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.
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).
Let U = {b1, b2, , bn} with n ≥ 3. Interpret the following algorithm in the context of urn problems. for i is in {1, 2, , n} do for j is in {i + 1, i + 2, , n} do for k is in {j + 1, j + 2, ..., n} do print bi, bj, bk How many lines does it print? It prints all the possible ways to draw three balls in sequence, without replacement. It prints P(n, 3) lines. It prints all the possible ways to draw an unordered set of three balls, without replacement. It prints P(n, 3) lines. It prints all the possible ways to draw three balls in sequence, with replacement. It prints P(n, 3) lines. It prints all the possible ways to draw an unordered set of three balls, without replacement. It prints C(n, 3) lines. It prints all the possible ways to draw three balls in sequence, with replacement. It prints C(n, 3) lines.
Answer:
Check the explanation
Explanation:
Kindly check the attached image for the first step
Note that the -print" statement executes n(n — I)(n — 2) times and the index values for i, j, and k can never be the same.
Therefore, the algorithm prints out all the possible ways to draw three balls in sequence, without replacement.
Now we need to determine the number of lines this the algorithm print. In this case, we are selecting three different balls randomly from a set of n balls. So, this involves permutation.
Therefore, the algorithm prints the total
P(n, 3)
lines.
Help its simple but I don't know the answer!!!
If you have a -Apple store and itunes gift card-
can you use it for in app/game purchases?
Answer:
yes you can do that it's utter logic
Answer:
Yes.
Explanation:
Itunes gift cards do buy you games/movies/In app purchases/ect.
You work at a cheeseburger restaurant. Write a program that determines appropriate changes to a food order based on the user’s dietary restrictions. Prompt the user for their dietary restrictions: vegetarian, lactose intolerant, or none. Then using if statements and else statements, print the cook a message describing how they should modify the order. The following messages should be used: - If the user enters "lactose intolerant", say "No cheese." - If the user enters "vegetarian", say "Veggie burger." - If the user enters "none", say "No alterations."
Answer:
See explaination
Explanation:
dietary_restrictions = input("Any dietary restrictions?: ")
if dietary_restrictions=="lactose intolerant":
print("No cheese")
elif dietary_restrictions == "vegetarian":
print("Veggie burger")
else:
print("No alteration")
A customer seeks to buy a new computer for private use at home. The customer primarily needs the computer to use the Microsoft PowerPoint application for the purpose of practicing presentation skills. As a salesperson what size hard disc would you recommend and why?
Answer:
500gbs of ssd storage. Microsoft also allows for multi platform cloud cross-saving.
Explanation:
Write a functionvector merge(vector a, vector b)that merges two vectors, alternating elements from both vectors. If one vector isshorter than the other, then alternate as long as you can and then append the remaining elements from the longer vector. For example, if a is 1 4 9 16and b is9 7 4 9 11then merge returns the vector1 9 4 7 9 4 16 9 1118. Write a predicate function bool same_elements(vector a, vector b)that checks whether two vectors have the same elements in some order, with the same multiplicities. For example, 1 4 9 16 9 7 4 9 11 and 11 1 4 9 16 9 7 4 9 would be considered identical, but1 4 9 16 9 7 4 9 11 and 11 11 7 9 16 4 1 would not. You will probably need one or more helper functions.19. What is the difference between the size and capacity of a vector
Answer:
see explaination for code
Explanation:
CODE
#include <iostream>
#include <vector>
using namespace std;
vector<int> merge(vector<int> a, vector<int> b) {
vector<int> result;
int k = 0;
int i = 0, j = 0;
while (i < a.size() && j < b.size()) {
if (k % 2 == 0) {
result.push_back(a[i ++]);
} else {
result.push_back(b[j ++]);
}
k ++;
}
while (i < a.size()) {
result.push_back(a[i ++]);
}
while(j < b.size()) {
result.push_back(b[j ++]);
}
return result;
}
int main() {
vector<int> a{1, 4, 9, 16};
vector<int> b{9, 7, 4, 9, 11};
vector<int> result = merge(a, b);
for (int i=0; i<result.size(); i++) {
cout << result[i] << " ";
}
}
