Amanda indicates that she wants to do some advertising to make people in the local area aware of the development, but she admits that her advertising budget is limited, so she wants to spend her advertising dollars wisely by reaching her target audience -- adults age 55+.She asks you for your recommendations on the following advertising media. Which of the following advertising media would you recommend as being most useful in promoting Act Two Retirement Community to the target market?

Television
Radio
AARP Magazine
Direct mail

Answer:

See explaination

Explanation:

import java.util.Scanner;

public class BubbleBubbleStarter {

public static void main(String[] args) {

Scanner sc = new Scanner(System.in);

double arr[] = new double[10];

System.out.println("Enter 10 GPA values: ");

for (int i = 0; i < 10; i++)

arr[i] = sc.nextDouble();

sc.close();

System.out.println("My list before sorting is: ");

printlist(arr);

bubbleSort(arr);

System.out.println("My list after sorting is: ");

printlist(arr);

}

static void bubbleSort(double[] list) {

boolean changed = true;

do {

changed = false;

for (int j = 0; j < list.length - 1; j++) {

if (list[j] > list[j + 1]) {

double temp = list[j];

list[j] = list[j + 1];

list[j + 1] = temp;

changed = true;

}

}

} while (changed);

}

static void printlist(double list[]) {

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

System.out.println(list[j]);

}

}

}

Answer:

mysql -t < c:\mysql_files\users.sql > c:\mysql_files\users.txt

Explanation:

MySQL is a system software that is written in programming language such as c++ and c, the software was initially released on the 23rd day of the month of May, in the year 1995 by Oracle corporation. It is a popular software in companies that are commerce related or orientated since it deals with things related to web database.

So, to answer the question we are given that the file that will results of the query is users.txt and the directory is C:\mysql_files, therefore, the command that should you use to execute the query is; mysql -t < c:\mysql_files\users.sql > c:\mysql_files\users.txt

Answer:

Following are the code to this question can be described as follows:

c= input('Input triangle_char: ') #defining variable c that input character value  

length = int(input('Enter triangle_height: ')) # Enter total height of triangle

for i in range(length): # loop for column values

   for j in range(i+1): #loop to print row values

       print(c,end=' ') #print value  

   print()# for new line

Output:

please find the attachment.

Explanation:

In the above python code, two variable "c and length" variables are declared, in variable c is used to input char variable value, in the next line, length variable is defined, that accepts total height from the user.

The question asks how to modify a program to print a right triangle using a character and height defined by the user. The solution is to use a nested loop, wherein an outer loop specifies the number of rows equal to the triangle's height and the inner loop iterates over each row to print the specified character. The range of the inner loop increases with each outer loop iteration.

To modify the given program to output a right triangle using the user-specified character and of a specified height, we would implement a nested loop in the program. An outer loop would control the number of rows, which equals the triangle's height, and an inner loop would handle the printing of the user-specified character per line. The inner loop's range would increase with each iteration of the outer loop. Here's an example in Python:

The triangle_char variable is the character used to generate the triangle. The triangle_height variable determines the number of rows. The range() function in the loops determines how many times the loop executes.

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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.

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Answer:

  A. cd /home/hnewman/discreteStructures/spring/2011

Explanation:

Nothing with 2009 in the pathname will get you where you want to go. The only reasonable answer choice is the first one:

  cd /home/hnewman/discreteStructures/spring/2011

Bob can use a hash chain method with a cryptographic hash function to send a message daily, validating his aliveness without risk of replay attacks. Each day he reveals the previous hash in the chain, which can be verified by hashing to obtain the initially shared value.

In order for Bob to prove he is still alive and has not been captured, without the risk of replay attacks, a strategy called hash chain can be employed using the cryptographic hash function. Bob starts with his secret random number x and computes a hash of it, H(x). He then repeatedly applies the hash function to this result a number of times equal to the number of days he needs to send a message (in this case, 7 times since he is there for a week), resulting in y = H(H(H(H(H(H(H(x))))))).

