Answer:
250 wpm for almost an hour straight. This was recorded by Barbara Blackburn in 1946.
In this lab, you use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.
// HouseSign.cpp - This program calculates prices for custom made signs.
#include
#include
using namespace std;
int main()
{
// This is the work done in the housekeeping() function
// Declare and initialize variables here
// Charge for this sign
// Color of characters in sign
// Number of characters in sign
// Type of wood
// This is the work done in the detailLoop() function
// Write assignment and if statements here
// This is the work done in the endOfJob() function
// Output charge for this sign
cout << "The charge for this sign is $" << charge << endl;
return(0);
}
Here is the complete question.
In this lab, you use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.
start input testScore,
classRank if testScore >= 90 then if classRank >= 25 then output "Accept"
else output "Reject" endif else if testScore >= 80
then if classRank >= 50 then output "Accept" else output "Reject" endif
else if testScore >= 70
then if classRank >= 75 then output "Accept"
else output "Reject"
endif else output "Reject"
endif
endif
endif
stop
Study the pseudocode in picture above. Write the interactive input statements to retrieve: A student’s test score (testScore) A student's class rank (classRank) The rest of the program is written for you. Execute the program by clicking "Run Code." Enter 87 for the test score and 60 for the class rank. Execute the program by entering 60 for the test score and 87 for the class rank.
[comment]: <> (3. Write the statements to convert the string representation of a student’s test score and class rank to the integer data type (testScore and classRank, respectively).)
Function: This program determines if a student will be admitted or rejected. Input: Interactive Output: Accept or Reject
*/ #include using namespace std; int main() { // Declare variables
// Prompt for and get user input
// Test using admission requirements and print Accept or Reject
if(testScore >= 90)
{ if(classRank >= 25)
{ cout << "Accept" << endl; }
else
cout << "Reject" << endl; }
else { if(testScore >= 80)
{ if(classRank >= 50)
cout << "Accept" << endl;
else cout << "Reject" << endl; }
else { if(testScore >= 70)
{ if(classRank >=75) cout << "Accept" << endl;
else cout << "Reject" << endl; }
else cout << "Reject" << endl; } } } //End of main() function
Answer:
Explanation:
The objective here is to use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.
PROGRAM:
#include<iostream>
using namespace std;
int main(){
// Declare variables
int testScore, classRank;
// Prompt for and get user input
cout<<"Enter test score: ";
cin>>testScore;
cout<<"Enter class rank: ";
cin>>classRank;
// Test using admission requirements and print Accept or Reject
if(testScore >= 90)
{ if(classRank >= 25)
{ cout << "Accept" << endl; }
else
cout << "Reject" << endl;
}
else { if(testScore >= 80)
{ if(classRank >= 50)
cout << "Accept" << endl;
else cout << "Reject" << endl; }
else { if(testScore >= 70)
{ if(classRank >=75) cout << "Accept" << endl;
else cout << "Reject" << endl;
}
else cout << "Reject" << endl; }
}
return 0;
} //End of main() function
OUTPUT:
See the attached file below:
For dinner, a restaurant allows you to choose either Menu Option A: five appetizers and three main dishes or Menu Option B: three appetizers and four main dishes. There are six kinds of appetizer on the menu and five kinds of main dish.
How many ways are there to select your menu, if...
a. You may not select the same kind of appetizer or main dish more than once.
b. You may select the same kind of appetizer and/or main dish more than once.
c. You may select the same kind of appetizer or main dish more than once, but not for all your choices, For example in Menu Option A, it would be OK to select four portions of 'oysters' and one portion of 'pot stickers', but not to select all five portions of 'oysters'.)
In each case show which formula or method you used to derive the result.
Answer:
The formula used in this question is called the probability of combinations or combination formula.
Explanation:
Solution
Given that:
Formula applied is stated as follows:
nCr = no of ways to choose r objects from n objects
= n!/(r!*(n-r)!)
The Data given :
Menu A : 5 appetizers and 3 main dishes
Menu B : 3 appetizers and 4 main dishes
Total appetizers - 6
Total main dishes - 5
Now,
Part A :
Total ways = No of ways to select menu A + no of ways to select menu B
= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)
= 6C5*5C3 + 6C3*5C4
= 6*10 + 20*5
= 160
Part B :
Since, we can select the same number of appetizers/main dish again so the number of ways to select appetizers/main dishes will be = (total appetizers/main dishes)^(no of appetizers/main dishes to be selected)
Total ways = No of ways to select menu A + no of ways to select menu B
= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)
= (6^5)*(5^3) + (6^3)*(5^4)
= 7776*125 + 216*625
= 1107000
Part C :
No of ways to select same appetizers and main dish for all the options
= No of ways to select menu A + no of ways to select menu B
= (no of ways to select appetizers in A)*(no of ways to select main dish in A) + (no of ways to select appetizers in B)*(no of ways to select main dish in B)
=(6*5) + (6*5)
= 60
Total ways = Part B - (same appetizers and main dish selected)
= 1107000 - 60
= 1106940
what should i name my slideshow if its about intellectual property, fair use, and copyright
Answer:
You should name it honest and sincere intention
Explanation:
That is a good name for what you are going to talk about in your slideshow.
3.34 LAB: Mad Lib - loops in C++
Mad Libs are activities that have a person provide various words, which are then used to complete a short story in unexpected (and hopefully funny) ways.
Write a program that takes a string and integer as input, and outputs a sentence using those items as below. The program repeats until the input is quit 0.
Ex: If the input is:
apples 5
shoes 2
quit 0
the output is:
Eating 5 apples a day keeps the doctor away.
Eating 2 shoes a day keeps the doctor away.
Make sure your answer is in C++.
