1],key,keySize);

  readPlaneText(argv[2],text,size);

  encrypt(text,size,key,keySize,ctext);

  // output

  printKey(key,keySize);

  cout<<endl<<endl;

  printText("Plaintext",text,size);

  cout<<endl<<endl;

  printText("Ciphertext",ctext,size);

  return 0;

}

cheers i hope this helped !!!

Explanation:

search engine because it helps to search find what you are expecting to search in the google.

Answer:

Please follow the code indentation for the python program.

Explanation:

Answer:

B - The user is no longer able to log on to the network

Explanation:

Answer:

Explanation:

(a) The recurrence relation for the given problem is :

T(n) = T(n-1) + T(n-4) + 1

(b) The O(n) time recursive algorithm with memoization for the above recurrence is given below :

Create a 1-d array 'memo' of size, n (1-based indexing) and initialize its elements with -1.

func : a recursive function that accepts the cost array and startingJobNo and returns the minimum cost for doing the jobs from startingJobNo to n.

Algorithm :

func(costArr[], startingJobNo){

if(startingJobNo>n)

then return 0

END if

if(memo[startingJobNo] != -1)

then return memo[startingJobNo];

END if

int ans1 = func(costArr, startingJobNo+1) + costArr[startingJobNo]

int ans2 = func(costArr, startingJobNo+4) + h

memo[startingJobNo] = min(ans1,ans2);

return memo[startingJobNo];

}

(c)

First, Create a 1-d array 'dp' of size, N+1.

dp[0] = 0

bottomUp(int c[]){

for  i=1 till i = n

DO

dp[i] = min(dp[i-1] + c[i], dp[max(0,i-4)] + h);

END FOR

return dp[n];

}

(d)

Modifying the algorithm given in part (b) as follows to know which job to do yourself and in which jobs we need to hire a handyman.

First, Create a 1-d array 'memo' of size, n (1-based indexing) and initialize its elements with -1.

Next, Create another 1-d array 'worker' of size,n (1-based indexing) and initialize its elements with character 'y' representing yourself.

Algorithm :

func(costArr[], startingJobNo){

if(startingJobNo>n)

then return 0

END if

if(memo[startingJobNo] != -1)

then return memo[startingJobNo];

END if

int ans1 = func(costArr, startingJobNo+1) + costArr[startingJobNo]

int ans2 = func(costArr, startingJobNo+4) + h

if(ans2 < ans1)

THEN

for (i = startingJobNo; i<startingJobNo+4 and i<=n; i++)

DO

// mark worker[i] with 'h' representing that we need to hire a mechanic for that job

worker[i] = 'h';

END for

END if

memo[startingJobNo] = min(ans1,ans2);

return memo[startingJobNo];

}

//the worker array will contain 'y' or 'h' representing whether the ith job is to be done 'yourself' or by 'hired man' respectively.