What are the programming concepts (within or outside the scope of IT210) that you would like to strengthen and delve into further after this course. Why? What is your plan to learn more about the concept(s) you identify? Identify the course concept(s) that you would like this course to provide more content about. Why? Is there any topic and / or concept for which you need more explanation? In addition to Java, what are the programming languages you are interested to learn? Why?

Answer:

The description for the given question is described in the explanation section below.

Explanation:

Since deviance constitutes a breach of a people's current standard. I believe Erickson's psychological concept that Deviance becomes a trait that is commonly considered to need social control agencies' intervention, i.e. 'Something must being done'.

Answer:

The cpp program for the given scenario is shown below.

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //variables to hold both the given values

   double normal_rate = 0.60;

   double rate_1000 = 0.45;

   //variable to to hold user input

   double kwh;

   //variable to hold computed value

   double bill;

   std::cout << "Enter the number of kilowatt hours used: ";

   cin>>kwh;

   std::cout<<std::endl<<"===== ELECTRIC BILL ====="<< std::endl;

   //bill computed and displayed to the user

   if(kwh<1000)

   {

       bill = kwh*normal_rate;

       std::cout<< "Total kilowatt hours used: "<<kwh<< std::endl;

       std::cout<< "Rate for the given usage: $"<<normal_rate<< std::endl;

       std::cout<< "Total bill: $" <<bill<< std::endl;

   }

   else  

   {

       bill = kwh*rate_1000;

       std::cout<< "Total kilowatt hours used: "<<kwh<< std::endl;

       std::cout<< "Rate for the given usage: $"<<rate_1000<< std::endl;

       std::cout<< "Total bill: $" <<bill<< std::endl;

   }

   std::cout<<std::endl<< "===== BILL FOR GIVEN VALUES ====="<< std::endl;

   //computing bill for given values of kilowatt hours

   double bill_900 = 900*normal_rate;

   std::cout << "Total bill for 900 kilowatt hours: $"<< bill_900<< std::endl;

   double bill_1754 = 1754*rate_1000;

   std::cout << "Total bill for 1754 kilowatt hours: $"<< bill_1754<< std::endl;

   double bill_10000 = 10000*rate_1000;

   std::cout << "Total bill for 10000 kilowatt hours: $"<< bill_10000<< std::endl;

   return 0;

}

OUTPUT

Enter the number of kilowatt hours used: 555

===== ELECTRIC BILL =====

Total kilowatt hours used: 555

Rate for the given usage: $0.6

Total bill: $333

===== BILL FOR GIVEN VALUES =====

Total bill for 900 kilowatt hours: $540

Total bill for 1754 kilowatt hours: $789.3

Total bill for 10000 kilowatt hours: $4500

Explanation:

1. The program takes input from the user for kilowatt hours used.

2. The bill is computed based on the user input.

3. The bill is displayed with three components, kilowatt hours used, rate and the total bill.

4. The bill for the three given values of kilowatt hours is computed and displayed.

Answer:

See explaination

Explanation:

/* reading a text file Character by Character

* Enter the character to Queue

* Search Specific character and Printingits occurrence

* otherwise Lack thereof

*/

// Include header file for File Reading

#include <iostream>

#include <fstream>

// Maximum no of character in a file

# define N 50

using namespace std;

// Queue Data Structure

typedef struct

{

char arr[N];

int front;

int rear;

int size;

}Queue;

// Function Prototypes

void initialize(Queue*);

void Enqueue(Queue*,char);

char Dequeue(Queue*);

int isEmpty(Queue*);

int isFull(Queue*);

int Qsize(Queue*);

int main () {

// Variable Declaration

char ch;

char searchchar;

// Allocating memeory for the Queue.