Answer:
A Program was written to carry out some set activities. below is the code program in C++ in the explanation section
Explanation:
Solution
CODE
#include <iostream>
using namespace std;
int main() {
string name; // variables
int number;
cin >> name >> number; // taking user input
while(number != 0)
{
// printing output
cout << "Eating " << number << " " << name << " a day keeps the doctor away." << endl;
// taking user input again
cin >> name >> number;
}
}
Note: Kindly find an attached copy of the compiled program output to this question.
In this exercise we have to use the knowledge in computational language in C++ to describe a code that best suits, so we have:
The code can be found in the attached image.
What is looping in programming?In a very summarized way, we can describe the loop or looping in software as an instruction that keeps repeating itself until a certain condition is met.
To make it simpler we can write this code as:
#include <iostream>
using namespace std;
int main() {
string name; // variables
int number;
cin >> name >> number; // taking user input
while(number != 0)
{
// printing output
cout << "Eating " << number << " " << name << " a day keeps the doctor away." << endl;
// taking user input again
cin >> name >> number;
}
}
See more about C++ at brainly.com/question/19705654
What is blue L.E.D. light
5.14 ◆ Write a version of the inner product procedure described in Problem 5.13 that uses 6 × 1 loop unrolling. For x86-64, our measurements of the unrolled version give a CPE of 1.07 for integer data but still 3.01 for both floating-point data. A. Explain why any (scalar) version of an inner product procedure running on an Intel Core i7 Haswell processor cannot achieve a CPE less than 1.00. B. Explain why the performance for floating-point data did not improve with loop unrolling.
Answer:
(a) the number of times the value is performs is up to four cycles. and as such the integer i is executed up to 5 times. (b)The point version of the floating point can have CPE of 3.00, even when the multiplication operation required is either 4 or 5 clock.
Explanation:
Solution
The two floating point versions can have CPEs of 3.00, even though the multiplication operation demands either 4 or 5 clock cycles by the latency suggests the total number of clock cycles needed to work the actual operation, while issues time to specify the minimum number of cycles between operations.
Now,
sum = sum + udata[i] * vdata[i]
in this case, the value of i performs from 0 to 3.
Thus,
The value of sum is denoted as,
sum = ((((sum + udata[0] * vdata[0])+(udata[1] * vdata[1]))+( udata[2] * vdata[2]))+(udata[3] * vdata[3]))
Thus,
(A)The number of times the value is executed is up to 4 cycle. And the integer i performed up to 5 times.
Thus,
(B) The floating point version can have CPE of 3.00, even though the multiplication operation required either 4 or 5 clock.
g Create your own data file consisting of integer, double or String values. Create your own unique Java application to read all data from the file echoing the data to standard output. After all data has been read, display how many data were read. For example, if 10 integers were read, the application should display all 10 integers and at the end of the output, print "10 data values were read" Demonstrate your code compiles and runs without issue. Respond to other student postings by enhancing their code to write the summary output to a file instead of standard output.
Answer:
See explaination
Explanation:
//ReadFile.java
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
public class ReadFile
{
public static void main(String[] args)
{
int elementsCount=0;
//Create a File class object
File file=null;
Scanner fileScanner=null;
String fileName="sample.txt";
try
{
//Create an instance of File class
file=new File(fileName);
//create a scanner class object
fileScanner=new Scanner(file);
//read file elements until end of file
while (fileScanner.hasNext())
{
double value=fileScanner.nextInt();
elementsCount++;
}
//print smallest value
System.out.println(elementsCount+" data values are read");
}
//Catch the exception if file not found
catch (FileNotFoundException e)
{
System.out.println("File Not Found");
}
}
}
Sample output:
sample.txt
10
20
30
40
50
output:
5 data values are read
D-H public key exchange Please calculate the key for both Alice and Bob.
Alice Public area Bob
Alice and Bob publicly agree to make
N = 50, P = 41
Alice chooses her Bob
picks his
Private # A = 19 private #
B= ?
------------------------------------------------------------------------------------------------------------------------------------------
I am on Alice site, I choose my private # A = 19.
You are on Bob site, you pick up the private B, B and N should have no common factor, except 1.
(Suggest to choose B as a prime #) Please calculate all the steps, and find the key made by Alice and Bob.
The ppt file of D-H cryptography is uploaded. You can follow the steps listed in the ppt file.
Show all steps.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Assume the existence of an UNSORTED ARRAY of n characters. You are to trace the CS111Sort algorithm (as described here) to reorder the elements of a given array. The CS111Sort algorithm is an algorithm that combines the SelectionSort and the InsertionSort following the steps below: 1. Implement the SelectionSort Algorithm on the entire array for as many iterations as it takes to sort the array only to the point of ordering the elements so that the last n/2 elements are sorted in increasing (ascending) order. 2. Implement the InsertionSort Algorithm to sort the first half of the resulting array elements so that these elements are sorted in decreasing (descending) order.