Each day, Bob unveils the previous hash in the chain. For example, on the first day, he sends H(H(H(H(H(H(x)))))), which anyone can hash once to verify it leads to y, the value his boss already knows. On the second day, he sends H(H(H(H(H(x))))), and so on. This proves that each message can only have come from someone with knowledge of the previous day's secret -- presumably, Bob himself. Because hash functions are one-way, even if someone knows the value H(x), they cannot reverse-engineer to find x.

This method ensures that even if the Cyberian Intelligence Agency (CIA) intercepted a message, they could not fake a future message since they would not be able to produce the next hash in the sequence without knowing the current one.

Answer:

see explaination

Explanation:

import re

def emails(document):

"""

function to find email in given doc using regex

:param document:

:return: set of emails

"""

# regex for matching emails

pattern = r'[\w\.-]+at[\w\.-]+\.\w+'

# finding all emails

emails_in_doc = re.findall(pattern, document)

# returning set of emails

return set(emails_in_doc)

# Testing the above function

print(emails('random text ertatxyz.com yu\npopatxyz.com random another'))

Answer:

amount = int(input())

#Check if input is less than 1

if amount<=0:

    print("No change")

else: #If otherwise

    #Convert amount to various coins

    dollar = int(amount/100) #Convert to dollar

    amount = amount % 100 #Get remainder after conversion

    quarter = int(amount/25) #Convert to quarters

    amount = amount % 25 #Get remainder after conversion

    dime = int(amount/10) #Convert to dimes

    amount = amount % 10 #Get remainder after conversion

    nickel = int(amount/5) #Convert to nickel

    penny = amount % 5 #Get remainder after conversion

    #Print results

    if dollar >= 1:

          if dollar == 1:

                print(str(dollar)+" Dollar")

          else:

                print(str(dollar)+" Dollars")

    if quarter >= 1:

          if quarter == 1:

                print(str(quarter)+" Quarter")

          else:

                print(str(quarter)+" Quarters")

    if dime >= 1:

          if dime == 1:

                print(str(dime)+" Dime")

          else:

                print(str(dime)+" Dimes")

    if nickel >= 1:

          if nickel == 1:

                print(str(nickel)+" Nickel")

          else:

                print(str(nickel)+" Nickels")

    if penny >= 1:

          if penny == 1:

                print(str(penny)+" Penny")

          else:

                print(str(penny)+" Pennies")

Explanation:

The first answer is almost right, but this should give you the complete answer with the proper starting point and correct capitalizations.

The program is an illustration of conditional statements.

Conditional statements are statements whose execution depends on its truth value.

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

#This gets input for the amount

amount = int(input("Enter Amount:  "))

#This checks if the amount is less than 1

if amount<=0:  

    print("No Change")

else: #If otherwise

    #The next lines convert amount to various coins  

    dollar = int(amount/100) #Convert to dollar

    amount = amount % 100 #Get remainder after conversion

    quarter = int(amount/25) #Convert to quarters

    amount = amount % 25 #Get remainder after conversion

    dime = int(amount/10) #Convert to dimes

    amount = amount % 10 #Get remainder after conversion

    nickel = int(amount/5) #Convert to nickel

    penny = amount % 5 #Get remainder after conversion

    #The following if statements print the change

    #The prints dollars  

    if dollar >= 1:  

          if dollar == 1:

                print(str(dollar)+" dollar")

          else:

                print(str(dollar)+" dollars")

    #The prints quarters

    if quarter >= 1:

          if quarter == 1:

                print(str(quarter)+" quarter")

          else:

                print(str(quarter)+" quarters")

    #The prints dimes

    if dime >= 1:

          if dime == 1:

                print(str(dime)+" dime")

          else:  

                print(str(dime)+" dimes")

    #The prints nickels

    if nickel >= 1:

          if nickel == 1:

                print(str(nickel)+" nickel")

          else:

                print(str(nickel)+" nickels")

    #The prints pennies

    if penny >= 1:

          if penny == 1:

                print(str(penny)+" penny")

          else:

                print(str(penny)+" pennies")

Read more about similar programs at:

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Answer:

see explaination

Explanation:

python code

def find_max(random_list):

list_max=0

for num in random_list:

if num > list_max:

list_max=num

return list_max

print(find_max([1,45,12,11,23]))

Answer:

I think the answer is going to be paragraph

The types of styles that can be applied to a word, phrase, or sentence are the character style and paragraph style. Thus, the correct options for this question are C and D.

