Problem 1 (10%): You are asked to design a database for an auto-shop that satisfies the following requirements: a. Each customer has a unique customer identification number, a name, address, day time telephone number, evening telephone number. The customer name includes first name, middle initial and last name. The address includes street address, city, state, and zip code. The telephone numbers include area code and the number. b. Each car type has a unique type name, the make, model, and year of make. c. The car has a unique license number, a car type, color, type of transmission, and the customer who brings the car to the auto-shop. d. The database keeps records of repairs. The repair record includes the date of the repair, the car that the repair was done, the name of the engineer who did the repair, and the cost of the repair.

Answer:

Following are the program to this question:

#include <iostream> //defining header file

using namespace std;

void reverse (string a) //defining method reverse  

{

for (int i=a.length()-1; i>=0; i--)  //defining loop that reverse string value  

 {

     cout << a[i];  //print reverse value

 }

}

int main() //defining main method                                                                                          

{

string name; // defining string variable

cout << "Enter any value: "; //print message

getline(cin, name); //input value by using getline method

reverse(name); //calling reverse method

return 0;

}

Output:

Enter any value: ABCD

DCBA

Explanation:

In the above program code, a reverse method is declared, that accepts a string variable "a" in its arguments, inside the method a for loop is declared, that uses an integer variable "i", in which it counts string value length by using length method and reverse string value, and prints its value.

Final answer:

The question involves creating an ArrayList that stores objects of type Plant and Flower, reading these objects from input, and printing their information using a defined method in Java.

Explanation:

The student's question pertains to object-oriented programming in Java and involves working with classes, inheritance, and polymorphism within an ArrayList. To solve this, you'll need to create an ArrayList that can store objects of both the Plant and Flower classes. The Plant class is the base class, while the Flower class is the derived class. You will then implement a printArrayList() method to iterate through the myGarden ArrayList and call the printInfo() method on each object.

Since both Plant and Flower objects can be added to myGarden, the printInfo() method needs to be defined in both the Plant and Flower classes, possibly overriding the method in Flower if it requires different behavior. The process of reading input and adding plants or flowers would continue until encountering a '-1', which signifies the end of input. After populating the ArrayList, the printArrayList() method will be called to display the information about each plant or flower in the garden.

Final answer:

The question involves creating an ArrayList in Java to hold Plant and Flower objects, utilizing polymorphism to store and process them. A printArrayList() method is employed to iterate through the ArrayList and call the printInfo() method on each Plant and Flower object.

Explanation:

The question revolves around object-oriented programming in Java, specifically dealing with the concept of polymorphism and ArrayLists. To accomplish the task, you will need to create an ArrayList that can hold objects of the Plant class as well as objects of the derived Flower class. This is possible because of polymorphism, where a Flower 'is-a' Plant, and thus can be treated as one.

To provide you with a starting point, here's an example of a possible printArrayList() method:

The actual main() is not provided here but would involve initializing the ArrayList, looping to read inputs, creating Plant or Flower objects based on the input, adding them to the myGarden ArrayList, and then calling the printArrayList() method to display the information for each plant or flower in the garden. Notice that the printInfo() method is called for each element; this method is expected to be defined in both the Plant and Flower classes, leveraging polymorphism for dynamic method dispatch.

Answer:

The program in cpp for the given scenario is shown below.

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <queue>

using namespace std;

int main()

{

   //object created of file stream

   ofstream out, out1;

   //file opened for writing

   out.open("words.txt");

   out1.open("word.txt");

   //queues declared for two files

   queue<string> first, second;

   //string variables declared to read from file

   string s1, s2;

   //object created of file stream

   ifstream one("words.txt");

   ifstream two("word.txt");

   //first file read into the que

   while (getline(one, s1)) {

       first.push(s1);

   }

   //second file read into the queue

   while (getline(two, s2)) {

       second.push(s2);

}

   //both files compared

   if(first!=second)

        cout<<"files are not the same"<<endl;

   else

        cout<<"files are identical"<<endl;

   //file closed

   out.close();

   return 0;

}

OUTPUT:

files are identical

Explanation:

1. The object of the file output stream are created for both the files.

2. The two files are opened using the objects created in step 1.

3. The objects of the file input stream are created for both the files.

4. The queues are declared for both the files.

5. Two string variables are declared.

6. Inside a while loop, the text from the first is read into the string variable. The value of this string variable is then inserted into the queue.

7. The loop continues till the string is read and end of file is not reached.

8. Inside another while loop, the text from the second file is read into the second string variable and this string is inserted into another queue.

9. The loop continues till the end of file is not reached.

10. Using if-else statement, both the queues are compared.

11. If any character in the first queue does not matches the corresponding character of the second queue, the message is displayed accordingly.