Answer:
class Main {
public static void main(String[] args) {
char arr[] = {'T','E','D','R','W','B','S','V','A'};
int n = arr.length;
System.out.println("Selection Sort:");
System.out.println("Iteration\tArray\tComparisons");
long comp1 = selectionSort(arr);
System.out.println("Total comparisons: "+comp1);
System.out.println("\nInsertion Sort:");
System.out.println("Iteration\tArray\tComparisons");
long comp2 = insertionSort(arr);
System.out.println("Total Comparisons: "+comp2);
System.out.println("\nOverall Total Comparisons: "+(comp1+comp2));
}
static long selectionSort(char arr[]) {
// applies selection sort for n/2 elements
// returns number of comparisons
int n = arr.length;
long comparisons = 0;
// One by one move boundary of unsorted subarray
for (int i = n-1; i>=n-n/2; i--) {
// Find the minimum element in unsorted array
int max_idx = i;
for (int j = i-1; j>=0; j--) {
// there is a comparison everytime this loop returns
comparisons++;
if (arr[j] > arr[max_idx])
max_idx = j;
}
// Swap the found minimum element with the first
// element
char temp = arr[max_idx];
arr[max_idx] = arr[i];
arr[i] = temp;
System.out.print(n-1-i+"\t");
printArray(arr);
System.out.println("\t"+comparisons);
}
return comparisons;
}
static long insertionSort(char arr[]) {
// applies insertion sort for n/2 elements
// returns number of comparisons
int n = arr.length;
n = n-n/2; // sort only the first n/2 elements
long comparisons = 0;
for (int i = 1; i < n; ++i) {
char key = arr[i];
int j = i - 1;
/* Move elements of arr[0..i-1], that are
greater than key, to one position ahead
of their current position */
while (j >= 0) {
// there is a comparison everytime this loop runs
comparisons++;
if (arr[j] > key) {
arr[j + 1] = arr[j];
} else {
break;
}
j--;
}
arr[j + 1] = key;
System.out.print(i-1+"\t");
printArray(arr);
System.out.println("\t"+comparisons);
}
return comparisons;
}
static void printArray(char arr[]) {
for (int i=0; i<arr.length; i++)
System.out.print(arr[i]+" ");
}
}
Explanation:
Explanation is in the answer.
The Taylor series expansion for ax is: Write a MATLAB program that determines ax using the Taylor series expansion. The program asks the user to type a value for x. Use a loop for adding the terms of the Taylor series. If cn is the nth term in the series, then the sum Sn of the n terms is . In each pass calculate the estimated error E given by . Stop adding terms when . The program displays the value of ax. Use the program to calculate: (a) 23.5 (b) 6.31.7 Compare the values with those obtained by using a calculator.
Answer: (a). 11.3137
(b). 22.849
Explanation:
Provided below is a step by step analysis to solving this problem
(a)
clc;close all;clear all;
a=2;x=3.5;
E=10;n=0;k=1;sn1=0;
while E >0.000001
cn=((log(a))^n)*(x^n)/factorial(n);
sn=sn1+cn;
E=abs((sn-sn1)/sn1);
sn1=sn;
n=n+1;
k=k+1;
end
fprintf('2^3.5 from tailor series=%6.4f after adding n=%d terms\n',sn,n);
2^3.5 from tailor series=11.3137 after adding n=15 terms
disp('2^3.5 using calculator =11.3137085');
Command window:
2^3.5 from tailor series=11.3137 after adding n=15 terms
2^3.5 using calculator =11.3137085
(b)
clc;close all;clear all;
a=6.3;x=1.7;
E=10;n=0;k=1;sn1=0;
while E >0.000001
cn=((log(a))^n)*(x^n)/factorial(n);
sn=sn1+cn;
E=abs((sn-sn1)/sn1);
sn1=sn;
n=n+1;
k=k+1;
end
fprintf('6.3^1.7 from tailor series=%6.4f after adding n=%d terms\n',sn,n);
disp('6.3^1.7 using calculator =22.84961748');
Command window:
6.3^1.7 from tailor series=22.8496 after adding n=16 terms
6.3^1.7 using calculator =22.84961748
cheers i hope this helped !!!
Consider the following relations:
Emp(eid: integer, ename: varchar, sal: integer, age: integer, did: integer) Dept(did: integer, budget: integer, floor: integer, mgr_eid: integer) Salaries range from $10,000 to $100,000, ages vary from 20 to 80, each department has about five employees on average, there are 10 floors, and budgets vary from $10,000 to $1 million. You can assume uniform distributions of values. For each of the following queries, which of the listed index choices would you choose to speed up the query.
1. Query: Print ename, age, and sal for all employees.
A) Clustered hash index on fields of Emp.
B) Unclustered hash index on fields of Emp.
C) Clustered B+ tree index on fields of Emp.
D) Unclustered hash index on fields of Emp.
E) No index.
2. Query: Find the dids of departments that are on the 10th floor and have a budget of less than $15,000.
A) Clustered hash index on the floor field of Dept.
B) Unclustered hash index on the floor field of Dept.
C) Clustered B+ tree index on fields of Dept.
D) Clustered B+ tree index on the budget field of Dept.
E) No index.
Answer:
Check the explanation
Explanation:
--Query 1)
SELECT ename, sal, age
FROM Emp;
--Query 2)
SELECT did
FROM Dept
WHERE floot = 10 AND budget<15000;
Box one:
logical
Date and time
Compatibility
Web
Box 2:
&
$
=
#
Box 3:
Not
If
Or
Sum
4. Write an interface ObjectWithTwoParameters which has a method specification: double area (double d1, double d2) which returns the area of a particular object. Write three classes RectangleClass, TriangleClass, and CylinderClass which implement the interface you created. Also, write a demo class which creates objects of RectangleClass, TriangleClass, and CylinderClass and call the corresponding methods
Answer:
The java program for the given scenario is as follows.
import java.util.*;
//interface with method area
interface ObjectWithTwoParameters
{
double area (double d1, double d2);
}
class RectangleClass implements ObjectWithTwoParameters
{
//overriding area()
public double area (double d1, double d2)
{
return d1*d2;
}
}
class TriangleClass implements ObjectWithTwoParameters
{
//overriding area()
public double area (double d1, double d2)
{
return (d1*d2)/2;
}
}
class CylinderClass implements ObjectWithTwoParameters
{
public double area (double d1, double d2)
{
return ( 2*3.14*d1*d1 + d2*(2*3.14*d1) );
}
}
public class Test
{
public static void main(String[] args)
{
//area displayed for all three shapes
ObjectWithTwoParameters r = new RectangleClass();
double arear = r.area(2, 3);
System.out.println("Area of rectangle: "+arear);
ObjectWithTwoParameters t = new TriangleClass();
double areat = t.area(4,5);
System.out.println("Area of triangle: "+areat);
ObjectWithTwoParameters c = new CylinderClass();
double areac = c.area(6,7);
System.out.println("Area of cylinder: "+areac);
}
}
OUTPUT
Area of rectangle: 6.0
Area of triangle: 10.0
Area of cylinder: 489.84
Explanation:
1. The program fulfils all the mentioned requirements.
2. The program contains one interface, ObjectWithTwoParameters, three classes which implement that interface, RectangleClass, TriangleClass and CylinderClass, and one demo class, Test, containing the main method.