Modify your "Star Search" program to read 100 scores from a text file, then execute the "Star Search" algorithm (drop the lowest and highest scores, then average the rest) and display the result, with three digits after the decimal point. As per the Star Search specifications, only scores from 0 - 10 will be accepted. Any score that is outside that range is to be rejected, and is not included in the average. That means that invalid scores can't be the high or low scores either. For example (using 10 scores): if the list is 6.1 -5.1 2.0 3.5 5.8 2.0 12.9 5.0 9.9 10.0 The -5.1 and 12.9 are rejected as invalid (not in the range 0-10). 2.0 and 10.0 are the low and high scores to be dropped. The remaining SIX scores are averaged. If you wish to use this data for testing, the answer is 5.383
Answer:
See explaination
Explanation:
#include <fstream>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <cstdlib>
#include <ctime>
#include <vector>
using namespace std;
void getJudgeData(double &score);
double calcScore(double scores[]);
int findLowest(double scores[]);
int findHighest(double scores[]);
int main() {
const int COL=5;
int row;
double score;
cout<<"How many students data you want to enter ? ";
cin>>row;
// Creating 2-D array Dynamically
double** scores = new double*[row];
for (int i = 0; i < row; ++i)
scores[i] = new double[COL];
for(int i=0;i<row;i++)
{
cout<<"\nPerformer#"<<(i+1)<<":"<<endl;
for(int j=0;j<COL;j++)
{
cout<<"Enter Score given by judge#"<<(j+1)<<":";
getJudgeData(scores[i][j]);
}
}
for(int i=0;i<row;i++)
{
cout<<"Performer#"<<(i+1)<<" score :"<<calcScore(scores[i])<<endl;
}
return 0;
}
void getJudgeData(double &score)
{
while(true)
{
cin>>score;
if(score<0 || score>10)
{
cout<<"Invalid.Must be between 0-10"<<endl;
cout<<"re-enter :";
}
else
{
break;
}
}
}
double calcScore(double scores[])
{
double sum=0;
int min= findLowest(scores);
int max= findHighest(scores);
for(int i=0;i<5;i++)
{
if(i!=min && i!=max)
{
sum+=scores[i];
}
}
return sum/3;
}
int findLowest(double scores[])
{
double min=scores[0];
int minIndex=0;
for(int i=0;i<5;i++)
{
if(min>scores[i])
{
min=scores[i];
minIndex=i;
}
}
return minIndex;
}
int findHighest(double scores[])
{
double max=scores[0];
int maxIndex=0;
for(int i=0;i<5;i++)
{
if(max<scores[i])
{
max=scores[i];
maxIndex=i;
}
}
return maxIndex;
}
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 !!!
How has the Aswan High Dem had an adverse effect on human health?
Answer:
How was the Aswan high dam had an adverse effect on human health? It created new ecosystems in which diseases such as malaria flourished. An environmental challenge facing the Great Man Made River is that pumping water near the coast. ... Water scarcity makes it impossible to grow enough food for the population.
Explanation:
Retail price data for n = 60 hard disk drives were recently reported in a computer magazine. Three variables were recorded for each hard disk drive: y = Retail PRICE (measured in dollars) x1 = Microprocessor SPEED (measured in megahertz) (Values in sample range from 10 to 40) x2 = CHIP size (measured in computer processing units) (Values in sample range from 286 to 486) A first-order regression model was fit to the data. Part of the printout follows: __________.
Answer:
Explanation:
Base on the scenario been described in the question, We are 95% confident that the price of a single hard drive with 33 megahertz speed and 386 CPU falls between $3,943 and $4,987
Write a program whose input is a character and a string, and whose output indicates the number of times the character appears in the string.
Ex: If the input is:
n Monday
the output is:
1
Ex: If the input is:
z Today is Monday
the output is:
0
Ex: If the input is:
n It's a sunny day
the output is:
2
Case matters.
Ex: If the input is:
n Nobody
the output is:
0
n is different than N.
This is what i have so far.
#include
#include
using namespace std;
int main() {
char userInput;
string userStr;
int numCount;
cin >> userInput;
cin >> userStr;
while (numCount == 0) {
cout << numCount << endl;
numCount = userStr.find(userInput);
}
return 0;
}
In this puzzle you have a list of n pitchers, with known capacities, c1, c2, ... cn, of holding water in gallons. You have a faucet that can be used as often and as much as the player needs. The goal is to measure exactly g gallons of water, using nothing other than these pitchers. Suppose the current amounts of water that these pitchers hold is w1, w2, ... wn,. These numbers are assumed to be 0 initially. The puzzle is solved as soon as wi = g for any i. A player can perform the following operations:
Fill pitcher i (from the faucet). The precondition of this operation is: ci > wi ≥ 0. The effect is: wi = ci.