Queue *Q=(Queue *)malloc(sizeof(Queue));

// Reading from the file

fstream fin("problem4.txt", fstream::in);

//initialize Queue with front rear and size

initialize(Q);

//Looping through the file and Enqueue it in Queue

while (fin >> noskipws >> ch) {

Enqueue(Q,ch);

}

// close the opened file.

fin.close();

int i=0;

int n=Qsize(Q);

int count=0;

//Asking user for Input

cout << "Please character to found ";

cin >> searchchar;

while (i<n) {

if(Dequeue(Q)==searchchar)

count++; // if found increase the count

i++;

}

// Total Count od character searched

cout << "Total Count of Character " << count << endl;

// Output Total Count of Character to a file named resultsp4.txt

ofstream outfile;

outfile.open("resultsp4.txt");

cout << "Writing to the file" << endl;

if (count>0)

outfile << "The total occurence of the " << searchchar <<" is "<< count <<endl;

else

outfile << "lack thereof " << searchchar <<endl;

// close the opened file.

outfile.close();

return 0;

}

void initialize(Queue *Q)

{

Q->front=-1;

Q->rear=-1;

Q->size=0;

}

void Enqueue(Queue * Q,char ch)

{

(Q->rear)+=1;

(Q->arr[Q->rear])=ch;

(Q->size)++;

}

char Dequeue(Queue * Q)

{

char ch=Q->arr[Q->front];

(Q->front)++;

(Q->size)--;

return ch;

}

int isEmpty(Queue * Q)

{

if ((Q->size)==0)

return 1;

else

return 0;

}

int isFull(Queue * Q)

{

if ((Q->size)==N)

return 1;

else

return 0;

}

int Qsize(Queue * Q)

{

return Q->size;

}

Answer:

Networking.

Explanation:

An operating system which was developed in the 1950s, is a software which acts as an intermediary between the computer hardware and end users.

The functions of an Operating System are; Memory, Device, Process, File, Secondary-Storage and Input/Output management.

The networking component of the computer's operating system ensures that a group of processors don't share memory, clock and hardware devices, instead the processors communicate with each other through the network.

Basically, the network Operating System (OS) runs on a server and provides the capability to serve to manage groups, user, application or program, data, security and any other networking functions.

Hence, the accountant couldn't access the invoices that the sales department placed on the shared drive because of a networking component problem of the computer’s operating system.

Answer:

Kindly go to the explanation for the answer.

Explanation:

Person.java

public class Person{

private String firstName, lastName;

private int zip;

public Person(String firstName, String lastName, int zip) {

super();

this.firstName = firstName;

this.lastName = lastName;

this.zip = zip;

}

public String toString() {

String line = "\tFirst Name = " + this.firstName + "\n\tLast Name = " + this.lastName;

line += "\n\tZip Code = " + this.zip + "\n";

return line;

}

}

Test.java

import java.util.ArrayList;

import java.util.Scanner;

public class Test{

public static void main(String args[]) {

Scanner in = new Scanner(System.in);

ArrayList<Person> al = new ArrayList<Person>();

for(int i = 0; i < 25; i++) {

System.out.print("Enter person details: ");

String line = in.nextLine();

String words[] = line.split("\t");

String fname = words[0];

String lname = words[1];

int numb = Integer.parseInt(words[2]);

Person p = new Person(fname, lname, numb);

al.add(p);

}

for(int i = 0; i < 25; i++) {

System.out.println("Person " + (i + 1) + ":");

System.out.println(al.get(i));

}

}

}

Answer:

Take Corrective Action

Answer:

See Explaination

Explanation:

public class Partitioner {

public static int partition(int[] values){

if(values==null || values.length==0)return 0;

// storing the pivot value

int pivot = values[values.length-1];

//sorting the array

for(int i=0;i<values.length-1;i++){

int m_index = i;

for (int j=i+1;j<values.length;j++)

if(values[j]<values[m_index])

m_index = j;

int tmp = values[m_index];

values[m_index] = values[i];

values[i] = tmp;

}

int i = 0;

// first finding the index of pivot

// value in sorted order and recording index in i

while (i<values.length){

if(pivot==values[i]){

if(i==values.length-1)break;

int j=0;

// finding the location for inserting the

while (j<values.length){

if(pivot<=values[j]){

// inserting the values

values[i] = values[j];

values[j] = pivot;

break;

}

j++;

}

break;

}

i++;

}

return i;

}

// main method for testing can be removable

public static void main(String[] args) {

int a[] = {4,1,6,2};

System.out.println(partition(a));

}// end of main

}

Question:

When composing an email message:

A) ideas should be organized inductively when the message contains good news or routine information.

B) just be direct, since such communications are routine.

C) present the information in the order it is likely needed or will be best received.

D) avoid repeating information that is in the subject line in the opening sentence.