Character style may be defined as a type of style that can be used for the word, phrase, or sentence in design. This type of style considerably remembers formatting for single characters, words, or phrases.

While Paragraph Style will remember the formatting applied to a whole paragraph. This type of style is most frequently applied to a section of text separated from other text by line breaks with the intention to fragment the literary work into short paragraphs.

Therefore, the character and paragraph styles are the types of styles that can be applied to a word, phrase, or sentence.

To learn more about Paragraph styles, refer to the link:

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Answer: I think a deadline?

The term that describes the results expected in a given time frame is the goal. The correct option is A.

The desired states that people want to achieve, maintain, or avoid are referred to as life goals. When we make goals, we commit to imagining, planning for, and obtaining these desired outcomes. The satisfaction with our careers is the satisfaction with our lives because we spend more than half of our waking hours at work.

The most significant professional goal we can set for ourselves is to identify our areas of love and make them into lifelong careers. An objective is something you hope to accomplish. It is the desired outcome that you or a group of people plan for and firmly resolve to attain. The phrase "goal" refers to the outcomes anticipated in a specific time range.

Therefore, the correct option is A, Goals.

To learn more about goals, refer to the link:

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The question is incomplete. Your most probably complete question is given below:

Goals

Mind maps

Purposes

Ideas

Answer:

Check the explanation

Explanation:

Linear search in JAVA:-

import java.util.Scanner;

class linearsearch

{

  public static void main(String args[])

  {

     int count, number, item, arr[];

     Scanner console = new Scanner(System.in);

     System.out.println("Enter numbers:");

     number = console.nextInt();

     arr = new int[number];

     System.out.println("Enter " + number + " ");

     for (count = 0; count < number; count++)

       arr[count] = console.nextInt();

     System.out.println("Enter search value:");

     item = console.nextInt();

     for (count = 0; count < number; count++)

     {

        if (arr[count] == item)

        {

          System.out.println(item+" present at "+(count+1));

          break;

        }

     }

     if (count == number)

       System.out.println(item + " doesn't found in array.");

  }

}

Kindly check the first attached image below for the code output.

Selection Sort in JAVA:-

public class selectionsort {

   public static void selectionsort(int[] array){

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

       {

           int ind = i;

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

               if (array[j] < array[ind]){

                   ind = j;

               }

           }

           int smaller_number = array[ind];  

           array[ind] = array[i];

           array[i] = smaller_number;

       }

   }

   public static void main(String a[]){

       int[] arr = {9,94,4,2,43,18,32,12};

       System.out.println("Before Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

       System.out.println();

       selectionsort(arr);

       System.out.println("After Selection Sort");

       for(int i:arr){

           System.out.print(i+" ");

       }

   }

}  

Kindly check the second attached image below for the code output.

Bubble Sort in JAVA:-

public class bubblesort {

   static void bubblesort(int[] array) {

       int num = array.length;

       int temp = 0;

        for(int i=0; i < num; i++){

                for(int j=1; j < (num-i); j++){

                         if(array[j-1] > array[j]){

                                temp = array[j-1];

                                array[j-1] = array[j];

                                array[j] = temp;

                        }

                }

        }

   }

   public static void main(String[] args) {

               int arr1[] ={3333,60,25,32,55,620,85};

               System.out.println("Before Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

               System.out.println();

               bubblesort(arr1);

               System.out.println("After Bubble Sort");

               for(int i=0; i < arr1.length; i++){

                       System.out.print(arr1[i] + " ");

               }

       }

}  

Kindly check the third attached image below for the code output.