Empty pitcher i. The precondition of this operation is: ci ≥ wi > 0. The effect is: wi = 0.
Pour pitcher i to pitcher j. The precondition of this operation is: (ci ≥ wi > 0) and (cj > wj ≥ 0). In words, pitcher i must have some water to pour, and pitcher j must have some unused capacity to receive it. The (partial) effect is: (wi = 0) or (wj = cj) or both. In words, the pour operation must continue until pitcher i becomes empty (and its content is added to pitcher j's content), or pitcher j becomes full (and pitcher i retains the remainder), whichever occurs first. They may occur simultaneously.
A Sample Run
Select the puzzle to solve:
1. Pitchers
2. Eight puzzle
Your selection: 1
Enter the number of pitchers: 3
Enter the capacities of the 3 pitchers (gallons): 2, 5, 10
Enter the goal (gallons): 1
Current configuration: [0, 0, 0]
Please select your next move from the following choices:
1. Fill pitcher 1
2. Fill pitcher 2
3. Fill pitcher 3
Your selection: 2
Current configuration: [0, 5, 0]
Please select your next move from the following choices:
1. Fill pitcher 1
2. Fill pitcher 3
3. Empty pitcher 2
4. Pour pitcher 2 to 1
5. Pour pitcher 2 to 3
Your selection: 4
Current configuration: [2, 3, 0]
Please select your next move from the following choices:
1. Fill pitcher 2
2. Fill pitcher 3
3. Empty pitcher 1
4. Empty pitcher 2
5. Pour pitcher 1 to 2
6. Pour pitcher 1 to 3
7. Pour pitcher 2 to 3
Your selection: 3 Current configuration: [0, 3, 0]
Please select your next move from the following choices:
1. Fill pitcher 1
2. Fill pitcher 2
3. Fill pitcher 3
4. Empty pitcher 2
5. Pour pitcher 2 to 1
6. Pour pitcher 2 to 3
Your selection: 5 Current configuration: [2, 1, 0]
Great! You have reached the goal in 4 moves. Bye.
Answer:
See explaination
Explanation:
#include <iostream>
#include <vector>
using namespace std;
// function to print configuration for each pitcher
void printConfig(vector<int> w, int n)
{
cout << "Current configuration: [" << w[0];
for(int i=1; i<n; i++)
{
cout << ", " << w[i];
}
cout << "]" << endl;
}
void getOptions(vector<int> c, vector<int> w, int n, vector<pair<int, pair<int, int>>> &options)
{
// options count
int count = 0;
// fill
for(int i=0; i<n; i++)
{
if(c[i] > w[i] && w[i] >= 0)
{
count ++;
cout << count << ". Fill pitcher " << (i+1) << endl;
// add options to list
options.push_back(make_pair(1, make_pair(i, 0)));
}
}
// empty
for(int i=0; i<n; i++)
{
if(c[i] >= w[i] && w[i] > 0)
{
count++;
cout << count << ". Empty pitcher " << (i+1) << endl;
// add options to list
options.push_back(make_pair(2, make_pair(i, 0)));
}
}
// Pour pitcher i to pitcher j
for(int i=0; i<n; i++)
{
if(c[i] >= w[i] && w[i] > 0)
{
for(int j=0; j<n; j++)
{
if(i!=j && c[j] > w[j] && w[j] >= 0)
{
count++;
cout << count << ". Pour pitcher " << (i+1) << " to " << (j+1) << endl;
// add options to list
options.push_back(make_pair(3, make_pair(i, j)));
}
}
}
}
}
void execute(vector<int> c, vector<int> &w, pair<int, pair<int, int>> option)
{
int i = option.second.first;
// for fill
if(option.first == 1)
{
w[i] = c[i];
}
// empty
else if(option.first == 2)
{
w[i] = 0;
}
// pour
else
{
int j = option.second.second;
if(w[i] >= c[j])
{
w[i] = w[i] - c[j];
w[j] = c[j];
}
else
{
w[j] = w[i];
w[i] = 0;
}
}
}
int main()
{
// required variables
int choice, n, temp, g, flag, moves = 0;
// vectors are dynamic sized arrays
vector<int> c, w;
vector<pair<int, pair<int, int>>> options;
// select puzzle
cout << "Select the puzzle to solve:\n";
cout << "1. Pitchers\n";
cout << "2. Eight puzzle\n";
cout << "Your selection: ";
cin >> choice;
if(choice == 1)
{
// input values required
cout << "Enter the number of pitchers: ";
cin >> n;
// array for pitchers
cout << "Enter the capacities of the " << n << " pitchers (gallons): ";
for(int i=0; i<n; i++)
{
cin >> temp;
c.push_back(temp);
w.push_back(0);
}
cout << "Enter the goal (gallons): ";
cin >> g;
// start game
while(true)
{
// print configuration
printConfig(w, n);
// check for goal state
flag = 0;
for(int i=0; i<n; i++)
{
if(w[i] == g)
{
flag = 1;
break;
}
}
if(flag)
{
cout << "Great! You have reached the goal in " << moves << " moves. Bye." << endl;
break;
}
// get and print options
//empty previous options
options.clear();
getOptions(c, w, n, options);
// ask for user's selection
cout << "Your selection: ";
cin >> choice;
// update moves
moves++;
// perform according to the selection
execute(c, w, options[choice-1]);
}
}
return 0;
}
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.
