Answer:
Check the explanation
Explanation:
solution a:
def Rotate(string) :
n = len(string)
temp = string + string
for i in range(n) :
for j in range(n) :
print(temp[i + j], end = "")
print()
string = ("abcde")
Rotate(string)
solution b:
nums = [3,9,5,8,2,4,7]
res = list(filter(lambda n : n < 5, nums))
print(res)
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.
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.
*Sometimes it is difficult to convince top management to commit funds to develop and implement a SIS why*
Step-by-step Explanation:
SIS stands for: The Student Information System (SIS).
This system (a secure, web-based accessible by students, parents and staff) supports all aspects of a student’s educational experience, and other information. Examples are academic programs, test grades, health information, scheduling, etc.
It is difficult to convince top management to commit funds to develop and implement SIS, this can be due to a thousand reasons.
The obvious is that the management don't see the need for it. They would rather have students go through the educational process the same way they did. Perhaps, they just don't trust the whole process, they feel more in-charge while using a manual process.
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 is blue L.E.D. light
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).
What do we call data that's broken down into bits and sent through a network?
Answer:
Bits
Explanation:
The cord of a bow string drill was used for
a. holding the cutting tool.
b. providing power for rotation.
c. transportation of the drill.
d. finding the center of the hole.
Answer:
I don't remember much on this stuff but I think it was B
Which of the following should be the first page of a report?
O Title page
Introduction
O Table of contents
Terms of reference
Answer:
Title page should be the first page of a report.
hope it helps!
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.
Computer A has an overall CPI of 1.3 and can be run at a clock rate of 600 MHz. Computer B has a CPI of 2.5 and can be run at a clock rate of 750 MHz. We have a particular program we wish to run. When compiled for computer A, this program has exactly 100,000 instructions. How many instructions would the program need to have when compiled for Computer B, in order for the two computers to have exactly the same execution time for this program
Answer:
Check the explanation
Explanation:
CPI means Clock cycle per Instruction
given Clock rate 600 MHz then clock time is Cー 1.67nSec clockrate 600M
Execution time is given by following Formula.
Execution Time(CPU time) = CPI*Instruction Count * clock time = [tex]\frac{CPI*Instruction Count}{ClockRate}[/tex]
a)
for system A CPU time is 1.3 * 100, 000 600 106
= 216.67 micro sec.
b)
for system B CPU time is [tex]=\frac{2.5*100,000}{750*10^6}[/tex]
= 333.33 micro sec
c) Since the system B is slower than system A, So the system A executes the given program in less time
Hence take CPU execution time of system B as CPU time of System A.
therefore
216.67 micro = =[tex]\frac{2.5*Instruction}{750*10^6}[/tex]
Instructions = 216.67*750/2.5
= 65001
hence 65001 instruction are needed for executing program By system B. to complete the program as fast as system A
The number of instructions that the program would need to have when compiled for Computer B is; 65000 instructions
What is the execution time?
Formula for Execution time is;
Execution time = (CPI × Instruction Count)/Clock Time
We are given;
CPI for computer A = 1.3
Instruction Count = 100000
Clock time = 600 MHz = 600 × 10⁶ Hz
Thus;
Execution time = (1.3 * 100000)/(600 × 10⁶)
Execution time(CPU Time) = 216.67 * 10⁻⁶ second
For CPU B;
CPU Time = (2.5 * 100000)/(750 × 10⁶)
CPU Time = 333.33 * 10⁻⁶ seconds
Thus, instructions for computer B for the two computers to have same execution time is;
216.67 * 10⁻⁶ = (2.5 * I)/(750 × 10⁶)
I = (216.67 * 10⁻⁶ * 750 × 10⁶)/2.5
I = 65000 instructions
Read more about programming instructions at; https://brainly.com/question/15683939
Let A be an array of n numbers. Recall that a pair of indices i, j is said to be under an inversion if A[i] > A[j] and i < j. Design a divide-and-conquer based algorithm to count the number of inversions in an array of n numbers. You can start by splitting into two subproblems. Answer clearly how you do the merge step, then obtain a recurrence relation that captures the run time of the algorithm. Solve the recurrence relation. Write the complete pseudocode.
Answer:
Check the explanation
Explanation:
#include <stdio.h>
int inversions(int a[], int low, int high)
{
int mid= (high+low)/2;
if(low>=high)return 0 ;
else
{
int l= inversions(a,low,mid);
int r=inversions(a,mid+1,high);
int total= 0 ;
for(int i = low;i<=mid;i++)
{
for(int j=mid+1;j<=high;j++)
if(a[i]>a[j])total++;
}
return total+ l+r ;
}
}
int main() {
int a[]={5,4,3,2,1};
printf("%d",inversions(a,0,4));
return 0;
}
Check the output in the below attached image.
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".In this question, we give two implementations for the function: def intersection_list(lst1, lst2) This function is given two lists of integers lst1 and lst2. When called, it will create and return a list containing all the elements that appear in both lists. For example, the call: intersection_list([3, 9, 2, 7, 1], [4, 1, 8, 2])could create and return the list [2, 1]. Note: You may assume that each list does not contain duplicate items. a) Give an implementation for intersection_list with the best worst-case runtime. b) Give an implementation for intersection_list with the best average-case runtime.
