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.

Answers

Answer 1

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.


Related Questions

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: __________.

Answers

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

Zoom Vacuum, a family-owned manufacturer of high-end vacuums, has grown exponentially over the last few years. However, the company is having difficulty preparing for future growth. The only information system used at Zoom is an antiquated accounting system. The company has one manufacturing plant located in Iowa; and three warehouses, in Iowa, New Jersey, and Nevada. The Zoom sales force is national, and Zoom purchases about 25 percent of its vacuum parts and materials from a single overseas supplier. You have been hired to recommend the information systems Zoom should implement in order to maintain their competitive edge. However, there is not enough money for a full-blown, cross-functional enterprise application, and you will need to limit the first step to a single functional area or constituency. What will you choose, and why?

Answers

Answer:A TPS focusing on production and manufacturing to keep production costs low while maintaining quality, and for communicating with other possible vendors. The TPS would later be used to feed MIS and other higher level systems.

Explanation:

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;
}


Answers

Here is a Python program:

tmp = input().split(' ')
c = tmp[0]; s = tmp[1]
ans=0
for i in range(len(s)):
if s[i] == c: ans+=1

# the ans variable stores the number of occurrences
print(ans)

#define DIRECTN 100
#define INDIRECT1 20
#define INDIRECT2 5
#define PTRBLOCKS 200

typedef struct {
filename[MAXFILELEN];
attributesType attributes; // file attributes
uint32 reference_count; // Number of hard links
uint64 size; // size of file
uint64 direct[DIRECTN]; // direct data blocks
uint64 indirect[INDIRECT1]; // single indirect blocks
uint64 indirect2[INDIRECT2]; // double indirect

} InodeType;

Single and double indirect inodes have the following structure:

typedef struct
{
uint64 block_ptr[PTRBLOCKS];
}
IndirectNodeType;

Required:
Assuming a block size of 0x1000 bytes, write pseudocode to return the block number associated with an offset of N bytes into the file.

Answers

Answer:

WOW! that does not look easy!

Explanation:

I wish i could help but i have no idea how to do that lol

write the steps to insert picture water mark​

Answers

Answer:

Is there a picture of the question?

Explanation:

Double any element's value that is less than minValue. Ex: If minValue = 10, then dataPoints = {2, 12, 9, 20} becomes {4, 12, 18, 20}.
import java.util.Scanner;
public class StudentScores {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
final int NUM_Points = 4;
int[] dataPoints = new int[NUM_POINTS];
int minValue;
int i;
minValue = scnr.nextInt();
for (i = 0; i < dataPoints.length; ++i) {
dataPoints[i] = scnr.nextInt();
}
/* Your solution goes here */
for (i = 0; i < dataPoints.length; ++i) {
System.out.print(dataPoints[i] + " ");
}
System.out.println();
}
}

Answers

Answer:

Following are the code to this question:

for(i=0;i<dataPoints.length;++i) //define loop to count array element  

{

if(dataPoints[i]<minValue) // define condition that checks array element is less then minValue

{

dataPoints[i] = dataPoints[i]*2; //double the value

}

}

Explanation:

Description of the code as follows:

In the given code, a for loop is declared, that uses a variable "i", which counts all array element, that is input by the user. Inside the loop and if block statement is used, that check array element value is less then "minValue", if this condition is true.  So, inside the loop, we multiply the value by 2.

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.

Answers

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))

TRUE OR FALSE! HELP!!

Answers

Answer:

True

Explanation:

There's no one law that governs internet privacy.

Donnell backed up the information on his computer every week on a flash drive. Before copying the files to the flash drive, he always ran a virus scan against the files to ensure that no viruses were being copied to the flash drive. He bought a new computer and inserted the flash drive so that he could transfer his files onto the new computer. He got a message on the new computer that the flash drive was corrupted and unreadable; the information on the flash drive cannot be retrieved. Assuming that the flash drive is not carrying a virus, which of the following does this situation reflect?
a. Compromise of the security of the information on the flash drive
b. Risk of a potential breach in the integrity of the data on the flash drive
c. Both of the above
d. Neither of the above.

Answers

Answer:

b. Risk of a potential breach in the integrity of the data on the flash drive

Explanation:

The corrupted or unreadable file error is an error message generated if you are unable to access the external hard drive connected to the system through the USB port. This error indicates that the files on the external hard drive are no longer accessible and cannot be opened.

There are several reasons that this error message can appear:

Viruses and Malware affecting the external hard drive .Physical damage to external hard drive or USB memory .Improper ejection of removable drives.

A cloud provider is deploying a new SaaS product comprised of a cloud service. As part of the deployment, the cloud provider wants to publish a service level agreement (SLA) that provides an availability rating based on its estimated availability over the next 12 months. First, the cloud provider estimates that, based on historical data of the cloud environment, there is a 25% chance that the physical server hosting the cloud service will crash and that such a crash would take 2 days before the cloud service could be restored. It is further estimated that, over the course of a 12 month period, there will be various attacks on the cloud service, resulting in a total of 24 hours of downtime. Based on these estimates, what is the availability rating of the cloud service that should be published in the SLA?