12. Alternatively, if the contents of both the queues match, the message is displayed accordingly.

13. In the given example program, the message is displayed as "files are identical." This is because both the files are empty and the respective queues are considered identical since both the queues are empty.

14. Since queues are used, the queue header file is included in the program.

15. An integer value 0 is returned to indicate successful execution of the program.

Answer:

The correct answer to the following question will be "Wattage ".

Explanation:

The quantity of power needed for the operation of such an electrical system seems to be a Wattage.

Answer:

The following are the table attachment to this question.

Explanation:

In the given question a table is defined, in the column section, we assign "Particulars, Surgical Equipment, Surgical Supplies Rehab Equipment, Rehab Supplies, and total " and in the row section we assign values.

Answer:

A. Java Script

Explanation:

java gives languages

Answer:

A. Java Script

Explanation:

That is the only one they teach and is easier than others!

Answer:

Check the explanation

Explanation:

Script:

#!/usr/bin/ksh

#using awk to manipulated the input file

if [ $# != 1 ]

then

echo "Usage: 08-numMajors.sh <file-name>"

exit 1

fi

if [ -f $1 ]

then

awk '

BEGIN {

FS=","

i=0

flag=0

major[0]=""

output[0]=0

total=0

}

{

if ($3)

{

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

{

# printf("1-%s-%s\n", major[j], $3);

if(major[j] == $3)

{

flag=1

output[j] = output[j] + 1;

total++

}

}

if (flag == 0)

{

major[i]=$3

output[i] = output[i] + 1;

i++;

total++

}

else

{

flag=0

}

}

}

END {

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

{

if(output[j] > 1)

printf("There are %d `%s` majors\n", output[j], major[j]);

else

printf("There is %d `%s` major\n", output[j], major[j]);

}

printf("There are a total of %d students in this class\n", total);

}

' < $1

else

echo "Error: file $1 does not exist"

fi

Output:

USER 1> sh 08-numMajors.sh sample.txt

There are 4 ` Computer Information Systems` majors

There is 1 ` Marketing` major

There are 2 ` Computer Science` majors

There are a total of 7 students in this class

USER 1> sh 08-numMajors.sh sample.txtdf

Error: file sample.txtdf does not exist

USER 1> sh 08-numMajors.sh

Usage: 08-numMajors.sh <file-name>

USER 1> sh 08-numMajors.sh file fiel2

Usage: 08-numMajors.sh <file-name>

Final answer:

The question is about writing an Assembly language procedure to find the largest number in an array. The procedure should be named FindLargest and use the PROC directive and PROTO declaration. The student must implement it, considering register preservation and set up test cases with arrays containing negative numbers.

Explanation:

Create a Procedure in Assembly Language

The student's question involves creating a procedure named FindLargest, which will find the largest value in an array of signed integers. This is a task that would be typically given in an assembly language programming course.

The PROC directive is used to define a new procedure in assembly language, and a PROTO declaration is used to declare the procedure's interface before it’s actually implemented, allowing other parts of the program to call it properly.

Here is an example template for the FindLargest procedure, which must be adapted to the specific assembly language being used, such as x86:

FindLargest PROC, arrayPtr:PTR DWORD, count:DWORD
 ; Procedure code goes here
FindLargest ENDP

And the corresponding PROTO declaration might look like this:

FindLargest PROTO, arrayPtr:PTR DWORD, count:DWORD

The test program would include calls to FindLargest with different arrays, for example:

invoke FindLargest, ADDR array1, LENGTHOF array1
invoke FindLargest, ADDR array2, LENGTHOF array2
invoke FindLargest, ADDR array3, LENGTHOF array3

Note: The procedure's task is to return the largest value in EAX, and all registers except for EAX must be preserved to maintain the state of the calling environment.

To create the FindLargest procedure, define it with the PROC directive, include operations to identify the largest value, and create a PROTO declaration. Also, write a test program to call FindLargest with different arrays. This ensures the largest value is correctly found and returned.

To create a procedure named FindLargest that receives a pointer to a signed doubleword array and the count of the array's length, follow these steps:

Define the PROC directive with the parameter list:

Initialize registers and perform the required operations to find the largest value:

Create a PROTO declaration for FindLargest:

Write a test program to call FindLargest with three different arrays:

Answer:

see explaination

Explanation:

SIG{INT} = sub {

`mv unprocessed.txt processed.txt`;

print "\n";

exit;

};

`ps aux > unprocessed.txt`;

print "Press 'q' to exit or 'ctrl-c' to process the file and exit:\n";

$inp = <>;

if ($inp == 'q')

{

exit;

}

The 100th item in an array would be at index 99 in a zero-based indexing language and at index 100 in a one-based indexing language. Indexing helps in accessing elements within data structures like arrays.

