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!
The electric company gives a discount on electricity based upon usage. The normal rate is $.60 per Kilowatt Hour (KWH). If the number of KWH is above 1,000, then the rate is $.45 per KWH. Write a program (L4_Ex1.cpp) that prompts the user for the number of Kilowatt Hours used and then calculates and prints the total electric bill. Please put comment lines, same as in Lab3, at the beginning of your program. According to your program in Lab 4.1, how much will it cost for: 900 KWH? 1,754 KWH? 10,000 KWH?
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.
Which of the following should you NOT do when using CSS3 properties to create text columns for an article element? a. make the columns wide enough to read easily b. include a heading in the article element c. set rules between the columns of the article element d. justify the text
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.
Create a structure with variables for a character, a string, an integer, and a floating point number. [Note: Use Typedef way of creating the structure]. The structure string variable is a "char* stringp". In other words, the structure will have a pointer to a string. Do not initialize the structure at definition time.
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]);
}
Create a function average_temp(s) that accepts a file name s that contains temperature readings. Each line in the file contains a date followed by 24 hourly temperature readings in a comma-separated-value format, like this example:
2/3/2016,18,17,17,18,20,22,25,30,32,32,32,33,31,28,26,26,25,22,20,20,19,18,18,18
For each line in the file, the function should print out a line containing two items: the date, then comma, then the average temperature on that date, e.g.
3/5/2018, 58.24
3/6/2018, 60.11
3/7/2018, 57.55
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.
Create a public non-final class named Larger parameterized by a type T that implements Comparable. (Please use T or the test suite will fail.) You should provide a single instance method named larger that accepts an array of the parameterized type as the first argument and a single value of the parameterized type as the second argument. larger should return true if the second argument is larger than or equal to every value of the array and false otherwise. If either the array or the value are null you should throw an IllegalArgumentException. As an ungraded bonus challenge, see if you can make the compiler warning about unchecked operations go away… (Note that normally we would write this as a class method. Java does support type parameters for static methods, but we aren’t going to cover that in class. So we’ll use an instance method here instead.) Note also that this homework is not due until Friday but was accidentally released Thursday. It does rely on material we will cover Friday. Feel free to wait to complete it then.
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;
}
}
Extra Credit Programming Assignment(7points)Due May 1, midnightWrite a JavaFX application that presents 20 circles, each with a random radius and location. If a circle does not overlap any other circle, fill in the circle with black. Fill in overlapping circles with a translucent blue. Use an array to store the circle objects,and check each new circle to see if it overlaps any previous created circle. Two circles overlap is the distance between their center points is less than the sum of their radii
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);
}
}
Bob and Alice have agreed that Bob would answer Alice's invitation using ElGamal with the following parameters: ( prime p = 29, e1= 3, d = 5 and the random r = 7 ) find first the set of public and private keys. Bob replies in pairs of C1,C2 as follows: (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). Please decipher the response that Bob sent to Alice.
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:
Create a public non-final class named Partitioner. Implement a public static method int partition(int[] values) that returns the input array of ints partitioned using the last array value as the pivot. All values smaller than the pivot should precede it in the array, and all values larger than or equal to should follow it. Your function should return the position of the pivot value. If the array is null or empty you should return 0.
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
}
Use the Law of Sines to solve the triangle. Round your answers to two decimal places.
A = 99.7°, C = 20.4º, a = 27.4
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
please code this in c++
5.19 Farmer's market - files
Get the file name from the user and open it using code. The file has the product name and price/lb in each line.
Until the EOF is reached:
Read the product name from the file
prompt the user for entering the weight of that product. (Points will be taken off if you hard-code the product name inside your code).
Once the user enters the weight(0 if user does not buy that item), calculate the cost of that item by reading the price of that item from the file and multiplying by the weight entered by the user.
Maintain a running total of the cost until all the items have been entered by user.
Once EOF is reached, display the total cost of the purchase
product.txt // contains the following
apple 1.59
orange 0.99
banana 0.69
grapes 2.99
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;
}
The pseudo-class selectors for links let you use CSS to change the formatting for all but one of the following. Which one is it?
a.
a link that has the focus or is being hovered over
b.
a link that is inactive
c.
a link that hasn’t been visited
d.
a link that has been visited
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:
Networking.
Explanation:
An operating system which was developed in the 1950s, is a software which acts as an intermediary between the computer hardware and end users.
The functions of an Operating System are; Memory, Device, Process, File, Secondary-Storage and Input/Output management.
The networking component of the computer's operating system ensures that a group of processors don't share memory, clock and hardware devices, instead the processors communicate with each other through the network.
Basically, the network Operating System (OS) runs on a server and provides the capability to serve to manage groups, user, application or program, data, security and any other networking functions.
Hence, the accountant couldn't access the invoices that the sales department placed on the shared drive because of a networking component problem of the computer’s operating system.
Which of the following is a possible disadvantage of recursion? Question 10 options: Recursive solutions can be less efficient than their iterative counterparts Recursive solutions tend to be longer than their iterative counterparts Recursive solutions are more likely to go into infinite loops than their iterative counterparts Recursive solutions tend to have more local variables than their iterative counterparts
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.
Write a program that opens a text file (name the textfile "problem4.txt") and reads its contents into a queue of characters. The user must then enter a character they are looking for. The program should then dequeue each character and count the number of characters that are equal to what the user is looking for. Output the count of the character or lack thereof in a second file (name the textfile "resultsp4.txt").
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;
}