A user in the accounting department reports he or she cannot access the invoices that the sales department has placed on the shared drive. This points toward a possible problem with which component of the computer’s operating system? Networking Time-sharing Interrupts Device Driver

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:

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:

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:

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:

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;

}

}

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:

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.

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:

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

}

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