Answer:

The correct answer is C)

When writing emails, it helps to put ones self in the shoes of the recipient. This helps us to present our thoughts in the way that the recipient will best receive them.

In addition to the above, one must ensure that they go directly to the point, use a courteous tone, and ensure that the message is free from typographical errors whenever he or she is writing an email.

Cheers!

Answer:

def average_temp(s): f = open("s.txt","r") for line in f: myList = line.split(",") print(myList[0],end=",") t=0 for i in range(1,25,1): t += int(myList[i]) t /= 24 print(t) f.close()

def average_temp(s):

f = open("s.txt","r")

for line in f:

myList = line.split(",")

print(myList[0],end=",")

t=0

for i in range(1,25,1):

t += int(myList[i])

t /= 24

print(t)

f.close()

Explanation:

I used Python for the solution.

Answer:

Recursive solutions can be less efficient than their iterative counterparts

Explanation:

Recursion can be defined or described as a method of solving a problem where the solution depends on solutions to smaller instances of the same problem.

It entails using iteration to ensure that smaller parts of a solution are satisfied towards solving thw overall problem.

Ita major disadvantage seems to be that it seem to be less efficient than their iterative counterparts. This is as a result of concentrating on trying to solve a smaller instances.

Answer: 37.1

Explanation: The Law of sines states that there is a proportionality between a side of triangle and its correspondent angle, i.e.:

[tex]\frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}[/tex]

where:

a, b and c are sides

A, B and C are angles

In the question, there is a triangle with 27.4 as a measure of side a, angles A and C. So, it wants the side c:

[tex]\frac{a}{sinA} = \frac{c}{sinC}[/tex]

[tex]\frac{27.4}{sin(99.7)} = \frac{c}{sin(20.4)}[/tex]

c = [tex]\frac{27.4.sin(20.4)}{sin(99.7)}[/tex]

c = 37.1

The side c is 37.1

Answer:

See Explaination

Explanation:

// CircleOverlap.java

import java.util.Random;

import javafx.application.Application;

import static javafx.application.Application.launch;

import javafx.scene.Scene;

import javafx.scene.layout.Pane;

import javafx.scene.paint.Color;

import javafx.scene.paint.Paint;

import javafx.scene.shape.Circle;

import javafx.stage.Stage;

public class CircleOverlap extends Application {

atOverride //Replace the at with at symbol

public void start(Stage primaryStage) {

//creating a Random number generator object

Random random = new Random();

//setting window size

int windowWidth = 500;

int windowHeight = 500;

//initializing array of circles

Circle array[] = new Circle[20];

//looping for 20 times

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

//generating a value between 10 and 50 for radius

int radius = random.nextInt(41) + 10;

//generating a random x,y coordinates, ensuring that the circle fits

//within the window

int centerX = random.nextInt(windowWidth - 2 * radius) + radius;

int centerY = random.nextInt(windowHeight - 2 * radius) + radius;

//creating Circle object

Circle circle = new Circle();

circle.setCenterX(centerX);

circle.setCenterY(centerY);

circle.setRadius(radius);

//adding to array

array[i] = circle;

//flag to check if circle is overlapping any previous circle

boolean isOverlapping = false;

//looping through the previous circles to see if they are overlapping

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

//finding x, y and radius of current circle under check

double x2 = array[j].getCenterX();

double dx = x2 - centerX;

double y2 = array[j].getCenterY();

double dy = y2 - centerY;

double r2 = array[j].getRadius();

//finding distance between this circle and the circle under check

double distance = Math.sqrt((dx * dx) + (dy * dy));

//checking if distance<radius1+radius2

if (distance <= (radius + r2)) {

//overlapping, setting transclucent blue color

Paint c = new Color(0, 0, 1.0, 0.3);

array[i].setFill(c);

isOverlapping = true;

//also changing the color of other circle

array[j].setFill(c);

}

}

if (!isOverlapping) {

//not overlapping, setting black color

array[i].setFill(Color.BLACK);

}

}

//creating a pane and adding all circles

Pane root = new Pane();

root.getChildren().addAll(array);

Scene scene = new Scene(root, windowWidth, windowHeight);

primaryStage.setScene(scene);

primaryStage.setTitle("");

primaryStage.show();

