Write two scanf statements to get input values into birthMonth and birthYear. Then write a statement to output the month, a slash, and the year. End with newline. The program will be tested with inputs 1 2000 and then with inputs 5 1950. Ex: If the input is 1 2000, the output is: 1/2000 Note: The input values come from user input, so be sure to use scanf statements, as in scanf("%d", &birthMonth), to get those input values (and don't assign values directly, as in birthMonth = 1).

Answer:

The program is written in c++ , go to the explanation part for it, the output can be found in the attached files.

Explanation:

C++ Code:

#include <iostream>

using namespace std;

int main() {

int x;

cout<<"Enter a number: ";

cin>>x;

int largest = x;

int position = 1, count = 0;

while(x != 1)

{

count++;

cout<<x<<" ";

if(x > largest)

{

largest = x;

position = count;

}

if(x%2 == 0)

x = x/2;

else

x = 3*x + 1;

}

cout<<x<<endl;

cout<<"The largest number of the sequence is "<<largest<<endl;

cout<<"The position of the largest number is "<<position<<endl;

return 0;

}

The program prompts the user for an integer ( x ). It calculates ( k ), the number of iterations until ( x ) becomes 1 in a sequence generated by specific rules: if ( x ) is even, it's divided by 2; if odd, it's multiplied by 3 and incremented by 1. .

Here's a Python program that implements the described sequence and finds the integer ( k ) for the given ( x ) value:

```python

def find_k(x):

   k = 0

   while x != 1:

       if x % 2 == 0:

           x = x // 2

       else:

           x = 3 * x + 1

       k += 1

   return k

def generate_sequence(x):

   sequence = [x]

   while x != 1:

       if x % 2 == 0:

           x = x // 2

       else:

           x = 3 * x + 1

       sequence.append(x)

   return sequence

def main():

   test_values = [75, 111, 678, 732, 873, 2048, 65535]

   for x in test_values:

       k = find_k(x)

       sequence = generate_sequence(x)

       print(f"For x = {x}, k = {k}, sequence: {sequence}")

if __name__ == "__main__":

   main()

```

This program calculates the integer \( k \) for each given value of \( x \) and generates the sequence according to the rules provided. Then it prints the \( k \) value and the sequence for each test value. You can run this program to verify the results for the provided test values.

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:

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'.

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:

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:

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

Explanation:

I would like to reinforce in advanced or complex concepts such as documents as well as channels, internet programming, multi-threading, after that last lesson.

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.

https://brainly.com/question/35123300

#SPJ11

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:

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:

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:

Take Corrective Action

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:

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:

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:

/* 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:

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: 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:

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.

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:

Check the explanation

Explanation:

After a particular step the registers X, Y and Z values are as it is in the first attached image below.

Now calculate the key stream bit, s using the following formula:

key stream bit , s= x0 XOR y0 XOR z0

                       s= 1 XOR 1 XOR 1

           Hence, the 1st key bit stream ,s= 1

Now, for the next step we have to re calculate the contents of registers X, Y and Z as it is in the second attached image below.

For register X:

t= x5 XOR x2 XOR x1 XOR x0

= 0 XOR 1 XOR 0 XOR 1

t=0

For register Y:

t= y1 XOR y0

=1 XOR 1

t=0

For register Z:

t= z15 XOR z2 XOR z1 XOR z0

=1 XOR 1 XOR 1 XOR 1

t=0

Now, the contents of X, Y and X are as it is in the third attached image below.

Key stream bit, s= x0 XOR y0 XOR z0

                             S= 0 XOR 1 XOR 1

                             Hence the 2nd key stream bit, s= 0

The A5/1 algorithm generates keystream bits by shifting three LFSRs based on a majority bit mechanism and producing output bits through XOR operations. Following this process, the registers X, Y, and Z are updated, and the keystream is generated. The new register states and keystream give us the final output.

The A5/1 algorithm uses three Linear Feedback Shift Registers (LFSRs) named X, Y, and Z. Here are the initial states of the registers:

To generate the next 32 keystream bits and update the registers, the following steps are followed:

After generating 32 keystream bits, the contents of the registers and the keystream are:

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:

See Explaination

Explanation:

a) Assume there are n nits each in x and y.SO we divide them into n/2 and n/2 bits.

x = xL * 2^{n/2} + xR

y = yL * 2^{n/2} + yR

x.y = 2^{n}.(xL.yL) + 2^{n/2}.(xL.yR + xR.yL) +(xR.yR)

If you see there are 4 multiplications of size n/2 and we took all other additions and subtractions as O(n).

So T(n) = 4*T(n/2) + O(n)

Now lets find run time using master theorem.

T(n) = a* T(n/b) + O(n^{d})

a = 4

b = 2

d = 1

if a > b^{d}

T(n) = O(n^v) where v is log a base b

In our case T(n) = O(n^v) v = 2

=> T(n) = O(n^{2})

The splittinng method is not benefecial if we solve by this way as the run time is same even if go by the naive approach

b)

x.y = 2^{n}.(xL.yL) + 2^{n/2}.((xL+xR).(yL+yR)-(xL.yL) - (xR.yR)) +(xR.yR)

Here we are doing only three multipliactions as we changed the term.

So T(n) = 3*T(n/2) + O(n)

a = 3

b = 2

d = 1

if a > b^{d}

T(n) = O(n^v) where v is log a base b

In our case T(n) = O(n^v) v = log 3 base 2

v = 1.584

So T(n) = O(n^{1.584})

As we can see this is better than O(n^{2}).Therefore this algorithm is better than the naive approach.

Answer:

See explaination

Explanation:

Taking a look at the The Logic function, which states that an output action will become TRUE if either one “OR” more events are TRUE, but the order at which they occur is unimportant as it does not affect the final result. For example, A + B = B + A.

Alternatively the Most significant bit which is also known as the alt bit, high bit, meta bit, or senior bit, the most significant bit is the highest bit in binary.

See the attached file for those detailed logic functions designed with relation to the questions asked.

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:

#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

Final answer:

To draw a circle using a Turtle object in Python, you can define a function called drawCircle that takes in the Turtle object, the center point coordinates, and the radius as arguments. This code uses the turtle module in Python to draw a circle.

Explanation:

Draw a Circle using a Turtle in Python

To draw a circle using a Turtle object in Python, you can define a function called drawCircle that takes in the Turtle object, the center point coordinates, and the radius as arguments. Here's an example of how the function can be implemented:

This code uses the turtle module in Python to draw a circle. It sets the pen size to 5 pixels and the color to yellow. The circle is drawn by turning 3 degrees and moving a distance of 2.0 * math.pi * radius / 120.0. After drawing the circumference, the function fills the circle with a blue color and hides the turtle.

Final answer:

The drawCircle function is designed for the Python Turtle library to draw a colored circle based on specified parameters. The Turtle's pen is set to yellow and the pen width to 5 pixels before the circle is drawn and filled with blue. After drawing, the Turtle is hidden.

Explanation:

The function drawCircle is intended to work with the Python Turtle graphics library to draw a circle on the screen. The function will expect a Turtle object, the coordinates of the circle's center point, and the circle's radius as arguments. Below is a corrected version of the function that follows the stated requirements:

Make sure to create a Turtle object and pass it to the drawCircle function along with the center point's coordinates and the radius of the circle you wish to draw.

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:

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:

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.