In a language with zero-based indexing, such as Python or Java, the 100th item in an array would be at index 99. This is because the first item in an array is at index 0, and the index increases by 1 for each subsequent item. Therefore, the index can be found by subtracting one from the item number (i.e., 100 - 1 = 99).

For a language with one-based indexing, like MATLAB or R, the 100th item would be at index 100 because the indexing starts from 1. Here, the index of the 100th item is the same as the item number.

Understanding the concept of array indexing is vital as it helps in accessing the elements in the data structure efficiently. The for loop can be set up with different starting indices and conditions depending on the language's indexing convention. For example, you might initialize the index i with 0 or 1, and the loop could run while i < 101 for zero-based or i <= 100 for one-based indexing.

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the solution to the above question.

Answer:

See explaination

Explanation:

// Complex.h

#ifndef COMPLEX_H

#define COMPLEX_H

#include <iostream>

using namespace std;

class Complex

{

friend ostream &operator<<(ostream &, const Complex &);

friend istream &operator>>(istream &, Complex &);

private:

// Declare variables

double real;

double imaginary;

public:

// Constructor

Complex(double = 0.0, double = 0.0);

// Addition operator

Complex operator+(const Complex&) const;

// Subtraction operator

Complex operator-(const Complex&) const;

// Multiplication operator

Complex operator*(const Complex&) const;

// Equal operator

Complex& operator=(const Complex&);

bool operator==(const Complex&) const;

// Not Equal operator

bool operator!=(const Complex&) const;

};

#endif

// Complex.cpp

// Header file section

#include "Complex.h"

#include <iostream>

using namespace std;

// Constructor

Complex::Complex( double realPart, double imaginaryPart )

: real( realPart ),

imaginary( imaginaryPart )

{

// empty body

}

// Addition (+) operator

Complex Complex::operator+( const Complex &operd2 ) const

{

return Complex( real + operd2.real,

imaginary + operd2.imaginary );

}

// Subtraction (-) operator

Complex Complex::operator-( const Complex &operd2 ) const

{

return Complex( real - operd2.real,

imaginary - operd2.imaginary );

}

// Multiplication (*) operator

Complex Complex::operator*( const Complex &operd2 ) const

{

return Complex(

( real * operd2.real ) + ( imaginary * operd2.imaginary ),

( real * operd2.imaginary ) + ( imaginary * operd2.real ) );

}

// Overloaded = operator

Complex& Complex::operator=( const Complex &right )

{

real = right.real;

imaginary = right.imaginary;

return *this; // enables concatenation

} // end function operator=

bool Complex::operator==( const Complex &right ) const

{

return ( right.real == real ) && ( right.imaginary == imaginary )

? true : false;

}

bool Complex::operator!=( const Complex &right ) const

{

return !( *this == right );

} // end function operator!=

ostream& operator<<( ostream &output, const Complex &complex )

{

output << "(" << complex.real << ", " << complex.imaginary << ")";

return output;

}

istream& operator>>( istream &input, Complex &complex )

{

input.ignore();

input >> complex.real;

input.ignore( 2 );

input.ignore();

return input;

}

// ComplexMain.cpp

// Header file section

#include <iostream>

#include "Complex.h"

using namespace std;

// main method

int main()

{

// Declare complex numbers

Complex x, y(4.3, 8.2), z(3.3, 1.1), k;

cout << "Enter a complex number in the form: (a, b): ";

// Demonstrating overloaded

// Accept a complex number

cin >> k;

// Display complex numbers

cout << "x: " << x

<< "\ny: " << y

<< "\nz: " << z

<< "\nk: " << k << '\n';

// Operator overloading

// Adding two complex numbers

x = y + z;

cout << "\nx = y + z:\n"

<< x << " = "

<< y << " + " << z << '\n';

// Subtract two complex numbers

x = y - z;

cout << "\nx = y - z:\n"

<< x << " = " << y << " - " << z << '\n';

// Multiply two complex numbers

x = y * z;

cout << "\nx = y * z:\n"

<< x << " = " << y << " * " << z << "\n\n";

if (x != k)

{

cout << x << " != " << k << '\n';

}

cout << '\n';

x = k;

if (x == k)

{

cout << x << " == " << k << '\n';

}

return 0;

}

Answer:

import java.util.Scanner;

public class nu3 {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       System.out.println("Enter number of cycles");

       int numCycles = in.nextInt();

       //Call the method

       printShampooInstructions(numCycles);

   }

   public static void printShampooInstructions(int numCycles){

                if (numCycles<1){

              System.out.println("Too few");

          }

          if (numCycles>4){

              System.out.println("Too Many");

          }

          else {

              for (int i = 1;i <= numCycles; i++) {

                  System.out.println( numCycles+ ": Lather and rinse");

                  numCycles--;

              }

              System.out.println("     Done");

          }

   }

}

Explanation:

This is solved using Java programming language

The method printShampooInstructions() is in bold in the answer section

I have also provided a complete program that request user to enter value for number of cycles, calls the method and passes that value to it

The logic here is using if .... else statements to handle the different possible values of Number of cycles.