}

public static void main(String[] args) {

launch(args);

}

}

Answer:

See explaination

Explanation:

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

//we are creataing a structure and naming it 'datatype' using typedef

typedef struct structure

{

int n;

char ch;

char *stringp;

float f;

} datatype;

void main()

{

//declaring 5 'datatype' type pointers using array

datatype *dataArray[5];

int i;

char str[500];

//dynamically allocating the structure pointers

for(i = 0; i < 5; i++)

dataArray[i] = (datatype *)malloc(sizeof(datatype));

//loop for data input

for(i = 0; i < 5; i++)

{

printf("\nEnter Data for structure %d:\n", i + 1);

printf("Enter an integer: ");

scanf("%d", &dataArray[i]->n);

printf("Enter a single character: ");

//we need fflush to clear the input stream in order to be able

// to take new values

fflush(stdin);

//notice the blankspace before %c, this makes scanf ignore the preceding '\n' character

scanf(" %c", &dataArray[i]->ch);

//we need fflush to clear the input stream in order to be able

// to take new values

fflush(stdin);

printf("Enter a string: ");

gets(str);

//dynamically allocating the size of stringp to fit the input string

//perfectly

dataArray[i]->stringp = (char *)malloc(sizeof(char) * (strlen(str) + 1));

strcpy(dataArray[i]->stringp, str);

printf("Enter a float: ");

scanf("%f", &dataArray[i]->f);

}

//output loop 1

for(i = 0; i < 5; i++)

{

printf("\n\nStructure %d", i + 1);

printf("\nStructure %d pointer: %p", i + 1, dataArray[i]);

printf("\nCharacter: %c", dataArray[i]->ch);

printf("\nInteger: %d", dataArray[i]->n);

printf("\nString: %s", dataArray[i]->stringp);

printf("\nFloating Point: %.1f", dataArray[i]->f);

}

//freeing the 5 pointers of memory

for(i = 0; i < 5; i++) free(dataArray[i]);

//output loop 2

printf("\n\nAfter free the malloc - the pointer are: ");

for(i = 0; i < 5; i++)

printf("\nStructure %d pointer: %p", i + 1, dataArray[i]);

}

Answer:

Explanation:

Let's use Python for this. We will start prompting the user for input number, until it get 0, count only if it's positive:

input_number = Input("Please input an integer")

positive_count = 0

while input_number != 0:

    if input_number > 0:

         positive_count += 1

    input_number = Input("Please input an integer")

print(positive_count)

Answer:

see explaination

Explanation:

class Larger<T extends Comparable<T>> {

public boolean larger(T[] arr, T item) {

if (arr == null || item == null)

throw new IllegalArgumentException();

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

if (item.compareTo(arr[i]) < 0) {

return false;

}

}

return true;

}

}

Following are the correct python code to this question:

Program Explanation:

Program:

n1 = float(input('Input first number: '))#input first number

n2 = float(input('Input second number: '))#input second number

n3 = float(input('Input third number: '))#input third number

n4 = float(input('Input fourth number: '))#input fourth number

average = (n1+n2+n3+n4)/4 #calculate input number average

product = n1*n2*n3*n4 # calculate input number product

print('product: {:.0f} average: {:.0f}'.format(round(product),round(average))) #print product and average using round function

print('product: {:.3f} average: {:.3f}'.format(product,average)) #print product and average value

Output:

Please find the attachment.

Learn more:

brainly.com/question/14689516

The program calculates the product and average of four floating-point numbers, and outputs them as integers (rounded) and as floating-point numbers with three digits after the decimal point, using specified string formatting expressions.

Here's the Python code to achieve that:

# Input

num1 = float(input())

num2 = float(input())

num3 = float(input())

num4 = float(input())

# Calculate product and average

product = num1 * num2 * num3 * num4

average = (num1 + num2 + num3 + num4) / 4

# Output rounded integers

print('{:.0f} {:.0f}'.format(product, average))

# Output floating-point numbers with three digits after the decimal point

print('{:.3f} {:.3f}'.format(product, average))

This code snippet ensures the desired output format by using the specified string formatting expressions.