Algorithmic Complexity: what is the asymptotic complexity (Big-O) of each code section? Identify the critical section of each.\ Line 1: for (int i=0; i<532; i++) { f(n) = O( ) Line 2: for (int j=1; j
Answer:
Check the explanation
Explanation:
1) f(n) = O( 1 ), since the loops runs a constant number of times independent of any input size
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
2) f(n) = O( log n! ), the outer loop runs for n times, and the inner loop runs log k times when i = k,ie the total number of print will be – log 1 + log2 +log3 +log4+…...+ log n = log (1 . 2 . 3 . 4 . ……. . n ) =log n!
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
Note : Log (m *n) = Log m + Log n : this is property of logarithm
3) f(n) = [tex]O( n^2 )[/tex], since both outer and inner loop runs n times hence , the total iterations of print statement will be : n +n+n+…+n
for n times, this makes the complexity – n * n = n2
there is no critical section in the code, as a critical section is some part of code which is shared by multiple threads or even processes to modify any shared variable.This code does not contain any variable which can be shared.
Write a statement that calls the recursive method backwardsAlphabet() with parameter startingLetter.
import java.util.Scanner; public class RecursiveCalls { public static void backwardsAlphabet(char currLetter) { if (currLetter == 'a') { System.out.println(currLetter); } else { System.out.print(currLetter + " "); backwardsAlphabet((char)(currLetter - 1)); } } public static void main (String [] args) { Scanner scnr = new Scanner(System.in); char startingLetter; startingLetter = scnr.next().charAt(0); /* Your solution goes here */ } }
Answer:
Following are the code to method calling
backwardsAlphabet(startingLetter); //calling method backwardsAlphabet
Output:
please find the attachment.
Explanation:
Working of program:
In the given java code, a class "RecursiveCalls" is declared, inside the class, a method that is "backwardsAlphabet" is defined, this method accepts a char parameter that is "currLetter". In this method a conditional statement is used, if the block it will check input parameter value is 'a', then it will print value, otherwise, it will go to else section in this block it will use the recursive function that prints it's before value. In the main method, first, we create the scanner class object then defined a char variable "startingLetter", in this we input from the user and pass its value into the method that is "backwardsAlphabet".Which of the following can be created when two tables contain a common field?
Referential key
Primary key
Key template
Relationship
Answer:
Referential key is the correct answer to the given question .
Explanation:
The Referential key is also known as foreign key .In the Referential key the foreign key table field must be match with the field of primary key table in another table .
The main objective of Referential key providing the interaction between the two tables also the Referential key is provides the concept of referential integrity .
Primary key is used find the unique record in the database they are not created when the two tables contain the common field so that's why it is incorrect option.Key template and Relationship are not providing the linking between the two tables that's why it is incorrect option .What is blue L.E.D. light
(34+65+53+25+74+26+41+74+86+24+875+4+23+5432+453+6+42+43+21)°
Create a macro named mReadInt that reads a 16- or 32-bit signed integer from standard input and returns the value in an argument. Use conditional operators to allow the macro to adapt to the size of the desired result. Write a program that tests the macro, passing it operands of various sizes.
Answer:
;Macro mReadInt definition, which take two parameters
;one is the variable to save the number and other is the length
;of the number to read (2 for 16 bit and 4 for 32 bit) .
%macro mReadInt 2
mov eax,%2
cmp eax, "4"
je read2
cmp eax, "2"
je read1
read1:
mReadInt16 %1
cmp eax, "2"
je exitm
read2:
mReadInt32 %1
exitm:
xor eax, eax
%endmacro
;macro to read the 16 bit number, parameter is number variable
%macro mReadInt16 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;macro to read the 32 bit number, parameter is number variable
%macro mReadInt32 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;program to test the macro.