In the Else Section a loop is used to decrementally print the number of cycles

Answer:

See Explaination

Explanation:

import java.util.Arrays;

/**

*

* atauthor xxxx //replace at with the at symbol

*/

public class Test {

public static void main(String[] args) {

java.util.Scanner input = new java.util.Scanner(System.in);

// Enter values for list1

System.out.print("Enter list1: ");

int size1 = input.nextInt();

int[] list1 = new int[size1];

for (int i = 0; i < list1.length; i++)

list1[i] = input.nextInt();

// Enter values for list2

System.out.print("Enter list2: ");

int size2 = input.nextInt();

int[] list2 = new int[size2];

for (int i = 0; i < list2.length; i++)

list2[i] = input.nextInt();

if (equal(list1, list2)) {

System.out.println("Two lists are identical");

}

else {

System.out.println("Two lists are not identical");

}

}

public static boolean equal(int[] list1, int[] list2) {

if(list1.length == list2.length)

Arrays.sort(list1);

return true;

else

Arrays.sort(list2);

return false;

// Hint: (1) first check if the two have the same size.

// (2) Sort list1 and list2 using the sort method.

// (3) Compare the corresponding elements from list1 and list2.

// return false, if not match. Return true if all matches.

}

}

Answer:

a)  [tex]P(X=5) = 15.63 \%[/tex]

b) [tex]P(X=9) = 6.88 \%[/tex]

c) [tex]P(X<3) = 67.65 \%[/tex]

Explanation:

a) What is the probability that 5 messages are received in a given minute?

The Poisson distribution is given by

[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}[/tex]

Where x is the number of messages received in some time interval t.

mean rate = λ = 4 per minute

number of messages = x = 5

[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=5) = e^{-4}\frac{4^{5}}{5!}\\\\P(X=5) = 0.1563\\\\P(X=5) = 15.63 \%[/tex]

b) What is the probability that 9 messages are received in 1.5 minutes?

mean rate = λ = 4*1.5 = 6

number of messages = x = 9

[tex]P(X=x) = e^{-\lambda}\frac{\lambda^{x}}{x!}\\\\P(X=9) = e^{-6}\frac{6^{9}}{9!}\\\\P(X=9) = 0.0688\\\\P(X=9) = 6.88 \%[/tex]

c) What is the probability that fewer than 3 messages are received in a period of 30 seconds?

mean rate = λ = 0.5*4 = 2

number of messages = x < 3

[tex]P(X<3) = P(X = 0) + P(X = 1) + P(X = 2)\\\\P(X=0) = e^{-2}\frac{2^{0}}{0!} = 0.1353\\\\ P(X=1) = e^{-2}\frac{2^{1}}{1!} = 0.2706\\\\ P(X=2) = e^{-2}\frac{2^{2}}{2!} = 0.2706\\\\ P(X<3) =0.1353+0.2706+ 0.2706\\\\P(X<3) = 0.6765\\\\P(X<3) = 67.65 \%[/tex]

A) The probability that 5 messages are received in a given minute is; 15.629%

B) The probability that 9 messages are received in 1.5 minutes is; 6.884%

C) The probability that fewer than 3 messages are received in a period of 30 seconds is; 67.668%

We are told that the number of hits follows a poisson distribution with mean of 4 per minutes.

Formula for poisson distribution is;

P(X = x) = [tex]e^{-\lambda}(\frac{\lambda^{x}}{x!})[/tex]

where;

x is number of items in counting

λ is the mean rate

A) We want to find the probability that 5 messages are received in a given minute. Thus;

λ = 4 × 1 = 4

x = 5

Thus;

P(X = 5) = [tex]e^{-4}(\frac{4^{5}}{5!})[/tex]

P(X = 5) = 0.15629

P(X = 5) = 15.629%

B) We want to find the probability that 9 messages are received in 1.5 minutes. Thus;

λ = 4 × 1.5 = 6

x = 9

Thus;

P(X = 9) = [tex]e^{-6}(\frac{6^{9}}{9!})[/tex]

P(X = 9) = 0.06884

P(X = 9) = 6.884%

C) We want to find the probability that 3 messages are received in 30 seconds. 30 seconds is 0.5 minutes. Thus;

λ = 4 × 0.5 = 2

x = 3

Thus;

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

From online Poisson distribution calculator, we have;

P(X < 3) = 0.13534 + 0.27067 + 0.27067

P(X < 3) = 0.67668

P(X < 3) = 67.668%

Read more about Poisson distribution at; https://brainly.com/question/7879375

Answer:

See explaination

Explanation:

def remove_italicized_text(x):

ans = ""

n = len(x) # length of string

i = 0

while i < n:

if x[i:(i+3)] == "<i>": # check for opening italic html tag

i += 3

cur = ""

while i < n:

if x[i:(i+4)] == "</i>": # check for closing italic html tag

break

cur += x[i]

i += 1

i += 4

ans += cur # add the text between two tags into output string

ans += " " # adding a space between two next

ans += cur # add the text again

else:

ans += x[i]

i += 1

return ans

sentence = "<i>Stony Brook</i>"

print("sentence = " + sentence)

print("Return value: " + remove_italicized_text(sentence))

print("\n")

sentence = "I <i>love</i> SBU, yes I do!"

print("sentence = " + sentence)

print("Return value: " + remove_italicized_text(sentence))

print("\n")

sentence = "Hey <i>Wolfie</i>, he's so fine, we hug him <i>all the time</i>!"

print("sentence = " + sentence)

print("Return value: " + remove_italicized_text(sentence))

print("\n")

sentence = "This text has no italics tags."

print("sentence = " + sentence)

print("Return value: " + remove_italicized_text(sentence))

print("\n")

Answer:

C. spectrum used by most mobile phone firms that does not travel well through solid objects.

Explanation:

Obstructions on the path of radio waves can result to poor network coverage. This is also called Phone reception or coverage black spot. Just as an obstruction on the path of light waves causes the emergence of shadows, so also an obstruction from waves coming from the cell tower. Common culprits are windows, doors, roofing materials, and walls. These solid objects cause an interruption of network for network providers inside a building.

The use of antennas connected from outside the building to inside can be helpful in reducing the effect of this obstruction. Problems resulting could be low signal, difficulty hearing a caller at the other end, slow internet connection, etc.

Answer:

See explaination

Explanation:

import java.io.File;

import java.io.IOException;

import java.util.Scanner;

import java.util.StringTokenizer;

public class SecretMessage {

public static void main(String[] args)throws IOException

{

File file = new File("secret.txt");

StringBuilder stringBuilder = new StringBuilder();

String str; char ch; int numberOfTokens = 1; // Changed the count to 1 as we already consider first workd as 1

if(file.exists())

{

Scanner inFile = new Scanner(file);

StringTokenizer line = new StringTokenizer(inFile.nextLine()); // Since the secret.txt file has only one line we dont need to loop through the file

ch = line.nextToken().toUpperCase().charAt(0); // Storing the first character of first word to string builder as mentioned in problem

stringBuilder = stringBuilder.append(ch);

while(line.hasMoreTokens()) { // Looping through each token of line read using Scanner.

str= line.nextToken();

numberOfTokens += 1; // Incrementing the numberOfTokens by one.

if(numberOfTokens == 5) { // Checking if it is the fifth word

ch = str.toUpperCase().charAt(0);

stringBuilder = stringBuilder.append(ch);

numberOfTokens =0;

}

}

System.out.println("----Secret Message----"+ stringBuilder);

}

}

}

Answer:

import random

def return_random_int(lower_bound, upper_bound):

   return random.randrange(lower_bound, upper_bound)

print(return_random_int(3, 8))

Explanation:

Import the random module

Create a function called return_random_int takes two parameters, lower_bound and upper_bound

Return a number between lower_bound and upper_bound using random.range()

Call the function with given inputs, and print the result

Answer:

Ans1.

double calc_a;

calc_a=Math.pow(3.0,2.0)+Math.sqrt(9);

Ans2.

double calc_b;

calc_b=((12.0/3.0)-(2.0*4.0));

Ans 3.

double calc_c;

calc_c=(Math.sqrt(16.0)*(7.0+9.0));

Ans 4.

double calc_d;

calc_d=Math.pow(7.0,2.0)/Math.sqrt(49.0);

Explanation:

The expressions are done with Java in answer above.

To solve these calculations, we can use mathematical functions on a calculator.

To solve these calculations, we can use the mathematical functions on a calculator.

Calculation A:

To add the square of 3 to the square root of 9, we can write it as follows:

calc_a = [tex]3^2 + sqrt(9)[/tex]

calc_a = 9 + 3

calc_a = 12

Calculation B:

To subtract the division of 12 by 3 from the multiplication of 2 and 4, we can write it as follows:

calc_b = (2 * 4) - (12 / 3)

calc_b = 8 - 4

calc_b = 4

Calculation C:

To multiply the square root of 16 by the sum of 7 and 9, we can write it as follows:

calc_c = sqrt(16) * (7 + 9)

calc_c = 4 * 16

calc_c = 64

Calculation D:

To divide the square of 7 by the square root of 49, we can write it as follows:

calc_d = [tex](7^2) / sqrt(49)[/tex]

calc_d = 49 / 7

calc_d = 7

Answer:

Explanation:

PiLatinWithMultipleWords.java

import java.util.Scanner;

public class PiLatinWithMultipleWords {

public static void main(String[] args) {

Scanner scan = new Scanner(System.in);

System.out.print("Enter a string to be translated: ");

String s = scan.nextLine();

String words[] = s.split("\\s+");

for(String word: words){

System.out.print(pigLatin(word)+" ");

}

System.out.println();

}

public static String pigLatin (String s){

s = s.toLowerCase();

s = s.substring(1,s.length())+ s.charAt(0) + "ay";

return s;

}

}

Output

Enter a string to be translated: the fox jumps over the lazy dog

hetay oxfay umpsjay veroay hetay azylay ogday

Answer:

hi your question lacks the necessary matrices attached to the answer is the complete question

1024 bytes

Explanation:

A) The minimum size of the cache to take advantage of blocked execution

 The minimum size of the cache is approximately 1 kilo bytes

There are 128 elements( 64 * 2 ) in the preceding simple implementation and this because there are two matrices and every matrix contains 64 elements .

note:  8 bytes is been occupied by every element therefore the minimum size of the cache to take advantage of blocked execution

= number of elements * number of bytes

= 128 * 8 = 1024 bytes ≈ 1 kilobytes

Answer:

Check the explanation

Explanation:

// Clothing.java

public class Clothing {

//Declaring instance variables

private int quantity;

private String color;

//Zero argumented constructor

public Clothing() {

this.quantity = 0;

this.color = "";

}

//Parameterized constructor

public Clothing(int quantity, String color) {

this.quantity = quantity;

this.color = color;

}

// getters and setters

public int getQuantity() {

return quantity;

}

public void setQuantity(int quantity) {

this.quantity = quantity;

}

public String getColor() {

return color;

}

public void setColor(String color) {

this.color = color;

}

public double calculatePrice()

{

return 0.0;

}

}

_____________________________

// Pants.java

public class Pants extends Clothing {

//Declaring instance variables

private int waist;

private int inseam;

//Zero argumented constructor

public Pants() {

this.waist = 0;

this.inseam = 0;

}

//Parameterized constructor

public Pants(int quantity, String color, int waist, int inseam) {

super(quantity, color);

setWaist(waist);

setInseam(inseam);

}

// getters and setters

public int getWaist() {

return waist;

}

public void setWaist(int waist) {

if (waist > 0)

this.waist = waist;

}

public int getInseam() {

return inseam;

}

public void setInseam(int inseam) {

if (inseam > 0)

this.inseam = inseam;

}

public double calculatePrice() {

double tot = 0;

if (waist > 48 || inseam > 36) {

tot = 65.50;

} else {

tot = 50.0;

}

return tot;

}

}

__________________________

// Shirt.java

public class Shirt extends Clothing {

//Declaring instance variables

private String size;

//Zero argumented constructor

public Shirt() {

this.size = "";

}

//Parameterized constructor

public Shirt(int quantity, String color, String size) {

super(quantity, color);

this.size = size;

}

// getters and setters

public String getSize() {

return size;

}

public void setSize(String size) {

this.size = size;

}

public double calculatePrice() {

double tot = 0;

if (size.equalsIgnoreCase("S")) {

tot = getQuantity() * 11.00;

} else if (size.equalsIgnoreCase("M")) {

tot = getQuantity() * 12.50;

} else if (size.equalsIgnoreCase("L")) {

tot = getQuantity() * 15.00;

} else if (size.equalsIgnoreCase("XL")) {

tot = getQuantity() * 16.50;

} else if (size.equalsIgnoreCase("XXL")) {

tot = getQuantity() * 18.50;

}

return tot;

}

}

___________________________

//Test.java

import java.util.ArrayList;

public class Test {

public static void main(String[] args) {

int totShirts=0,totPants=0;

double sprice=0,pprice=0,totWaist=0,totinseam=0,avgWaist=0,avginseam=0;

int cnt=0;

ArrayList<Clothing> clothes=new ArrayList<Clothing>();

Shirt s=new Shirt(8,"Green","XXL");

Pants p1=new Pants(6,"Brown",48,30);

Pants p2=new Pants(4,"Blue",36,34);

clothes.add(s);

clothes.add(p1);

clothes.add(p2);

for(int i=0;i<clothes.size();i++)

{

if(clothes.get(i) instanceof Shirt)

{

Shirt s1=(Shirt)clothes.get(i);

totShirts+=s1.getQuantity();

sprice+=s1.calculatePrice();

}

else if(clothes.get(i) instanceof Pants)