3. The method in the interface has no access specifier.
4. The overridden methods in the three classes have public access specifier.
5. No additional variables have been declared.
6. The test class having the main() method is declared public.
7. The area of the rectangle, triangle and the cylinder have been computed as per the respective formulae.
8. The interface is similar to a class which can have only declarations of both, variables and methods. No method can be defined inside an interface.
9. The other classes use the methods of the interface by implementing the interface using the keyword, implements.
10. The object is created using the name of the interface as shown.
ObjectWithTwoParameters r = new RectangleClass();
Design and implement the Heap.h header using the given Heap class below:
template
class Heap
public:
Heap();
Heap(const T elements[], int arraySize); //
Remove the root from the heap and maintain the heap property
T remove() throw (runtime_error);
// Insert element into the heap and maintain the heap property
void add(const T& element);
// Get the number of element in the heap
int getSize() const;
private:
vector v;
Removing the root in a heap - after the root is removed, the tree must be rebuilt to maintain the heap property:
Move the last node to replace the root;
Let the root be the current node;
While (the current node has children and the current node is smaller than one of its children) Swap the current node with the larger of its children; The current node now is one level down;}
Adding a new node - to add a new node to the heap, first add it to the end of the heap and then rebuild the tree as follows:
Let the last node be the current node;
While (the current node is greater than its parent)
{Swap the current node with its parent; The current node now is one level up:}
To test the header file, write the heapSort function and use the following main program:
#include
#include "Heap.h"
using namespace std;
template
void heapSort(T list[], int arraySize) 1/
your code here .. p
int main()
const int SIZE = 9;
int list[] = { 1, 2, 3, 4, 9, 11, 3, 1, 2 }; heapSort(list, SIZE);
cout << "Sorted elements using heap: \n";
for (int i = 0; i < SIZE; i++) cout << list[i] << " ";
cout << endl;
system("pause");
return;
Sample output:
Sorted elements using heap:
1 1 2 3 3 4 7 9 11
Answer:
See explaination
Explanation:
/* Heap.h */
#ifndef HEAP_H
#define HEAP_H
#include<iostream>
using namespace std;
template<class T>
class Heap
{
public:
//constructor
Heap()
{
arr = nullptr;
size = maxsize = 0;
}
//parameterized constructor
Heap(const T elements[], int arraySize)
{
maxsize = arraySize;
size = 0;
arr = new T[maxsize];
for(int i=0; i<maxsize; i++)
{
add(elements[i]);
}
}
//Remove the root from the heap and maintain the heap property
T remove() //code is altered
{
T item = arr[0];
arr[0] = arr[size-1];
size--;
T x = arr[0];
int parent = 0;
while(1)
{
int child = 2*parent + 1;
if(child>=size) break;
if(child+1 < size &&arr[child+1]>arr[child])
child++;
if(x>=arr[child]) break;
arr[parent] = arr[child];
parent = child;
}
arr[parent] = x;
return item;
}
//Insert element into the heap and maintain the heap property
void add(const T&element)
{
if(size==0)
{
arr[size] = element;
size++;
return;
}
T x = element;
int child = size;
int parent = (child-1)/2;
while(child>0 && x>arr[parent])
{
arr[child] = arr[parent];
child = parent;
parent = (child-1)/2;
}
arr[child] = x;
size++;
}
//Get the number of element in the heap
int getSize() const
{
return size;
}
private:
T *arr; //code is altered
int size, maxsize;
};
#endif // HEAP_H
///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
/* main.cpp */
#include<iostream>
#include "Heap.h"
using namespace std;
template <typename T>
void heapSort(T list[], int arraySize)
{
Heap<T> heap(list, arraySize);
int n = heap.getSize();
for(int i=n-1; i>0; i--)
{
list[i] = heap.remove();
}
}
int main()
{
const int SIZE = 9;
int list[] = {1, 7, 3, 4, 9, 11, 3, 1, 2};
heapSort<int>(list, SIZE);
cout << "Sorted elements using heap: \n";
for (int i = 0; i < SIZE; i++)
cout << list[i] << " ";
cout <<endl;
system("pause");
return 0;
}
Write a program named RectangleArea to calculate the area of a rectangle with a length of 15.8 and a width of 7.9. The program should have two classes, one is the RectangleArea holding the Main(), and the other one is Rectangle. The Rectangle class has three members: a property of length, a property of width, and a method with a name of CalArea() to calculate the area. You can invoke the auto-provided constructor or write a self-defined constructor to initialize the rectangle. Calculate and display the area of the rectangle by first initialize an object named myRectangle and then calculate its area using the CalArea() method.
Answer:
public class RectangleArea {
public static void main(String[] args) {
Rectangle myRectangle = new Rectangle(15.8, 7.9);
double calArea = myRectangle.calcArea(15.8, 7.9);
System.out.println("The area is "+ calArea);
}
}
class Rectangle{
private double length;
private double width;
//Defined Constructor
public Rectangle(double length, double width) {
this.length = length;
this.width = width;
}
public double calcArea(double len, double wi){
return len* wi;
}
}
Explanation:
This is solved with Java programming languageStart by creating both classes RectangleArea and and RectangleSInce both classes are in the same file, only one can be publicClass Rectangle contains the three members are decribed by the questionprivate double length, private double width and the method calcArea(). It also has a sefl-declared constructor.In the class RectangleArea, an object of Rectangle is created and initialized.The method calcArea is called using the created object and the value is printed.This represents a group of Book values as a list (named books). We can then dig through this list for useful information and calculations by calling the methods we're going to implement. class Library: Define the Library class. • def __init__(self): Library constructor. Create the only instance variable, a list named books, and initialize it to an empty list. This means that we can only create an empty Library and then add items to it later on.