Answer:

#include<iostream>

using namespace std;

double sumMajorDiagonal(double n,double sum);

int main()

{

int N;

double a[10][10],n,sum=0; //declare and define matrix of size 10x10

cout<<"Enter the size of the matrix: ";

cin>>N; //input the size of the matrix

if(N<2) //check condition wheather size of the matrix is greater than or equal to 2

{

cout<<"Size of matrix must be 2 or more!"<<"\n";

return 0;

}

for(int i=0;i<N;i++) //loop for Row

{

for(int j=0;j<N;j++) //loop for column

{

cout<<"Enter element: ";

cin>>n; //input the element in matrix

a[i][j] = n;

if(i==j) //check for major diagonal condition

{

sum = sumMajorDiagonal(a[i][j],sum); //call sunMajorDiagonal funtion if condition statisfied

}

}

}

for(int i=0;i<N;i++) //loop for row

{

for(int j=0;j<N;j++) //loop for column

{

cout<<a[i][j]<<" "; //print elements of matrix

}

cout<<"\n"; //next line for new row

}

cout<<"Sum of the elements of major diagonal is: "<<sum<<"\n"; //print sum of the major row calculated

}

double sumMajorDiagonal(double n,double sum)

{

return sum+n; //add the major diagonal elements and return the value

}

Explanation:

See attached image for output

Answer:

See explaination

Explanation:

#include<iostream>

#include<fstream>

using namespace std;

int main(){

double price, totalPrice = 0, weight;

string product, filename;

cout<<"Enter filename: ";

cin>>filename;

ifstream fin;

fin.open(filename.c_str());

while(fin>>product>>price){

cout<<"Enter weight for "<<product<<": ";

cin>>weight;

totalPrice+=price*weight;

}

cout<<"\nThe total cost of the purchase: $"<<totalPrice<<endl;

return 0;

}

Answer:

Store the application data in an EBS volume

Explanation:

EC2 Instance types determines the hardware of the host computer used for the instance.

Each instance type offers different compute, memory, and storage capabilities and are grouped in instance families based on these capabilities

EC2 provides each instance with a consistent and predictable amount of CPU capacity, regardless of its underlying hardware.

EC2 dedicates some resources of the host computer, such as CPU, memory, and instance storage, to a particular instance.

EC2 shares other resources of the host computer, such as the network and the disk subsystem, among instances. If each instance on a host computer tries to use as much of one of these shared resources as possible, each receives an equal share of that resource. However, when a resource is under-utilized, an instance can consume a higher share of that resource while it’s available

Answer:

540, 380,260,440, 0, 40, 500, 380, 20, 440, 60, 460, 20 and 80.

Explanation:

So, we are given the following parameters or data or information which is going to assist us in solving the question above, they are;

(1). "prime p = 29, e1= 3, d = 5 and the random r = 7"

(2). C1C2 reply; "(12, 27), (12, 19), (12, 13), (12, 22), (12, 0), (12, 2), (12, 25), (12, 19), (12, 1), (12, 22), (12, 3), (12, 23), (12, 1), (12, 4)".

So, let us delve into the solution to the question;

Step one: determine the primitive modulo 29.

These are; 2, 3, 8, 10, 11, 14, 15, 18, 19, 21, 26, 27.

Step two: Compute V = k ^c mod p.

Say k = 2.

Then;

V = 2^7 mod 29 = 128 mod 29.

V = 12.

Step three: determine the Public key.

Thus, (p,g,y) = (29,2,12)

Private key = c = 7.

Step four: decipher.

Thus for each code pair we will decided it by using the formula below;

(1). (12,27).

W = j × b^(p - 1 - c) mod p.

W= 27 × 12^(29 -1 -7) mod 29. = 540

(2). (12, 19).

19 × 12^(29 - 1 - 7) mod 29.

( 12^(29 - 1 - 7) mod 29 = 20).

= 19 × 20 = 380.

(3).(12, 13) = 13× 20 = 260.

(4). (12, 22) = 22 × 20 = 440

(5). (12, 0) = 0 × 20 = 0.

(6). (12, 2) = 2× 20= 40.

(7). (12, 25) = 25 × 20 = 500.

(8). (12, 19) = 19 × 20 = 380.

(9).(12, 1) = 1 × 20 = 20.

(10). (12, 22) = 22 × 20 = 440.

(11). (12, 3) = 3× 20 = 60.