{

Pants pa=(Pants)clothes.get(i);

totPants+=pa.getQuantity();

pprice+=pa.calculatePrice();

totinseam+=pa.getInseam();

totWaist+=pa.getWaist();

cnt++;

}

}

System.out.println("Total number of shirts :"+totShirts);

System.out.println("Total price of Shirts :"+sprice);

System.out.println("Total number of Pants :"+totPants);

System.out.println("Total price of Pants :"+pprice);

System.out.printf("Average waist size is :%.1f\n",totWaist/cnt);

System.out.printf("Average inseam length is :%.1f\n",totinseam/cnt);

}

}

_________________________

Output:

Total number of shirts :8

Total price of Shirts :148.0

Total number of Pants :10

Total price of Pants :100.0

Average waist size is :42.0

Average inseam length is :32.0

Answer:

Check the explanation

Explanation:

IPSec-based VPN is configured in two different modes namely

IPSec Tunnel mode and IPSec Transport mode so here we are using IPsec transport mode which is used for end to end communications between a client and a server in this original IP header is remain intact except the IP protocol field is changed and the original protocol field value is saved in the IPSec trailer to be restored when the packet is decrypted on the other side due to this arrangement you need to use application based firewall because there are certain specific rules that can address the particular field (explained above-change in IP protocol field) which at the other end need to be same so at the type of decryption, original IP protocol field can be matched.

Answer:

see explaination

Explanation:

Function Real calcDiscountPrice(Real price, Real percentage) // Calculate the discount.

Declare Real discount= (price * percentage) / 100.0 // Subtract the discount from the price.

//dividing by 100.0 because of % concept

Declare Real discountPrice = price - discount // Return the discount price.

Return discount End Function

Final answer:

The calcDiscountPrice function's issue is the incorrect calculation of the discount as the percentage is not converted into its decimal form before multiplication.

Explanation:

The problem with the provided pseudocode in calcDiscountPrice function is that the calculation for the discount is incorrect because the percentage is not converted into its decimal form before the calculation. The correct method to calculate the discount should be taking the percentage in decimal (by dividing the percentage by 100) and then multiplying it by the price. Hence, the corrected line of code should be:

Declare Real discount = price * (percentage / 100)

By doing so, the function will correctly calculate the discount and subtract it from the original price to get the discounted price. If the percentage is not divided by 100, you are not calculating a percentage of the price; you are instead calculating a multiple of the price, which is not the intended behavior for a discount.

Answer:

ROOT

Explanation:

Correct me if I'm wrong but I'm pretty sure it's called a "Root Folder" meaning it is the very first folder in the list

Answer:

See explaination

Explanation:

#include<iostream>

#include<vector>

#include<string>

#include<iostream>

#include<fstream>

using namespace std;

class BibleBooks{

public:

BibleBooks(string bk_title, int chapters, double percentage){

title=bk_title;

numChapters=chapters;

amountRead=percentage;

}

string getTitle(){

return title;

}

int getChapters(){

return numChapters;

}

double getPercentage(){

return amountRead;

}

private:

string title;

int numChapters;

double amountRead;

};

int main(){

char filename[200];

cout<<"Enter filename: ";

cin.get(filename,199);

vector<BibleBooks> bibleBooks;

ifstream infile(filename);

if(infile.is_open()){

string title; int chapters; double percentage;

while(infile>>title>>chapters>>percentage){

BibleBooks bb = BibleBooks(title,chapters,percentage);

bibleBooks.push_back(bb);

}

infile.close();

}else{

cout<<"Unable to read data from file: "<<filename;

}

//Your program will then iterate through the vector,

// displaying all of the information about each object.

int index_lowest_read=0;

for(int i=0; i<bibleBooks.size();i++){

cout<<bibleBooks.at(i).getTitle()<<" "

<<bibleBooks.at(i).getChapters()<<" "

<<bibleBooks.at(i).getPercentage()<<endl;

if(bibleBooks.at(index_lowest_read).getPercentage()>

bibleBooks.at(index_lowest_read).getPercentage()){

index_lowest_read=i;

}

}

cout<<"The book with the lowest amount read is "

<<bibleBooks.at(index_lowest_read).getTitle()<<endl;

}

Answer:

(a) The block size = 16 words (b) the format for the sizes of the tag, block, and word fields are stated below:

Tag Block Offset  

11 bits 5 bits 4 bits

(c)The memory address 0DB63 will map to 22nd block and 3 byte of a block. (d)  The main advantage of the cache is, when the cache block is made larger or bigger then, there are fewer misses. this occur when the data in the block is used.

One of the disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful. here there is also a larger miss penalty.

Explanation:

Solution

(a)The Cache of 32 blocks and the memory is word addressable  

The main memory size is not stated, so let's consider it as 2^20  

The main memory size = 2^20  

The block size = 16 words  

The Number of blocks = main memory size/ block size = 2^20/16. = 2^16= 64 K words.