Binary search in JAVA:-

public class binarysearch {

  public int binarySearch(int[] array, int x) {

     return binarySearch(array, x, 0, array.length - 1);

  }

  private int binarySearch(int[ ] arr, int x,

        int lw, int hg) {

     if (lw > hg) return -1;

     int middle = (lw + hg)/2;

     if (arr[middle] == x) return middle;

     else if (arr[middle] < x)

        return binarySearch(arr, x, middle+1, hg);

     else

        return binarySearch(arr, x, lw, middle-1);

  }

  public static void main(String[] args) {

     binarysearch obj = new binarysearch();

     int[] ar =

       { 22, 18,12,14,36,59,74,98,41,23,

        34,50,45,49,31,53,74,56,57,80,

        61,68,37,12,58,79,904,56,99};

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

        System.out.print(obj.binarySearch(ar,

           ar[i]) + " ");

     System.out.println();

     System.out.print(obj.binarySearch(ar,19) +" ");

     System.out.print(obj.binarySearch(ar,25)+" ");

     System.out.print(obj.binarySearch(ar,82)+" ");

     System.out.print(obj.binarySearch(ar,19)+" ");

     System.out.println();

  }

}

Kindly check the fourth attached image below for the code output

Answer:

See explaination

Explanation:

EndOfSentenceException.java

//Create a class EndOfSentenceException that

//extends the Exception class.

class EndOfSentenceException extends Exception

{

//Construtor.

EndOfSentenceException(String str)

{

System.out.println(str);

}

}

CommaException.java

//Create a class CommaException that extends the class

//EndOfSentenceException.

class CommaException extends EndOfSentenceException

{

//Define the constructor of CommaException.

public CommaException(String str)

{

super(str);

}

}

PunctuationException.java

//Create a class PunctuationException that extends the class

//EndOfSentenceException.

class PunctuationException extends EndOfSentenceException

{

//Constructor.

public PunctuationException(String str)

{

super(str);

}

}

Driver.java

//Include the header file.

import java.util.Scanner;

//Define the class Driver to check the sentence.

public class Driver {

//Define the function to check the sentence exceptions.

public String checkSentence(String str)

throws EndOfSentenceException

{

//Check the sentence ends with full stop,

//exclamation mark

//and question mark.

if(!(str.endsWith(".")) && !(str.endsWith("!"))

&& !(str.endsWith("?")))

{

//Check the sentence is ending with comma.

if(str.endsWith(","))

{

//Throw the CommaException.

throw new CommaException("You can't "

+ "end a sentence in a comma.");

}

//Otherwise.

else

{

//Throw PunctuationException.

throw new PunctuationException("The sentence "

+ "does not end correctly.");

}

}

//If the above conditions fails then

//return this message to the main function.

return "The sentence ends correctly.";

}

//Define the main function.

public static void main(String[] args)

{

//Create an object of Scanner

Scanner object = new Scanner(System.in);

//Prompt the user to enter the sentence.

System.out.println("Enter the sentence:");

String sentence=object.nextLine();

//Begin the try block.

try {

//Call the Driver's check function.

System.out.println(new

Driver().checkSentence(sentence));

}

//The catch block to catch the exception.

catch (EndOfSentenceException e)

{}

}

}

Answer:

See explaination

Explanation:

Looking at telecommunications network, a link is a communication channel that connects two or more devices for the purpose of data transmission. The link is likely to be a dedicated physical link or a virtual circuit that uses one or more physical links or shares a physical link with other telecommunications links.

Please check attachment for further solution.

Answer:

The java program is given below.

import java.util.*;

class GradeBook  

{

//variables to hold all the given values

static int m11, m12, m13;

static int m21, m22, m23;

static String name1, name2;

//variables to hold the computed values

static double avg1, avg2;

//constructor initializing the variables with the given values

GradeBook()

{

name1 = "Jack";

m11 = 33;

m12 = 55;

m13 = 77;

name2 = "Jill";

m21 = 99;

m22 = 66;

m23 = 33;

}

//method to compute and display the student having highest average grade

static void computeAvg()

{

avg1=(m11+m12+m13)/3;

avg2=(m21+m22+m23)/3;

if(avg1>avg2)

System.out.println("Student with highest average is "+ name1);

else

System.out.println("Student with highest average is "+ name2);

}  

}

//contains only main() method

public class Teacher{  

public static void main(String[] args)

{

//object created

GradeBook ob = new GradeBook();

//object used to call the method

ob.computeAvg();

}

}

OUTPUT

Student with highest average is Jill

Explanation:

1. The class GradeBook declares all the required variables as static with appropriate data types, integer for grades and double for average grade.

2. Inside the constructor, the given values are assigned to the variables.

3. The method, computeAvg(), computes the average grade of both the students. The student having highest average grade is displayed using System.out.println() method.