(13). (12, 23) = 23 × 20 = 460.

(14). (12, 1) =1 × 20 = 20.

(15). (12, 4) = 4 × 20 = 80.

Answer:

Answer:

540, 380,260,440, 0, 40, 500, 380, 20, 440, 60, 460, 20 and 80.

Explanation:

So, we are given the following parameters or data or information which is going to assist us in solving the question above, they are;

(1). "prime p = 29, e1= 3, d = 5 and the random r = 7"

(2). C1C2 reply; "(12, 27), (12, 19), (12, 13), (12, 22), (12, 0), (12, 2), (12, 25), (12, 19), (12, 1), (12, 22), (12, 3), (12, 23), (12, 1), (12, 4)".

So, let us delve into the solution to the question;

Step one: determine the primitive modulo 29.

These are; 2, 3, 8, 10, 11, 14, 15, 18, 19, 21, 26, 27.

Step two: Compute V = k ^c mod p.

Say k = 2.

Then;

V = 2^7 mod 29 = 128 mod 29.

V = 12.

Step three: determine the Public key.

Thus, (p,g,y) = (29,2,12)

Private key = c = 7.

Step four: decipher.

Thus for each code pair we will decided it by using the formula below;

(1). (12,27).

W = j × b^(p - 1 - c) mod p.

W= 27 × 12^(29 -1 -7) mod 29. = 540

(2). (12, 19).

19 × 12^(29 - 1 - 7) mod 29.

( 12^(29 - 1 - 7) mod 29 = 20).

= 19 × 20 = 380.

(3).(12, 13) = 13× 20 = 260.

(4). (12, 22) = 22 × 20 = 440

(5). (12, 0) = 0 × 20 = 0.

(6). (12, 2) = 2× 20= 40.

(7). (12, 25) = 25 × 20 = 500.

(8). (12, 19) = 19 × 20 = 380.

(9).(12, 1) = 1 × 20 = 20.

(10). (12, 22) = 22 × 20 = 440.

(11). (12, 3) = 3× 20 = 60.

(13). (12, 23) = 23 × 20 = 460.

(14). (12, 1) =1 × 20 = 20.

(15). (12, 4) = 4 × 20 = 80.

Explanation:

Answer:

See explaination

Explanation:

import java.io.BufferedReader;

import java.io.FileInputStream;

import java.io.FileNotFoundException;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.ArrayList;

public class WordSearch {

public static void main(String[] args) {

testReadDictionary();

testReadWordSearch();

}

/**

* Opens and reads a dictionary file returning a list of words. Example: dog cat

* turtle elephant

*

* If there is an error reading the file, such as the file cannot be found, then

* the following message is shown: Error: Unable to read file with replaced with

* the parameter value.

*

* atparam dictionaryFilename The dictionary file to read.

* atreturn An ArrayList of words.

*/

public static ArrayList readDictionary(String dictionaryFilename) {

ArrayList<String> animals = new ArrayList<>();

FileInputStream filestream;

try {

filestream = new FileInputStream(dictionaryFilename);

BufferedReader br = new BufferedReader(new InputStreamReader(filestream));

String str = br.readLine();

while (str != null) {

animals.add(str);

str = br.readLine();

}

br.close();

} catch (FileNotFoundException e) {

System.out.println(" Unable to read file " + dictionaryFilename);

} catch (IOException e) {

e.printStackTrace();

}

return animals;

}

/**

* Opens and reads a wordSearchFileName file returning a block of characters.

* Example: jwufyhsinf agzucneqpo majeurnfyt

*

* If there is an error reading the file, such as the file cannot be found, then

* the following message is shown: Error: Unable to read file with replaced with

* the parameter value.

*

* atparam wordSearchFileName The dictionary file to read.

* atreturn A 2d-array of characters representing the block of letters.