Answer:
class Library: def __init__(self): self.books = [] lib1 = Library()lib1.books.append("Biology") lib1.books.append("Python Programming Cookbook")Explanation:
The solution code is written in Python 3.
Firstly, we can create a Library class with one constructor (Line 2). This constructor won't take any input parameter value. There is only one instance variable, books, in the class (Line 3). This instance variable is an empty list.
To test our class, we can create an object lib1 (Line 5). Next use that object to add the book item to the books list in the object (Line 6-8).
Write the interface (.h file) of a class Counter containing: A data member counter of type int. A data member named counterID of type int. A static int data member named nCounters. A constructor that takes an int argument. A function called increment that accepts no parameters and returns no value. A function called decrement that accepts no parameters and returns no value. A function called getValue that accepts no parameters and returns an int. A function named getCounterID that accepts no parameters and returns an int.
Explanation:
See the attached image for The interface (.h file) of a class Counter
Describe the Software Development Life Cycle. Describe for each phase of the SDLC how it can be used to create software for an Employee Payroll System that allows employees to log the number of hours completed in a work period and then generate their pay. Use Microsoft Word to complete this part of the assessment. Save your file as LASTNAME FIRSTNAME M08FINAL1 were LASTNAME is your lastname and FIRSTNAME is your first name. Upload your answer to this question; do not submit this assessment until you have completed all questions. This question is worth 25 points.
Answer:
Check the explanation
Explanation:
SDLC
SDLC stands for Software Development Life Cycle
It provides steps for developing a software product
It also checks for the quality and correctness of a software
The main aim of SDLC is to develop a software, which meets a customer requirements
SDLC involves various phases to develop a high-quality product for its end user
Phases of SDLC
There are seven phases in a software development life cycle as follows,
Requirement Analysis Feasibility Study Design Coding Testing Deployment Maintenance
kindly check the attached image below
Requirement analysis
This is the first stage in a SDLC.
In this phase, a detailed and precise requirements are collected from various teams by senior team members
It gives a clear idea of the entire project scope, opportunities and upcoming issues in a project
This phase also involves, planning assurance requirements and analyze risk involved in the project
Here, the timeline to finish the project is finalized
Feasibility Study
In this stage, the SRS document, that is “Software Requirement Specification” document is defined, which includes everything to be designed and developed during the life cycle of a project
Design
In this phase as the name indicates, software and system design documents are prepared, based on the requirement specification document
This enable us to define the whole system architecture
Two types of design documents are prepared in this phase as follows,
High-Level Design
Low-Level Design
Coding
In this phase, the coding of project is done in a selected programming language
Tasks are divided into many units and given to various developers
It is the longest phase in the SDLC
Testing
In this stage, the developed coding is tested by the testing team
The testing team, checks the coding in the user point of view
Installation/Deployment
In this stage, a product is released by the company to the client
Maintenance
Once the product is delivered to the client, the maintenance phase work will start by fixing errors, modifying and enhancing the product.
Employee Payroll System
Requirement analysis
Gather information from various team for requirement
Analyze the number of employees to handle the project
Timeline to finish the project, eg.1 month
Feasibility Study
The system requirements such as how many systems are required
Decide which software to be installed to handle the project.
For example, if we are going to develop the software using C language, then software for that has to be installed
The SRS document has to be prepared
Design
In this the HLD and LLD is prepared, the overall software architecture is prepared
The modules involved in coding are decided
Here the modules can be employee, payroll manager, Employee Log time, account details
The input and output are designed, such as employee name is the input and output is the salary for him, based on working hours
Coding
Here the coding is divided into various sections and given to 2 or more employees
One may code employee detail, one will code working hours of employees and one may code the banking details of employee
Testing
The coding is tested for syntax, declaration, data types and various kinds of error
The testing team will test the coding with various possible values
They may even check with wrong values and analyze the output for that
Installation
Now the software is installed in the client system and the client is calculating the payroll for an employee based on working hours in a month
Maintenance
If any error occurs, the team will clear the issue.
When a new employee joins, then the employee data will added to the database and the database is updated
Also if the client asks for any new features, it will done in this phase.
Write a class called DisArray with methods to convert a 1-dimensional array to a 2-dimensional array. The methods' name should be convert2D. You should create methods to convert int[] and String[], that can be tested against the following class. Your convert2D methods should choose the closest possible square-ish size for the 2D array. For example, if your input array is [10], its 2D conversion should be [3][4] or [4][3] -- you decide if you want to favor rows over columns. Your method should place the elements of the one-dimensional array into the two-dimensional array in row-major order, and fill the remaining elements with 0 (for integer arrays) or null (for String arrays). The process of filling unused elements with 0 or null is called padding. If the input array's length is a perfect square, e.g., [16], then your output should a square array, i.e., [4][4]. For any other size, your objective is to minimize the number of padded elements. For example, if your input is [10] you should opt for a [3][4] array instead of a [4][4]. The former will have only 2 padded elements; the latter 6.