;data section, defining the user messages and lenths
section .data
userMsg db 'Please enter the 32 bit number: '
lenUserMsg equ $-userMsg
userMsg1 db 'Please enter the 16 bit number: '
lenUserMsg1 equ $-userMsg1
dispMsg db 'You have entered: '
lenDispMsg equ $-dispMsg
;.bss section to declare variables
section .bss
;num to read 32 bit number and num1 to rad 16-bit number
num resb 5
num1 resb 3
;.text section
section .text
;program start instruction
global _start
_start:
;Displaying the message to enter 32bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h
;calling the micro to read the number
mReadInt num, 4
;Printing the display message
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Printing the 32-bit number
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 4
int 80h
;displaying message to enter the 16 bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg1
mov edx, lenUserMsg1
int 80h
;macro call to read 16 bit number and to assign that number to num1
;mReadInt num1,2
;calling the display mesage function
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Displaying the 16-bit number
mov eax, 4
mov ebx, 1
mov ecx, num1
mov edx, 2
int 80h
;exit from the loop
mov eax, 1
mov ebx, 0
int 80h
Explanation:
For an assembly code/language that has the conditions given in the question, the program that tests the macro, passing it operands of various sizes is given below;
;Macro mReadInt definition, which take two parameters
;one is the variable to save the number and other is the length
;of the number to read (2 for 16 bit and 4 for 32 bit) .
%macro mReadInt 2
mov eax,%2
cmp eax, "4"
je read2
cmp eax, "2"
je read1
read1:
mReadInt16 %1
cmp eax, "2"
je exitm
read2:
mReadInt32 %1
exitm:
xor eax, eax
%endmacro
;macro to read the 16 bit number, parameter is number variable
%macro mReadInt16 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;macro to read the 32 bit number, parameter is number variable
%macro mReadInt32 1
mov eax, 3
mov ebx, 2
mov ecx, %1
mov edx, 5
int 80h
%endmacro
;program to test the macro.
;data section, defining the user messages and lenths
section .data
userMsg db 'Please enter the 32 bit number: '
lenUserMsg equ $-userMsg
userMsg1 db 'Please enter the 16 bit number: '
lenUserMsg1 equ $-userMsg1
dispMsg db 'You have entered: '
lenDispMsg equ $-dispMsg
;.bss section to declare variables
section .bss
;num to read 32 bit number and num1 to rad 16-bit number
num resb 5
num1 resb 3
;.text section
section .text
;program start instruction
global _start
_start:
;Displaying the message to enter 32bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg
mov edx, lenUserMsg
int 80h
;calling the micro to read the number
mReadInt num, 4
;Printing the display message
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Printing the 32-bit number
mov eax, 4
mov ebx, 1
mov ecx, num
mov edx, 4
int 80h
;displaying message to enter the 16 bit number
mov eax, 4
mov ebx, 1
mov ecx, userMsg1
mov edx, lenUserMsg1
int 80h
;macro call to read 16 bit number and to assign that number to num1
;mReadInt num1,2
;calling the display mesage function
mov eax, 4
mov ebx, 1
mov ecx, dispMsg
mov edx, lenDispMsg
int 80h
;Displaying the 16-bit number
mov eax, 4
mov ebx, 1
mov ecx, num1
mov edx, 2
int 80h
;exit from the loop
mov eax, 1
mov ebx, 0
int 80h
For this problem, you will write a function standard_deviation that takes a list whose elements are numbers (they may be floats or ints), and returns their standard deviation, a single number. You may call the variance method defined above (which makes this problem easy), and you may use sqrt from the math library, which we have already imported for you. Passing an empty list to standard_deviation should result in a ZeroDivisionError exception being raised, although you should not have to explicitly raise it yourself.
Answer:
import math def standard_deviation(aList): sum = 0 for x in aList: sum += x mean = sum / float(len(aList)) sumDe = 0 for x in aList: sumDe += (x - mean) * (x - mean) variance = sumDe / float(len(aList)) SD = math.sqrt(variance) return SD print(standard_deviation([3,6, 7, 9, 12, 17]))Explanation:
The solution code is written in Python 3.
Firstly, we need to import math module (Line 1).