Answers

Answer:

99.6

Explanation:

The cloud provider  provides the certain modules of the cloud service like infrastructure as a service , software as a service ,etc to other companies .The cloud provider implements a new SaaS service that includes a cloud platform also the cloud provider intends to disclose the Service Level Agreement that is score based on its approximate accessibility during the next 12 months.

The Cloud providers predict that, depending on past cloud system data, it is a 25 % risk that a physical server holding the cloud platform will fail therefore such a failure will require 2 days to recover the cloud service so 99.6 is the Cloud storage accessibility ranking, to be released in the SLA.

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.

Answers

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

Explain possible ways that Darla can communicate with her coworker Terry, or her manager to make sure Joe receives great customer service?

Answers

Answer:

They can communicate over the phone or have meetings describing what is and isn't working for Joe. It's also very important that Darla makes eye contact and is actively listening to effectively handle their customer.

Explanation:

"Write pseudocode that outputs the contents of parallel arrays. You do NOT have to write the entire program. The first array will hold phone numbers The second array will hold company names You do NOT need to load the arrays. Your code can assume they are already loaded with data. The arrays are named phone[ ] and company[ ] Output the phone number and associated company name for every entry in the array."

Answers

Answer:

Check the explanation

Explanation:

PSEUDO CODE:

for(int i=0,j=0;i<phone.length,j<company.length;i++,j++)

{

print(phone[i]," - ",company[j]);

}

this for loop will print each phone number associated with the company names.

Create a Binary Expressions Tree Class and create a menu driven programyour program should be able to read multiple expressions from a file and create expression trees for each expression, one at a timethe expression in the file must be in "math" notation, for example x+y*a/b.display the preorder traversal of a binary tree as a sequence of strings each separated by a tabdisplay the postorder traversal of a binary tree in the same form as aboveWrite a function to display the inorder traversal of a binary tree and place a (before each subtree and a )after each subtree. Don’t display anything for an empty subtree. For example, the expression tree should would be represented as ( (x) + ( ( (y)*(a) )/(b) ) )

Answers

Answer:

Explanation:

Program:

#include<iostream>

#include <bits/stdc++.h>

using namespace std;

//check for operator

bool isOperator(char c)

{

switch(c)

{

case '+': case '-': case '/': case '*': case '^':

return true;

}

return false;

}

//Converter class

class Converter

{

private:

string str;

public:

//constructor

Converter(string s):str(s){}

//convert from infix to postfix expression

string toPostFix(string str)

{

stack <char> as;

int i, pre1, pre2;

string result="";

as.push('(');

str = str + ")";

for (i = 0; i < str.size(); i++)

{

char ch = str[i];

if(ch==' ') continue;

if (ch == '(')

as.push(ch);

else if (ch == ')')

{

while (as.size() != 0 && as.top() != '('){

result = result + as.top() + " ";

as.pop();

}

as.pop();

}

else if(isOperator(ch))

{

while (as.size() != 0 && as.top() != '(')

{

pre1 = precedence(ch);

pre2 = precedence(as.top());

if (pre2 >= pre1){

result = result + as.top() + " ";

as.pop();

}

else break;

}

as.push(ch);

}

else

{

result = result + ch;

}

}

while(as.size() != 0 && as.top() != '(') {

result += as.top() + " ";

as.pop();

}

return result;

}

//return the precedence of an operator

int precedence(char ch)

{

int choice = 0;

switch (ch) {

case '+':

choice = 0;

break;

case '-':

choice = 0;

break;

case '*':

choice = 1;

break;

case '/':

choice = 1;

break;

case '^':

choice = 2;

default:

choice = -999;

}

return choice;

}

};

//Node class

class Node

{

public:

string element;

Node *leftChild;

Node *rightChild;

//constructors

Node (string s):element(s),leftChild(nullptr),rightChild(nullptr) {}

Node (string s, Node* l, Node* r):element(s),leftChild(l),rightChild(r) {}

};

//ExpressionTree class

class ExpressionTree

{

public:

//expression tree construction

Node* covert(string postfix)

{

stack <Node*> stk;

Node *t = nullptr;

for(int i=0; i<postfix.size(); i++)

{

if(postfix[i]==' ') continue;

string s(1, postfix[i]);

t = new Node(s);

if(!isOperator(postfix[i]))

{

stk.push(t);

}

else

{

Node *r = nullptr, *l = nullptr;

if(!stk.empty()){

r = stk.top();

stk.pop();

}

if(!stk.empty()){

l = stk.top();

stk.pop();