=======================================================
//Class header definition
public class DisArray {
//First method with int array as parameter
public static void convert2D(int[] oneD) {
//1. First calculate the number of columns
//a. get the length of the one dimensional array
int arraylength = oneD.length;
//b. find the square root of the length and typecast it into a float
float squareroot = (float) Math.sqrt(arraylength);
//c. round off the result and save in a variable called row
int row = Math.round(squareroot);
//2. Secondly, calculate the number of columns
//a. if the square of the number of rows is greater than or equal to the
//length of the one dimensional array,
//then to minimize padding, the number of
//columns is the same as the number of rows.
//b. otherwise, the number of columns in one more than the
// number of rows
int col = ((row * row) >= arraylength) ? row : row + 1;
//3. Create a 2D int array with the number of rows and cols
int [ ][ ] twoD = new int [row][col];
//4. Place the elements in the one dimensional array into
//the two dimensional array.
//a. First create a variable counter to control the cycle through the one
// dimensional array.
int counter = 0;
//b. Create two for loops to loop through the rows and columns of the
// two dimensional array.
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
//if counter is less then the length of the one dimensional array,
//then copy the element at that position into the two dimensional
//array. And also increment counter by one.
if (counter < oneD.length) {
twoD[i][j] = oneD[counter];
counter++;
}
//Otherwise, just pad the array with zeros
else {
twoD[i][j] = 0;
}
}
}
//You might want to create another pair of loop to print the elements
//in the populated two dimensional array as follows
for (int i = 0; i < twoD.length; i++) {
for (int j = 0; j < twoD[i].length; j++) {
System.out.print(twoD[i][j] + " ");
}
System.out.println("");
}
} //End of first method
//Second method with String array as parameter
public static void convert2D(String[] oneD) {
//1. First calculate the number of columns
//a. get the length of the one dimensional array
int arraylength = oneD.length;
//b. find the square root of the length and typecast it into a float
float squareroot = (float) Math.sqrt(arraylength);
//c. round off the result and save in a variable called row
int row = Math.round(squareroot);
//2. Secondly, calculate the number of columns
//a. if the square of the number of rows is greater than or equal to the length of
//the one dimensional array, then to minimize padding, the number of
//columns is the same as the number of rows.
//b. otherwise, the number of columns in one more than the
//number of rows.
int col = (row * row >= arraylength) ? row : row + 1;
//3. Create a 2D String array with the number of rows and cols
String[][] twoD = new String[row][col];
//4. Place the elements in the one dimensional array into the two
// dimensional array.
//a. First create a variable counter to control the cycle through the one
// dimensional array.
int counter = 0;
//b. Create two for loops to loop through the rows and columns of the
//two dimensional array.
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
//if counter is less then the length of the one dimensional array,
//then copy the element at that position into the two dimensional
//array. And also increment counter by one.
if (counter < oneD.length) {
twoD[i][j] = oneD[counter];
counter++;
}
//Otherwise, just pad the array with null values
else {
twoD[i][j] = null;
}
}
}
//You might want to create another pair of loop to print the elements
//in the populated two dimensional array as follows:
for (int i = 0; i < twoD.length; i++) {
for (int j = 0; j < twoD[i].length; j++) {
System.out.print(twoD[i][j] + " ");
}
System.out.println("");
}
} // End of the second method
//Create the main method
public static void main(String[] args) {
//1. Create an arbitrary one dimensional int array
int[] x = {23, 3, 4, 3, 2, 4, 3, 3, 5, 6, 5, 3, 5, 5, 6, 3};
//2. Create an arbitrary two dimensional String array
String[] names = {"abc", "john", "dow", "doe", "xyz"};
//Call the respective methods
convert2D(x);
System.out.println("");
convert2D(names);
} // End of the main method
} // End of class definition
=========================================================
==========================================================
Sample Output23 3 4 3
2 4 3 3
5 6 5 3
5 5 6 3
abc john dow
doe xyz null
==========================================================
Explanation:The above code has been written in Java and it contains comments explaining each line of the code. Please go through the comments. The actual executable lines of code are written in bold-face to distinguish them from the comments.
Which of the following is an example of the rewards & consequences characteristic of an organization?
OOO
pay raise
time sheets
pay check
pay scale
Answer:
Pay raise is an example of the rewards & consequences characteristic of an organization.
Write the definition of a function words_typed, that receives two parameters. The first is a person's typing speed in words per minute (an integer greater than or equal to zero). The second is a time interval in seconds (an integer greater than zero). The function returns the number of words (an integer) that a person with that typing speed would type in that time interval.
//Method definition of words_typed
//The return type is int
//Takes two int parameters: typingSpeed and timeInterval
public static int words_typed(int typingSpeed, int timeInterval) {
//Get the number of words typed by
//finding the product of the typing speed and the time interval
//and then dividing the result by 60 (since the typing speed is in "words
// per minute" and the time interval is in "seconds")
int numberOfWords = typingSpeed * timeInterval / 60;
//return the number of words
return numberOfWords;
} //end of method declaration
Explanation:The code above has been written in Java and it contains comments explaining each of the lines of the code. Please go through the comments.
Answer:
I am writing the program in Python.
def words_typed(typing_speed,time_interval):
typing_speed>=0
time_interval>0
no_of_words=int(typing_speed*(time_interval/60))
return no_of_words
output=words_typed(20,30)
print(output)
Explanation:
I will explain the code line by line.
First the statement def words_typed(typing_speed,time_interval) is the definition of a function named words_typed which has two parameters typing_speed and time_interval.
typing_speed variable of integer type in words per minute.
time_interval variable of int type in seconds
The statements typing_speed>=0 and time_interval>0 means that value of typing_speed is greater than or equal to zero and value of time_interval is greater than zero as specified in the question.