Next, create a function standard_deviation that takes one input parameter, which is a list (Line 3). In the function, calculate the mean for the value in the input list (Line 4-8). Next, use the mean to calculate the variance (Line 10-15). Next, use sqrt method from math module to get the square root of variance and this will result in standard deviation (Line 16). At last, return the standard deviation (Line 18).
We can test the function using a sample list (Line 20) and we shall get 4.509249752822894
If we pass an empty list, a ZeroDivisionError exception will be raised.
Write a program with total change amount as an integer input that outputs the change using the fewest coins, one coin type per line. The coin types are dollars, quarters, dimes, nickels, and pennies. Use singular and plural coin names as appropriate, like 1 penny vs. 2 pennies. Ex: If the input is: 0 or less, the output is: no change Ex: If the input is: 45 the output is: 1 quarter 2 dimes
Answer:.
// Program is written in C++.
// Comments are used for explanatory purposes
// Program starts here..
#include<iostream>
using namespace std;
int main()
{
// Declare Variables
int amount, dollar, quarter, dime, nickel, penny;
// Prompt user for input
cout<<"Amount: ";
cin>>amount;
// Check if input is less than 1
if(amount<=0)
{
cout<<"No Change";
}
else
{
// Convert amount to various coins
dollar = amount/100;
amount = amount%100;
quarter = amount/25;
amount = amount%25;
dime = amount/10;
amount = amount%10;
nickel = amount/5;
penny = amount%5;
// Print results
if(dollar>=1)
{
if(dollar == 1)
{
cout<<dollar<<" dollar\n";
}
else
{
cout<<dollar<<" dollars\n";
}
}
if(quarter>=1)
{
if(quarter== 1)
{
cout<<quarter<<" quarter\n";
}
else
{
cout<<quarter<<" quarters\n";
}
}
if(dime>=1)
{
if(dime == 1)
{
cout<<dime<<" dime\n";
}
else
{
cout<<dime<<" dimes\n";
}
}
if(nickel>=1)
{
if(nickel == 1)
{
cout<<nickel<<" nickel\n";
}
else
{
cout<<nickel<<" nickels\n";
}
}
if(penny>=1)
{
if(penny == 1)
{
cout<<penny<<" penny\n";
}
else
{
cout<<penny<<" pennies\n";
}
}
}
return 0;
}
Any set of logic-gate types that can realize any logic function is called a complete set of logic gates. For example, 2-input AND gates, 2- input OR gates, and inverters are a complete set, because any logic function can be expressed as a sum of products of variables and their complements, and AND and OR gates with any number of inputs can be made from 2-input gates. Do 2-input NAND gates form a complete set of logic gates? Prove your answer.
Answer:
Explanation:
We can use the following method to solve the given problem.
The two input NAND gate logic expression is Y=(A*B)'
we can be able to make us of this function by using complete set AND, OR and NOT
we take one AND gate and one NOT gate
we are going to put the inputs to the AND gate it will give us the output = A*B
we collect the output from the AND and we put this as the input to the NOT gate
then we will get the output as = (A*B)' which is needed
Assume the class Student implements the Speaker interface from the textbook. Recall that this interface includes two abstract methods, speak( ) and announce(String str). A Student contains one instance data, String class Rank. Write the Student class so that it implements Speaker as follows. The speak method will output "I am a newbie here" if the Student is a "Freshman", "I like my school" if the Student is either a "Sophomore" or a "Junior", or "I can not wait to graduate" if the student is a "Senior". The announce method will output "I am a Student, here is what I have to say" followed by the String parameter on a separate line. Finally, the class Rank is initialized in the constructor. Only implement the constructor and the methods to implement the Speaker interface.
Answer:
Check the explanation
Explanation:
Based on the above information, this the following resulting code.
The solution has namely one interface named Speaker, one class called Student and one Main class called StudentMain.
The StudentMain class just initialize various instances of the Student class with various classRank values and then subsequently calling in the speak() methods of all.
announce() method is called on one such instance of Student class to demonstrate its functioning.
Comments are added at required place to better help in understanding the logic.
Speaker.java
public interface Speaker {
abstract void speak();
abstract void announce(String str);
}
Student.java
public class Student implements Speaker {
/*
* Write the Student class so that it implements Speaker as follows.
The speak method will output "I am a newbie here" if the Student is a "Freshman",
"I like my school" if the Student is either a "Sophomore" or a "Junior",
or "I can not wait to graduate" if the student is a "Senior".
The announce method will output "I am a Student, here is what I have to say" followed by the String parameter on a separate line.