4. Inside the class, Teacher, the main() method only has 2 lines of code.

5. The object of the class GradeBook is created.

6. The object is used to call the method, computeAvg(), which displays the message on the screen.

7. All the methods are also declared static except the constructor. The constructor cannot have any return type nor any access specifier.

8. The program does not takes any user input.

9. The program can be executed with different values of grades and names of the students.

it would be C

hope this helps!

Answer:

See explaination

Explanation:

#include <iostream>

using namespace std;

struct contact

{

string name;

string phoneNumber;

string email;

bool available;

};

class Phonebook

{

private :

int numberOfContacts;

int size;

contact *myContacts;

public :

Phonebook(int size)

{

this->size=size;

this->numberOfContacts=0;

myContacts=new contact[size];

}

~Phonebook()

{

delete[] myContacts;

}

int getNumberOfContacts()

{

return numberOfContacts;

}

int getSize()

{

return size;

}

void addContact(string name,string phonenumber,string email)

{

contact c;

c.name=name;

c.phoneNumber=phonenumber;

c.email=email;

c.available=true;

myContacts[numberOfContacts]=c;

numberOfContacts++;

}

void removeContact(string name)

{

for(int i=0;i<numberOfContacts;i++)

{

if(myContacts[i].name.compare(name)==0)

{

this->myContacts[i].available=false;

contact temp=myContacts[numberOfContacts-1];

myContacts[numberOfContacts-1]=myContacts[i];

myContacts[i]=temp;

numberOfContacts--;

}

}

}

Phonebook(const Phonebook& pb)

{

this->myContacts=pb.myContacts;

this->size=pb.size;

this->numberOfContacts=numberOfContacts;

}

Phonebook& operator=(const Phonebook & pb)

{

myContacts=new contact[size];

this->size=pb.size;

this->numberOfContacts=pb.numberOfContacts;

for(int i=0;i<numberOfContacts;i++)

{

this->myContacts[i]=pb.myContacts[i];

}

return *this;

}

void print()

{

for(int i=0;i<numberOfContacts;i++)

{

if(myContacts[i].available==true)

{

cout<<"Name :"<<myContacts[i].name<<endl;

cout<<"Phone Number :"<<myContacts[i].phoneNumber<<endl;

cout<<"Email :"<<myContacts[i].email<<endl;

cout<<endl;

}

}

}

};

int main(){

Phonebook pb1(10);

pb1.addContact("Williams","9845566778","williamatmail.com");

pb1.addContact("James","9856655445","jamesatmail.com");

cout<<"_____ Displaying Phonebook#1 contacts _____"<<endl;

pb1.print();

pb1.removeContact("Williams");

cout<<"_____ After removing a contact Displaying Phonebook#1 _____"<<endl;

pb1.print();

Phonebook pb2(5);

pb2.addContact("Billy","9845554444","billyatmail.com");

pb2.addContact("Jimmy","9834444444","jimmyatmail.com");

cout<<"_____ Displaying Phonebook#2 contacts _____"<<endl;

pb2.print();

pb2=pb1;

cout<<"____ After Assignment Operator Displaying Phonebook#2 contacts____"<<endl;

pb2.print();

return 0;

}

Nb: Replace the at with at symbol.

Answer:

See Explaination

See Explaination

Explanation:

d. If both players play perfectly, then too each player will play not let the other win. Let there be a state of the game in which the goal state can be achieved in just 2 more moves. If player 1 makes one move then the next move made player 1 will win or player 2 do the same.

Either way whatever the rule states, there must be a winner emerging.

Check attachment for option a to c.