The function words_typed is used to return the number of words that a person with typing speed would type in that time interval. In order to compute the words_typed, the following formula is used:
no_of_words=int(typing_speed*(time_interval/60))
The value of typing_speed is multiplied by value of time_interval in order to computer number of words and the result of the multiplication is stored in no_of_words. Here the time_interval is divided by 60 because the value of time_interval is in seconds while the value of typing_speed is in minutes. So to convert seconds into minutes the value of time_interval is divided by 60 because 1 minute = 60 seconds.
return no_of_words statement returns the number of words.
output=words_typed(20,30) statement calls the words_typed function and passed two values i.e 20 for typing_speed and 30 for time_interval.
print(output) statement displays the number of words a person with typing speed would type in that time interval, on the output screen.
6 things you should consider when planning a PowerPoint Presentation.
Answer: I would suggest you consider your audience and how you can connect to them. Is your presentation, well, presentable? Is whatever you're presenting reliable and true? Also, no more than 6 lines on each slide. Use colors that contrast and compliment. Images, use images. That pulls whoever you are presenting to more into your presentation.
Explanation:
Write an expression to detect that the first character of userinput matches firstLetter.
import java.util.Scanner; public class CharMatching { public static void main (String [] args) { Scanner scnr = new Scanner(System.in); String userInput; char firstLetter; userInput = scnr.nextLine(); firstLetter = scnr.nextLine().charAt(0); if (/* Your solution goes here */) { System.out.println("Found match: " + firstLetter); } else { System.out.println("No match: " + firstLetter); } return; } }
Answer: Provided in the explanation segment
Explanation:
Below is the code to run this program.
we have that the
Program
import java.util.Scanner;
public class CharMatching {
public static void main(String[] args) {
//Scanner object for keyboard read
Scanner scnr=new Scanner(System.in);
//Variable for user input string
String userInput;
//Variable for firstletter input
char firstLetter;
//Read user input string
userInput=scnr.nextLine();
//Read first letter from user
firstLetter=scnr.nextLine().charAt(0);
//Comparison without case sensititvity and result
if(Character.toUpperCase(firstLetter)==Character.toUpperCase(userInput.charAt(0))) {
System.out.println("Found match: "+firstLetter);
}
else {
System.out.println("No match: "+firstLetter);
}
}
}
Output
Hello
h
Found match: h
cheers i hope this helped !!!
3. You are working for a usability company that has recently completed a usability study of an airline ticketing system. They conducted a live AB testing of the original website and a new prototype your team developed. They brought in twenty participants into the lab and randomly assigned them to one of two groups: one group assigned to the original website, and the other assigned to the prototype. You are provided with the data and asked to perform a hypothesis test. Using the standard significance value (alpha level) of 5%, compute the four step t-test and report all important measurements. You can use a calculator or SPSS or online calculator to compute the t-test. (4 pts)
Answer:
Check the explanation
Explanation:
Let the population mean time on task of original website be [tex]\mu_1[/tex] and that of new prototype be [tex]\mu_2[/tex]
Here we are to test
[tex]H_0:\mu_1=\mu_2\;\;against\;\;H_1:\mu_1<\mu_2[/tex]
The data are summarized as follows:
Sample 1 Sample 2
Sample size n=20 n=20
Sample mean [tex]\bar{x}_1=243.4[/tex] [tex]\bar{x}_2=253.4[/tex]
Sample SD s1=130.8901 s2=124.8199
The test statistic is obtained from online calculator as
t=-0.23
The p-value is given by 0.821
As the p-value is more than 0.05, we fail to reject the null hypothesis at 5% level of significance and hence conclude that there is no statistically significant difference between the original website and the prototype developed by the team at 5% level of significance.
The equation of certain traveling waves is y(x.t) = 0.0450 sin(25.12x - 37.68t-0.523) where x and y are in
meters, and t in seconds. Determine the following:
(a) Amplitude. (b) wave number (C) wavelength. (d) angular frequency. (e) frequency: (1) phase angle, (g) the
wave propagation speed, (b) the expression for the medium's particles velocity as the waves pass by them, and (i)
the velocity of a particle that is at x=3.50m from the origin at t=21.os
Answer:
A. 0.0450
B. 4
C. 0.25
D. 37.68
E. 6Hz
F. -0.523
G. 1.5m/s
H. vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)
I. -1.67m/s.
Explanation:
Given the equation:
y(x,t) = 0.0450 sin(25.12x - 37.68t-0.523)
Standard wave equation:
y(x, t)=Asin(kx−ωt+ϕ)
a.) Amplitude = 0.0450
b.) Wave number = 1/ λ
λ=2π/k
From the equation k = 25.12
Wavelength(λ ) = 2π/25.12 = 0.25
Wave number (1/0.25) = 4
c.) Wavelength(λ ) = 2π/25.12 = 0.25
d.) Angular frequency(ω)
ωt = 37.68t
ω = 37.68
E.) Frequency (f)
ω = 2πf
f = ω/2π
f = 37.68/6.28
f = 6Hz
f.) Phase angle(ϕ) = -0.523
g.) Wave propagation speed :
ω/k=37.68/25.12=1.5m/s
h.) vy = ∂y/∂t = 0.045(-37.68) cos (25.12x - 37.68t - 0.523)
(i) vy(3.5m, 21s) = 0.045(-37.68) cos (25.12*3.5-37.68*21-0.523) = -1.67m/s.
A user has called the company help desk to inform them that their Wi-Fi enabled mobile device has been stolen. A support ticket has been escalated to the appropriate team based on the corporate security policy. The security administrator has decided to pursue a device wipe based on corporate security policy. However, they are unable to wipe the device using the installed MDM solution. What is one reason that may cause this to happen?
Answer:
The software can't locate the device
Explanation:
The device wasn't properly setup in order for the software to locate it.
(TCO 4) Give the contents of Register A after the execution of the following instructions:
Data_1 EQU $12
Data_2 EQU %01010101
LDAA #Data_1
ADDA #Data_2
A) $12
B) $67
C) $55
D) $57
What do we call data that's broken down into bits and sent through a network?