*/
String classRank;
// Based on the requirement taking in classRank String as a constructor parameter
public Student(String classRank) {
super();
this.classRank = classRank;
}
// Using the switch-case to code the above requirment
public void speak() {
switch(classRank) {
case "Freshman":
System.out.println("I am a newbie here");
break;
// This case would output the same result for the 2 case conditions
case "Sophomore":
case "Junior":
System.out.println("I like my school");
break;
case "Senior":
System.out.println("I can not wait to graduate");
break;
default:
System.out.println("Unknown classRank inputted");
}
}
//Based on the requirement, first line is printed first after second which is
// the inputted String parameter
public void announce(String str) {
System.out.println("I am a Student, here is what I have to say");
System.out.println(str);
}
}
StudentMain.java
public class StudentMain {
public static void main(String[] args) {
// Initializing all the Objects with required classRank
Student stu1=new Student("Freshman");
Student stu2=new Student("Sophomore");
Student stu3=new Student("Junior");
Student stu4=new Student("Senior");
Student stu5=new Student("Freshman");
//Calling the speak methods of the all the above create objects
stu1.speak();
stu2.speak();
stu3.speak();
stu4.speak();
stu4.speak();
stu4.announce("Wish u all the best");
}
}
Following is the output generated from the code run.
Create a program that generates a report that displays a list of students, classes they are enrolled in and the professor who teaches that class. There are 3 files that provide the input data: 1. FinalRoster.txt : List of students and professors ( S: student, P: professor) Student row: S,Student Name, StudentID Professor row: P, Professor Name, Professor ID, Highest Education 2. FinalClassList.txt: List of classes and professor who teach them (ClassID, ClassName, ID of Professor who teach that class) 3. FinalStudentClassList.txt: List of classes the students are enrolled in. (StudentID, ClassID) The output shall be displayed on screen and stored in a file in the format described in FinalOutput.txt You will need to apply all course concepts in your solution. 1. Student and Professor should be derived from a super class "Person" 2. Every class should implement toString method to return data in the format required for output 3. Exception handling (e.g. FileNotFound etc.) must be implemented 4. Source code must be documented in JavaDoc format 5. Do not hard code number of students, professors or classes. Submit source Java files, the output file and screenshot of the output in a zip format
Answer:
All the classes are mentioned with code and screenshots. . That class is shown at last.
Explanation:
Solution
Class.Java:
Code
**
* atauthor your_name
* This class denotes Class at the college.
*
*/
public class Class {
private String className,classID;
private Professor professor;
/**
* atparam className
* atparam classID
* atparam professor
*/
public Class(String className, String classID, Professor professor) {
this.classID=classID;
this.className=className;
this.professor=professor;
}
/**
* at return classID of the class
*/
public String getClassID() {
return classID;
}
/**
* Override toString() from Object Class
*/
public String toString() {
return classID+" "+className+" "+professor.getName()+" "+professor.getEducation();
}
}
Person.Java:
Code:
/**
* atauthor your_name
* This class represents Person
*
*/
public class Person {
protected String name;
/**method to fetch name
* at return
*/
public String getName() {
return name;
}
/**method to set name
* at param name
*/
public void setName(String name) {
this.name = name;
}
Professor.java:
Code:
import java.util.ArrayList;
import java.util.List;
/**
* at author your_name
*
*This class represents professors
*
*/
public class Professor extends Person{
private String professorID, education;
private List<Class> classes=new ArrayList<Class>();
/**
* at param name
* at param professorID
* at param education
*/
public Professor(String name,String professorID,String education) {
this.name=name;
this.professorID=professorID;
this.education=education;
}
/**
* at return
*/
public String getEducation() {
return this.education;
}
/**
* at return
*/
public String getprofessorID() {
return this.professorID;
}
/** to add classes
* at param Class
*/
public void addClass(Class c) {
classes.add(c);
}
/**
* Override toString() from Object Class
*/
public String toString() {
String result=this.getName()+" - "+professorID+" - "+education;
for(Class c:classes) {
result+=c.toString()+"\n";
}
return result;
}
}
}
Student.java:
Code:
import java.util.ArrayList;
import java.util.List;
/**
* This class represents students
* at author your_Name
*
*/
public class Student extends Person{
private String studentID;
private List<Class> classes=new ArrayList<Class>();
/**
* atparam name
* atparam studentID
*/
public Student(String name,String studentID) {
this.name=name;
this.studentID=studentID;
}
/**
* atreturn
*/
public String getStudentID() {
return studentID;
}
/**
* atparam c
*/
public void addClass(Class c) {
classes.add(c);
}
/**
* atreturn
*/
public int getClassCount() {
return classes.size();
}
/**
* Override toString() from Object Class
*/
public String toString() {
String result=this.getName()+" - "+studentID+"\n";
for(Class c:classes) {
result+=c.toString()+"\n";
}
return result;
}
}
NOTE: Kindly find an attached copy of screenshot of the output, which is a part of the solution to this question
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
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;
(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
Which statement is LEAST accurate? Select one: a. A common reason for ERP failure is that the ERP does not support one or more important business processes of the organization b. Implementing an ERP system has as much to do with changing the way an organization does business than it does with technology. c. The big-bang approach to ERP implementation is generally riskier than the phased in approach. d. To take full advantage of the ERP process, reengineering will need to occur.