The problem of solving two 8-puzzles involves having a specific goal state, making moves on a puzzle, and using an algorithm like Minimax for an adversarial setting. The reachable state space size of an 8-puzzle is 181,440. Given perfect play, an eventual win is guaranteed due to the finite number of puzzle configurations.

a. The problem of solving an 8-puzzle can be formulated as follows: The goal state is a specific configuration of tiles, with an empty space in a particular location. The actions are to move the empty space up, down, left, or right, swapping it with the adjacent tile. The transition model returns the new configuration of tiles resulting from the action. A cost function could return 1 for each move.

b. The size of reachable state space for an 8-puzzle is 9!/2 = 181,440.

c. In the adversarial setting described, a Minimax algorithm could be used to determine the best move.

d. An eventual win in this setting can be informally proven by noting that with each move, the current state of the puzzle moves closer to the goal state. Since there is a limited number of configurations (as determined in part b), the puzzle cannot be played indefinitely and eventually one player will reach the goal state.

https://brainly.com/question/33562838

#SPJ3

Answer:

Check the explanation

Explanation:

import java.io.File;

import java.io.FileNotFoundException;

import java.util.LinkedList;

import java.util.Scanner;

import static org.junit.Assert.assertEquals;

/**

* CS146 Assignment 3 Node class This class is used for undirected graphs

* represented as adjacency lists The areFriends() method checks if two people

* are friends by checking if an edge exists between the two

*

*/

public class NetworkAdjList {

   // Initialize array with max number of vertices taken from SNAP

   static int max_num_vertices = 88234;

   static LinkedList<Integer>[] adjacencyList = new LinkedList[max_num_vertices];

   public static void createAdjacencyList() {

       // Initialize array elements

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

           adjacencyList[j] = new LinkedList<Integer>();

       }

       // Get file path of the 3980.edges file

       String filePath = "C:\\Users\\shahd\\Documents\\CS146\\Assignment 3\\Question 3\\Question3\\src\\a3\\3980.edges";

       File f = new File(filePath);

       // Use Scanner to read edges from the file and put it into adjacency list

       int a;

       int b;

       try {

           Scanner fileIn = new Scanner(f);

           while (fileIn.hasNext()) {

               a = fileIn.nextInt();

               b = fileIn.nextInt();

               adjacencyList[a].add(b);

               adjacencyList[b].add(a); // We need to add the edges both ways

           }

       } catch (FileNotFoundException e) {

           e.printStackTrace();

       }

   }

   public static boolean areFriends(int A, int B) {

       // If the adjacency list contains (A, B) edge, then return true, else false

       if (adjacencyList[A].contains(B)) {

           return true;

       } else {

           return false;

       }

   }

  private static void bfsHelper(boolean visited[], int currentNode, int dis, int sourceNode) {

       dis++;

       if(!visited[currentNode]) {

           visited[currentNode] = true;

           for(int neighbor: adjacencyList[currentNode]) {

               System.out.println(neighbor + " is at a distance of " + dis + " from " + sourceNode);

               bfsHelper(visited, neighbor, dis++, sourceNode);

           }

       }

   }

   public static void BFStraversal(int start) {

       boolean visited[] = new boolean[max_num_vertices];

       bfsHelper(visited, start, 0, start);

   }

   public static void main(String[] args) {

       /**

        * These test cases assume the file 3980.edges was used

        */

       createAdjacencyList();

       System.out.println("Testing...");

       assertEquals(areFriends(4038, 4014), true);

       System.out.println("1 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(3982, 4037), true);

       System.out.println("2 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(4030, 4017), true);

       System.out.println("3 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(4030, 1), false);

       System.out.println("4 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(1, 4030), false);

       System.out.println("5 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(4003, 3980), true);

       System.out.println("6 of 7");

       System.out.println("Testing...");

       assertEquals(areFriends(3985, 4038), false);

       System.out.println("7 of 7");

       System.out.println("Success!");

   }

}

**************************************************

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:

1.6666 g/mol = 1 [tex]\frac{2}{3}[/tex] g/mol

Explanation:

Molar mass of NaOH= 23+16+1 =40g/mol

Mols in 15g = 15/40 mol

If this was dissolved in 225ml of water molarity of the solution is

[tex]\frac{15}{40}[/tex] ÷ 225 x 1000 = 1.6666 g/mol = 1 [tex]\frac{2}{3}[/tex] g/mol

Final answer:

The molarity of the sodium hydroxide solution is calculated by dividing the number of moles of NaOH (0.375 moles) by the volume of the solution in liters (0.225 L), resulting in 1.667 M.