Answer:
see explaination
Explanation:
a)Worst Case-time complexity=O(n)
def intersection_list(lst1, lst2):
lst3 = [value for value in lst1 if value in lst2]
return lst3
lst1 = []
lst2 = []
n1 = int(input("Enter number of elements for list1 : "))
for i in range(0, n1):
ele = int(input())
lst1.append(ele) # adding the element
n2 = int(input("Enter number of elements for list2 : "))
for i in range(0, n2):
ele = int(input())
lst2.append(ele) # adding the element
print(intersection_list(lst1, lst2))
b)Average case-time complexity=O(min(len(lst1), len(lst2))
def intersection_list(lst1, lst2):
return list(set(lst1) & set(lst2))
lst1 = []
lst2 = []
n1 = int(input("Enter number of elements for list1 : "))
for i in range(0, n1):
ele = int(input())
lst1.append(ele)
n2 = int(input("Enter number of elements for list2 : "))
for i in range(0, n2):
ele = int(input())
lst2.append(ele)
print(intersection_list(lst1, lst2))
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
Rewrite this if/else if code segment into a switch statement int num = 0; int a = 10, b = 20, c = 20, d = 30, x = 40; if (num > 101 && num <= 105) { a += 1; } else if (num == 208) { b += 1; x = 8; } else if (num > 208 && num < 210) { c = c * 3; } else { d += 1004; }
Answer:
public class SwitchCase {
public static void main(String[] args) {
int num = 0;
int a = 10, b = 20, c = 20, d = 30, x = 40;
switch (num){
case 102: a += 1;
case 103: a += 1;
case 104: a += 1;
case 105: a += 1;
break;
case 208: b += 1; x = 8;
break;
case 209: c = c * 3;
case 210: c = c * 3;
break;
default: d += 1004;
}
}
}
Explanation:
Given above is the equivalent code using Switch case in JavaThe switch case test multiple levels of conditions and can easily replace the uses of several if....elseif.....else statements.When using a switch, each condition is treated as a separate case followed by a full colon and the the statement to execute if the case is true.The default statement handles the final else when all the other coditions are false
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.
(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
Implement the function pairSum that takes as parameters a list of distinct integers and a target value n and prints the indices of all pairs of values in the list that sum up to n. If there are no pairs that sum up to n, the function should not print anything. Note that the function does not duplicate pairs of indices
Answer:
Following are the code to this question:
def pairSum(a,x): #find size of list
s=len(a) # use length function to calculte length of list and store in s variable
for x1 in range(0, s): #outer loop to count all list value
for x2 in range(x1+1, s): #inner loop
if a[x1] + a[x2] == x: #condition check
print(x1," ",x2) #print value of loop x1 and x2
pairSum([12,7,8,6,1,13],13) #calling pairSum method
Output:
0 4
1 3
Explanation:
Description of the above python can be described as follows:
In the above code, a method pairSum is declared, which accepts two-parameter, which is "a and x". Inside the method "s" variable is declared that uses the "len" method, which stores the length of "a" in the "s" variable. In the next line, two for loop is declared, in the first loop, it counts all variable position, inside the loop another loop is used that calculates the next value and inside a condition is defined, that matches list and x variable value. if the condition is true it will print loop value.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
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:
Select the correct navigational path to create the function syntax to use the IF function.
Click the Formula tab on the ribbon and look in the ???
'gallery
Select the range of cells.
Then, begin the formula with the ????? click ?????. and click OK.
Add the arguments into the boxes for Logical Test, Value_if_True, and Value_if_False.
Answer:
1. Logical
2.=
3.IF
Explanation:
just did the assignment
Dave owns a construction business and is in the process of buying a laptop. He is looking for a laptop with a hard drive that will likely continue to function if the computer is dropped. Which type of hard drive does he need?
Answer:
Solid state drive
Explanation:
The term solid-state drive is used for the electronic circuitry made entirely from semiconductors. This highlights the fact that the main storage form, in terms of a solid-state drive, is via semiconductors instead of a magnetic media for example a hard disk. In lieu of a more conventional hard drive, SSD is built to live inside the device. SSDs are usually more resistant to physical shock in comparison to the electro-mechanical drives and it functions quietly and has faster response time. Therefore, SSD will be best suitable for the Dave.
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
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();
Who is your favorite smite god in Hi-Rez’s “Smite”
Answer:
Variety
Explanation:
6. Why did he choose to install the window not totally plumb?
Answer:
Because then it would break
Explanation:
You achieve this by obtaining correct measurements. When measuring a window, plumb refers to the vertical planes, and level refers to the horizontal planes. So he did not install the window totally plumb
Caches are important to providing a high-performance memory hierarchy to processors. Below is a list of 32-bits memory address references given as word addresses. 0x03, 0xb4, 0x2b, 0x02, 0xbf, 0x58, 0xbe, 0x0e, 0xb5, 0x2c, 0xba, 0xfd For each of these references identify the binary word address, the tag, and the index given a direct mapped cache with 16 one-word blocks. Also list whether each reference is a hit or a miss, assuming the cache is initially empty.
Answer:
See explaination
Explanation:
please kindly see attachment for the step by step solution of the given problem.
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!!!
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