*/

public static char[][] readWordSearch(String wordSearchFileName) {

ArrayList<String> words = new ArrayList<>();

FileInputStream filestream;

try {

filestream = new FileInputStream(wordSearchFileName);

BufferedReader br = new BufferedReader(new InputStreamReader(filestream));

String str = br.readLine();

while (str != null) {

words.add(str);

str = br.readLine();

}

br.close();

} catch (FileNotFoundException e) {

System.out.println(" Unable to read file " + wordSearchFileName);

} catch (IOException e) {

e.printStackTrace();

}

char[][] charWords = new char[words.size()][];

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

charWords[i] = words.get(i).toCharArray();

}

return charWords;

}

public static void testReadDictionary() {

// ADD TEST CASES

ArrayList dictionaryWords;

System.out.println("Positive Test Case for Dictionary Serach");

String dictionaryFilePath = "C:\\PERSONAL\\LIBRARY\\JAVA\\file\\dictionary.txt";

dictionaryWords = readDictionary(dictionaryFilePath);

System.out.println("Number of words found : "+dictionaryWords.size());

System.out.println("They are : "+dictionaryWords.toString());

System.out.println("\nNegative Test Case for Dictionary Serach");

dictionaryFilePath = "C:\\PERSONAL\\LIBRARY\\JAVA\\file\\dictionaryDummy.txt";

dictionaryWords = readDictionary(dictionaryFilePath);

}

public static void testReadWordSearch() {

// ADD TEST CASES

char[][] wordsList;

System.out.println("\n\nPositive Test Case for Word Serach");

String wordFilePath = "C:\\PERSONAL\\LIBRARY\\JAVA\\file\\wordsearch.txt";

wordsList = readWordSearch(wordFilePath);

System.out.println("Number of words found : "+wordsList.length);

System.out.println("\nNegative Test Case for Word Serach");

wordFilePath = "C:\\PERSONAL\\LIBRARY\\JAVA\\file\\wordsearchDummy.txt";

wordsList = readWordSearch(wordFilePath);

}

}

Note: replace at with the at symbol

Answer:

Either Radio or AARP Magazine

Explanation:

Given the clue that Amanda's audience is aimed towards adults aged 55+, they were born before online mail and televisions were big on the market. Many seniors enjoy reading daily news papers while others enjoy getting their news from an audio channel like the radio.

To effectively promote Act Two Retirement Community to the target market of adults aged 55+, I would recommend focusing on advertising media that resonate with this demographic and offer cost-effective ways to reach them.

AARP Magazine: This publication is specifically tailored to the 50+ age group and offers a highly targeted platform to reach the desired audience. It provides an opportunity to showcase the retirement community's features and benefits in detail.

Direct Mail: Direct mail can be an effective way to reach the target audience directly in their homes. It allows for personalized, detailed information about the retirement community, making it suitable for a demographic that appreciates tangible information.

Read more about adverts here:

https://brainly.com/question/14227079

#SPJ3

Answer:

See explaination

Explanation:

Here we will use two semaphore variables to satisfy our goal

We will initialize s1=1 and s2=1 globally and they are accessed by all 3 processes and use up and down operations in following way

Code:-

s1,s2=1

P1 P2 P3

P(s1)

P(s2)

C1

V(s2) .

P(s2). .

. C2

V(s1) .

P(s1)

. . C3

V(s2)

Explanation:-

The P(s1) stands for down operation for semaphore s1 and V(s1) stands for Up operation for semaphore s1.

The Down operation on s1=1 will make it s1=0 and our process will execute ,and down on s1=0 will block the process

The Up operation on s1=0 will unblock the process and on s1=1 will be normal execution of process

Now in the above code:

1)If C1 is executed first then it means down on s1,s2 will make it zero and up on s2 will make it 1, so in that case C3 cannot execute because P3 has down operation on s1 before C3 ,so C2 will execute by performing down on s2 and after that Up on s1 will be done by P2 and then C3 can execute

So our first condition gets satisfied

2)If C1 is not executed earlier means:-

a)If C2 is executed by performing down on S2 then s2=0,so definitely C3 will be executed because down(s2) in case of C1 will block the process P1 and after C3 execute Up operation on s2 ,C1 can execute because P1 gets unblocked .

b)If C3 is executed by performing down on s1 then s1=0 ,so definitely C2 will be executed now ,because down on s1 will block the process P1 and after that P2 will perform up on s1 ,so P1 gets unblocked

So C1 will be executed after C2 and C3 ,hence our 2nd condition satisfied.

Answer:

b. include a heading in the article element

Explanation:

If you put a heading element in an article element where you've applied the CSS column property or similar, the heading will be forced into one of the columns.