Answer:
Bits
Explanation:
Write the method addItemToStock to add an item into the grocery stock array. The method will: • Insert the item with itemName add quantity items to stock. • If itemName is already present in stock, increase the quantity of the item, otherwise add new itemName to stock. • If itemName is not present in stock, insert at the first null position in the stock array. After the insertion all items are adjacent in the array (no null positions between two items). • Additionally, double the capacity of the stock array if itemName is a new item to be inserted and the stock is full.
Answer:
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
//varible to indicate size of the array
int size=20;
//arrays to hold grocery item names and their quantities
std::string grocery_item[size];
int item_stock[size];
//variables to hold user input
std::string itemName;
int quantity;
//variable to check if item is already present
bool itemPresent=false;
//variable holding full stock value
int full_stock=100;
for(int n=0; n<size; n++)
{
grocery_item[n]="";
item_stock[n]=0;
}
do
{
std::cout << endl<<"Enter the grocery item to be added(enter q to quit): ";
cin>>itemName;
if(itemName=="q")
{
cout<<"Program ends..."<<endl; break;
}
else
{
std::cout << "Enter the grocery item stock to be added: ";
cin>>quantity;
}
for(int n=0; n<size; n++)
{
if(grocery_item[n]==itemName)
{
itemPresent=true;
item_stock[n]=item_stock[n]+quantity;
}
}
if(itemPresent==false)
{
for(int n=0; n<size; n++)
{
if(grocery_item[n]=="")
{
itemPresent=true;
grocery_item[n]=itemName;
item_stock[n]=item_stock[n]+quantity;
}
if(item_stock[n]==full_stock)
{
item_stock[n]=item_stock[n]*2;
}
}
}
}while(itemName!="q");
return 0;
}
OUTPUT
Enter the grocery item to be added(enter q to quit): rice
Enter the grocery item stock to be added: 23
Enter the grocery item to be added(enter q to quit): bread
Enter the grocery item stock to be added: 10
Enter the grocery item to be added(enter q to quit): bread
Enter the grocery item stock to be added: 12
Enter the grocery item to be added(enter q to quit): q
Program ends...
Explanation:
1. The length of the array and the level of full stock has been defined inside the program.
2. Program can be tested for varying values of length of array and full stock level.
3. The variables are declared outside the do-while loop.
4. Inside do-while loop, user input is taken. The loop runs until user opts to quit the program.
5. Inside do-while loop, the logic for adding the item to the array and adding the quantity to the stock has been included. For loops have been used for arrays.
6. The program only takes user input for the item and the respective quantity to be added.
Write a program to read as many test scores as the user wants from the keyboard (assuming at most 50 scores). Print the scores in (1) original order, (2) sorted from high to low (3) the highest score, (4) the lowest score, and (5) the average of the scores. Implement the following functions using the given function prototypes: void displayArray(int array[], int size) - Displays the content of the array void selectionSort(int array[], int size) - sorts the array using the selection sort algorithm in descending order. Hint: refer to example 8-5 in the textbook. int findMax(int array[], int size) - finds and returns the highest element of the array int findMin(int array[], int size) - finds and returns the lowest element of the array double findAvg(int array[], int size) - finds and returns the average of the elements of the array
Answer: Provided in the explanation segment
Explanation:
Below is the code to carry out this program;
/* C++ program helps prompts user to enter the size of the array. To display the array elements, sorts the data from highest to lowest, print the lowest, highest and average value. */
//main.cpp
//include header files
#include<iostream>
#include<iomanip>
using namespace std;
//function prototypes
void displayArray(int arr[], int size);
void selectionSort(int arr[], int size);
int findMax(int arr[], int size);
int findMin(int arr[], int size);
double findAvg(int arr[], int size) ;
//main function
int main()
{
const int max=50;
int size;
int data[max];
cout<<"Enter # of scores :";
//Read size
cin>>size;
/*Read user data values from user*/
for(int index=0;index<size;index++)
{
cout<<"Score ["<<(index+1)<<"]: ";
cin>>data[index];
}
cout<<"(1) original order"<<endl;
displayArray(data,size);
cout<<"(2) sorted from high to low"<<endl;
selectionSort(data,size);
displayArray(data,size);
cout<<"(3) Highest score : ";
cout<<findMax(data,size)<<endl;
cout<<"(4) Lowest score : ";
cout<<findMin(data,size)<<endl;
cout<<"(5) Lowest scoreAverage score : ";
cout<<findAvg(data,size)<<endl;
//pause program on console output
system("pause");
return 0;
}
/*Function findAvg that takes array and size and returns the average of the array.*/
double findAvg(int arr[], int size)
{
double total=0;
for(int index=0;index<size;index++)
{
total=total+arr[index];
}
return total/size;
}
/*Function that sorts the array from high to low order*/
void selectionSort(int arr[], int size)
{
int n = size;
for (int i = 0; i < n-1; i++)
{
int minIndex = i;
for (int j = i+1; j < n; j++)
if (arr[j] > arr[minIndex])
minIndex = j;
int temp = arr[minIndex];
arr[minIndex] = arr[i];
arr[i] = temp;
}
}
/*Function that display the array values */
void displayArray(int arr[], int size)
{
for(int index=0;index<size;index++)
{
cout<<setw(4)<<arr[index];
}
cout<<endl;
}
/*Function that finds the maximum array elements */
int findMax(int arr[], int size)
{
int max=arr[0];
for(int index=1;index<size;index++)
if(arr[index]>max)
max=arr[index];
return max;
}
/*Function that finds the minimum array elements */
int findMin(int arr[], int size)
{
int min=arr[0];
for(int index=1;index<size;index++)
if(arr[index]<min)
min=arr[index];
return min;
}
cheers i hope this help!!!