(b) The Main memory is 2^20, which says it is 1 M words size and it requires 20 bits  

Now,

From which 32 cache block requires 2^5, 5 bits and block size means offset requires 2^4, 4 bits

Thus, the sizes of the tag, block, and word fields are stated below:

Tag Block Offset  

11 bits 5 bits 4 bits

(c)The hexadecimal address is 0DB63 , its binary equivalent is  

0000 1101 1011 0110 0011  

The tag = ( first 11 bits) = 0000 1101 101  

block = ( next 5 bits ) = 1 0110  

offset = ( next 4 bits ) = 0011  

Therefore, the memory address 0DB63 will map to 22nd block and 3 byte of a block.

(d) The main advantage of the cache is that, when the cache block is made larger or bigger then, there are fewer misses. this occur when the data in the block is used.

One of the disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful. here there is also a larger miss penalty.

In this exercise we want to use computer and C++ knowledge to write the code correctly, so it is necessary to add the following to the informed code:

(a) The block size = 16 words

(b) Tag Block Offset  

11 bits 5 bits 4 bits

(c)The memory address 0DB63 will map to 22nd block and 3 byte of a block.

(d)  The main advantage of the cache is, when the cache block is made larger or bigger then, there are fewer misses. One of the disadvantage is that if the data is not used before the cache block is removed from the cache, then it is no longer useful.

So analyzing the first question about the code we find that:

(a)The Cache of 32 blocks and the memory is word addressable :

(b) Says it is 1 M words size and it requires 20 bits. Now, from which 32 cache block requires 2^5, 5 bits and block size means offset requires 2^4, 4 bits. Thus, the sizes of the tag, block, and word fields are stated below:

(c)The hexadecimal address is 0DB63 , its binary equivalent is  

Therefore, the memory address 0DB63 will map to 22nd block and 3 byte of a block.

(d) The main favored position or circumstance of the cache happen that, when the cache block is created best or considerable therefore, there happen hardly any misses. this take place when the information in visible form in the block exist secondhand. One of the loss exist that if the data exist not secondhand before the cache block exist detached from the cache, then it exist not any more valuable. in this place there exist in addition to a best miss punishment.

See more about computer at brainly.com/question/950632

Answer:

well, you will have your account deleted/ arrested because of cyber bullying

Explanation:

so don't!!! (hint hint XD)

Answer:

1. they falsely believes he or she cannot control what is posted on his or her social networking site or who sends messages to him or her.

2. they  feel uncomfortable telling you what someone else wrote on your their wall because the comment is inappropriate. Therefore your they fears you may be disappointed in him or her.

3. they feels powerless to deal with cyberbullying

Function are collections of named code blocks, that are executed when called or evoked.

The setKthDigit function written in Python, where comments are used to explain each line is as follows

#This defines the function

def setKthDigit(n, k, d):

   #If the number of digits in n is greater than k

   if len(str(n)) > k:

       #This splits the number into a list

       myWord = list(str(n))

       #This reverses the list

       myWord.reverse()

       #This updates the kth digit

       myWord[k] = str(d)

       #This reverses the list

       myWord.reverse()

       #This initializes string newStr

       newStr = ''

       #This following loop creates the the number to return, as string

       for i in myWord:

           newStr+=i

   #If otherwise

   else:

       #This initializes string newStr

       newStr = ''

       #This following loop creates the the number to return, as string

       for i in range(1+k-len(str(n))):

           newStr+=str(d)

       newStr+=str(n)

   #This returns the updated number, as an integer

   return int(newStr)

Read more about functions at:

brainly.com/question/19360941

Final answer:

The function setKthDigit replaces the kth digit of an integer n with a digit d, considering the rightmost digit as the 0th. The implementation involves string manipulation and careful handling of the sign of the input number.

Explanation:

To implement the setKthDigit function in Python, you need to first understand how to manipulate individual digits of a number. This involves converting the number into a string to easily access and modify specific digits. Then, you will convert the string back to an integer to return the final result. Below is a Python code snippet that demonstrates how to accomplish this task:

def setKthDigit(n, k, d):
   n_str = str(abs(n)) # Convert the absolute value of n to a string
   if k >= len(n_str): # Check if k is within the bounds of the number
       return n # Return n as is if k is out of bounds
   modified_n_str = n_str[:-k-1] + str(d) + n_str[-k:] if k != 0 else n_str[:-1] + str(d)
   result = int(modified_n_str)
   return result if n >= 0 else -result # Keep the original sign of n

This function works by first converting the number n into a string to easily manipulate its digits. If the digit to be replaced is indeed in the bounds of the number, the string is modified by replacing the kth digit with d. The modified string is then converted back to an integer, taking care to preserve the original sign of n.