Explanation:

The molarity of a solution is determined by the number of moles of solute per liter of solution. To find the molarity of the solution, we need to know the molar mass of sodium hydroxide (NaOH), which is approximately 40 g/mol. First, we convert the mass of NaOH to moles:

Mass of NaOH = 15.0 g

Molar mass of NaOH = 40 g/mol

Moles of NaOH = Mass / Molar mass = 15.0 g / 40 g/mol = 0.375 moles

Second, we convert the volume of the solution from milliliters to liters:

Volume of solution = 225 mL

1 liter = 1000 mL

Volume in liters = 225 mL / 1000 = 0.225 liters

Finally, we calculate the molarity of the solution using the formula:

Molarity (M) = Moles of solute / Volume of solution in liters

Molarity (M) = 0.375 moles / 0.225 liters

Molarity (M) = 1.667 M

Therefore, the molarity of the sodium hydroxide solution is 1.667 M.

Answer:

See explaination

Explanation:

import java.util.Scanner;

public class SortArray {

public static void main(String[] args) {

// TODO Auto-generated method stub

Scanner sc = new Scanner(System.in);

System.out.println("Enter Size Of Array");

int size = sc.nextInt();

int[] arr = new int[size]; // creating array of size

read(arr); // calling read method

sort(arr); // calling sort method

print(arr); // calling print method

}

// method for read array

private static void read(int[] arr) {

Scanner sc = new Scanner(System.in);

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

System.out.println("Enter " + i + "th Position Element");

// read one by one element from console and store in array

arr[i] = sc.nextInt();

}

}

// method for sort array

private static void sort(int[] arr) {

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

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

if (arr[i] < arr[j]) {

// Comparing one element with other if first element is greater than second then

// swap then each other place

int temp = arr[j];

arr[j] = arr[i];

arr[i] = temp;

}

}

}

}

// method for display array

private static void print(int[] arr) {

System.out.print("Your Array are: ");

// display element one by one

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

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

}

}

}

See attachment

Answer:

See explaination

Explanation:

#include <iostream>

#include <string>

using namespace std;

int main()

{

string raptor_prompt_variable_zzyz;

?? standing;

?? dark;

raptor_prompt_variable_zzyz ="Is it dark?";

cout << raptor_prompt_variable_zzyz << endl;

cin >> DARK;

if (DARK=="Yes")

{

raptor_prompt_variable_zzyz ="Is someone standing on the doormat outside the front door?";

cout << raptor_prompt_variable_zzyz << endl;

cin >> STANDING;

if (STANDING=="Yes")

{

cout << "LAMP IS ON" << endl; }

else

{

cout << "LAMP IS OFF" << endl; }

}

else

{

cout << "LAMP IS OFF" << endl; }

return 0;

}

Here's a simple program in pseudo-code to demonstrate the behavior of the lamp based on the inputs from the light sensor and pressure sensor:

Ask user: "Is it dark?" (yes or no)

Read user's response and store it in variable 'is_dark'

Ask user: "Is someone standing on the doormat outside the front door?" (yes or no)

Read user's response and store it in variable 'is_someone_on_doormat'

if is_dark equals yes and is_someone_on_doormat equals yes:

   Turn on the lamp

   Display message: "Lamp turned on because it is dark and someone is standing on the doormat."

else if is_dark equals yes and is_someone_on_doormat equals no:

   Keep the lamp off

   Display message: "Lamp remains off because it is dark but no one is standing on the doormat."

else if is_dark equals no and is_someone_on_doormat equals yes:

   Turn on the lamp

   Display message: "Lamp turned on because someone is standing on the doormat."

else:

   Keep the lamp off

   Display message: "Lamp remains off because it is not dark and no one is standing on the doormat."

This program covers all possible scenarios based on the inputs from the light sensor and pressure sensor and determines whether the lamp should be turned on or off accordingly.