Answer:
d. To take full advantage of the ERP process, re-engineering will need to occur.
Explanation:
Justification:
ERP Implementation: ERP gives a change in technological ways for managing resources, the way to run the business is remained same but now Ft becomes well planned Diversified P units of Mich do not share common process. and data, generally employ phased-in approach of implementing ERP
• It is the way to overcome from the risk which is with the big bang approach, direct switching
• Perfect implementation of ERP needs re-engineering on the basis of performance
a. maintenance Sometimes ERP which is installed could not support one of the business process. due to lack of research before selecting it
• The statement that the diversified organizations. not. implement ERPs is least accurate because they implement it by employing phased-in approach Therefore, the correct option is d
Kitchen Gadgets sells a line of high-quality kitchen utensils and gadgets. When customers place orders on the company’s Web site or through electronic data interchange (EDI), the system checks to see if the items are in stock, issues a status message to the customer, and generates a shipping order to the warehouse, which fills the order. When the order is shipped, the customer is billed. The system also produces various reports.
1. List four elements used in DFDs, draw the symbols, and explain how they are used.
2. Draw a context diagram for the order system.
3. Draw a diagram 0 DFD for the order system.
4. Explain the importance of leveling and balancing. Your boss, the IT director, wants you to explain FDDs, BPM, DFDs, and UML to a group of company managers and users who will serve on a systems development team for the new marketing system.
Answer:
Ahhhhh suckkkkkkkkkkk
Design an application for Bob's E-Z Loans. The application accepts a client's loan amount and monthly payment amount. Output the customer's loan balance each month until the loan is paid off. b. Modify the Bob's E-Z Loans application so that after the payment is made each month, a finance charge of 1 percent is added to the balance.
Answer:
part (a).
The program in cpp is given below.
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
//variables to hold balance and monthly payment amounts
double balance;
double payment;
//user enters balance and monthly payment amounts
std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;
std::cout << "Enter the balance amount: ";
cin>>balance;
std::cout << "Enter the monthly payment: ";
cin>>payment;
std::cout << "Loan balance: " <<" "<< "Monthly payment: "<< std::endl;
//balance amount and monthly payment computed
while(balance>0)
{
if(balance<payment)
{ payment = balance;}
else
{
std::cout << balance <<"\t\t\t"<< payment << std::endl;
balance = balance - payment;
}
}
return 0;
}
part (b).
The modified program from part (a), is given below.
#include <stdio.h>
#include <iostream>
using namespace std;
int main()
{
//variables to hold balance and monthly payment amounts
double balance;
double payment;
double charge=0.01;
//user enters balance and monthly payment amounts
std::cout << "Welcome to Bob's E-Z Loans application!" << std::endl;
std::cout << "Enter the balance amount: ";
cin>>balance;
std::cout << "Enter the monthly payment: ";
cin>>payment;
balance = balance +( balance*charge );
std::cout << "Loan balance with 1% finance charge: " <<" "<< "Monthly payment: "<< std::endl;
//balance amount and monthly payment computed
while(balance>payment)
{
std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;
balance = balance +( balance*charge );
balance = balance - payment;
}
if(balance<payment)
{ payment = balance;}
std::cout << balance <<"\t\t\t\t\t"<< payment << std::endl;
return 0;
}
Explanation:
1. The variables to hold the loan balance and the monthly payment are declared as double.
2. The program asks the user to enter the loan balance and monthly payment respectively which are stored in the declared variables.
3. Inside a while loop, the loan balance and monthly payment for each month is computed with and without finance charges in part (a) and part (b) respectively.
4. The computed values are displayed for each month till the loan balance becomes zero.
5. The output for both part (a) and part (b) are attached as images.