Answer:

Transport layer

establishes, maintains and terminates network connections

Network layer

ensures the reliability of link

Data link layer

provides functions to guarantee reliable network link

Physical layer

controls transmission of the raw bit stream over the medium

Explanation:

see Answer

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:

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:

See explaination

Explanation:

package miscellaneous;

import java.text.NumberFormat;

import java.util.Currency;

import java.util.Locale;

public class sales {

public static void main(String[] args) {

double[][] sales= {

{1540.0,2010.0,2450.0,1845.0},//region1

{1130.0,1168.0,1847.0,1491.0},//region2

{1580.0,2305.0,2710.0,1284.0},//region3

{1105.0,4102.0,2391.0,1576.0}};//region4

//object for NumberFormat class

//needed in $

NumberFormat defaultFormat = NumberFormat.getCurrencyInstance(java.util.Locale.US);

System.out.println("The sales report application: ");

//the j(th) element in i(th) row in sales matrix contains sales value for

//sales in j(th) quarter

System.out.println("Sales by quarter: ");

System.out.println("Region\tQ1\t\tQ2\t\tQ3\t\tQ4");

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

System.out.print(i+1+"\t");

for(int j=0;j<sales[i].length;j++) {

System.out.print(defaultFormat.format(sales[i][j])+"\t");

}

System.out.println();

}

//i(th) row in the matrix has sales for i(th) region

System.out.println("Sales by region: \n");

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

System.out.print("Region "+(i+1)+":");

double region_sale=0;

for(int j=0;j<sales[i].length;j++) {

region_sale+=sales[i][j];

}

System.out.println(defaultFormat.format(region_sale));

}

System.out.println("\nSales by Quarter: \n");

//we have quarters so for adding up their sales we need 4 variables

double q1=0;

double q2=0;

double q3=0;

double q4=0;

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

for(int j=0;j<sales[i].length;j++) {

//j=0 implies the sales data for first quarter

if(j==0) {

q1+=sales[i][j];

}

//j=1 implies the sales data for second quarter

if(j==1) {

q2+=sales[i][j];

}

//j=2 implies the sales data for third quarter

if(j==2) {

q3+=sales[i][j];

}

//j=3 implies the sales data for fourth quarter

if(j==3) {

q4+=sales[i][j];

}

}

}

System.out.println("Q1: "+defaultFormat.format(q1));

System.out.println("Q2: "+defaultFormat.format(q2));

System.out.println("Q3: "+defaultFormat.format(q3));

System.out.println("Q4: "+defaultFormat.format(q4));

//with the help of 2 loops every sales data

//in the matrix can be accessed, which can be added

//to total_sales variable

double total_sales=0;

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

for(int j=0;j<sales[i].length;j++) {

total_sales+=sales[i][j];

}

}

System.out.println("\nTotal Sales: "+defaultFormat.format(total_sales));

}

}

Answer: provided in explanation segment

Explanation:

the code to carry out this program is thus;

import java.util.ArrayList;

import java.util.Scanner;

public class ArrayListMode {

public static int getMode(ArrayList<Integer>arr) {

      int maxValue = 0, maxCount = 0;

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

          int count = 0;

          for (int j = 0; j < arr.size(); ++j) {

              if (arr.get(j) == arr.get(i))

                  ++count;

          }

          if (count > maxCount) {

              maxCount = count;

              maxValue = arr.get(i);

          }

      }

      return maxValue;

  }

public static void main(String args[]){

  Scanner sc = new Scanner(System.in);

  ArrayList<Integer>list= new ArrayList<Integer>();

  int n;

  System.out.println("Enter sequence of numbers (negative number to quit): ");

  while(true) {

      n=sc.nextInt();

      if(n<=0)

          break;

      list.add(n);

  }

  System.out.println("Mode : "+getMode(list));

}

}

⇒ the code would produce Mode:6

cheers i hope this helps!!!!

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

The program will manage students' test scores using a struct that records each student's name, score, and grade. It involves functions to input data, calculate grades, and identify the highest scorer. Output is formatted as 'LastName, FirstName: Grade'.

Explanation:

Program Structure for Student Grade Records

To create a program for managing student test scores and grades, we would define a struct named studentType with components studentFName, studentLName, testScore, and grade. We'd use an array of studentType of size 20 to store each student's details. The program would include functions to read student data, assign grades, find the highest test score, and print students with the highest score. Grades would be assigned based on the test scores in a typical A-F scale.

The main function should be clean and comprise primarily of function calls to handle various operations such as reading data and processing grades.

The output format for each student's name and grade should be "LastName, FirstName: Grade" with the last name and first name left justified, following the requirements specified for the assignment.