}

t->leftChild = l;

t->rightChild = r;

stk.push(t);

}

}

return stk.top();

}

//inorder traversal

void infix(Node *root)

{

if(root!=nullptr)

{

cout<< "(";

infix(root->leftChild);

cout<<root->element;

infix(root->rightChild);

cout<<")";

}

}

//postorder traversal

void postfix(Node *root)

{

if(root!=nullptr)

{

postfix(root->leftChild);

postfix(root->rightChild);

cout << root->element << " ";

}

}

//preorder traversal

void prefix(Node *root)

{

if(root!=nullptr)

{

cout<< root->element << " ";

prefix(root->leftChild);

prefix(root->rightChild);

}

}

};

//main method

int main()

{

string infix;

cout<<"Enter the expression: ";

cin >> infix;

Converter conv(infix);

string postfix = conv.toPostFix(infix);

cout<<"Postfix Expression: " << postfix<<endl;

if(postfix == "")

{

cout<<"Invalid expression";

return 1;

}

ExpressionTree etree;

Node *root = etree.covert(postfix);

cout<<"Infix: ";

etree.infix(root);

cout<<endl;

cout<<"Prefix: ";

etree.prefix(root);

cout<<endl;

cout<< "Postfix: ";

etree.postfix(root);

cout<<endl;

return 0;

}

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 */ } }

Answers

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".

You can use this area to create your resume.

Answers

Answer:

YOUR NAME

YOUR CITY, STATE, AND ZIP CODE

YOUR PHONE NUMBER

YOUR EMAIL

Professional Summary

Reliable, top-notch sales associate with outstanding customer service skills and relationship-building strengths. Dedicated to welcoming customers and providing comprehensive service. In-depth understanding of sales strategy and merchandising techniques. Dependable retail sales professional with experience in dynamic, high-performance environments. Skilled in processing transactions, handling cash, using registers and arranging merchandise. Maintains high-level customer satisfaction by smoothly resolving customer requests, needs and problems. Seasoned Sales Associate with consistent record of exceeding quotas in sales environments. Delivers exceptional customer service and product expertise to drive customer satisfaction ratings. Proficient in use and troubleshooting of POS systems.

Skills

Returns and Exchanges

Adaptable and Flexible

Excellent Written and Verbal Communication

Meeting Sales Goals

Strong Communication and Interpersonal Skills

Time Management

Cash Handling

Reliable and Responsible

Work History

March 2020 to September 2021

Goodwill OF YOUR STATE

Retail Sales Associate    

Helped customers complete purchases, locate items and join reward programs.

Checked pricing, scanned items, applied discounts and printed receipts to ring up customers.

Folded and arranged merchandise in attractive displays to drive sales.

Greeted customers and helped with product questions, selections and purchases.

Organized store merchandise racks and displays to promote and maintain visually appealing environments.

Monitored sales floor and merchandise displays for presentable condition, taking corrective action such as restocking or reorganizing products.

Balanced and organized cash register by handling cash, counting change and storing coupons.

Trained new associates on cash register operations, conducting customer transactions and balancing drawer.

Answered questions about store policies and addressed customer concerns.

Issued receipts and processed refunds, credits or exchanges.

Maintained clean sales floor and straightened and faced merchandise.

Education

YOUR HIGH SCHOOL THAT YOU ATTEND  

Languages

Spanish

Explanation:

THIS IS MY RESUME IF YOU HAVE MORE WORK EXPERIENCE THEN ADD IT AFTER THE GOODWILL. I GOT A 100% ON EDGE.

1. Write a pair of classes, Square1 and Rectangle1. Define Square1 as a subclass of Rectangle1. In addition to setters and getters, provide such methods as computeArea and computePerimeter. Specify the preconditions, postconditions and class invariants, if any, as comments (

Answers

Answer:

The java program including classes is shown below.

import java.util.*;

import java.lang.*;

//base class

class Rectangle1

{

   //variables to hold dimensions and area

   static double l;

   static double b;

   static double a;

   //setter

   static void setWidth(double w)

   {

       b=w;

   }

   //getter

   static double getWidth()

   {

       return b;

   }

   //setter

   static void setLength(double h)

   {

       l=h;

   }

   //getter

   static double getLength()

   {

       return b;

   }

   //area calculation for rectangle

   static void computeArea()

   {

       a = getLength()*getWidth();

       System.out.println("Area of rectangle = " +a);

   }

}

//derived class

class Square1 extends Rectangle1

{

   //variables to hold dimensions and perimeter

   static double s;

   static double peri;

   //settter

   static void setSide(double d)

   {

       s=d;

   }

   //getter

   static double getSide()

   {

       return s;

   }

   //perimeter calculation for square

   static void computePerimeter()

   {

       peri = 2*(getSide()+getSide());

       System.out.println("Perimeter of square = "+peri);

   }

}

public class MyClass {

   public static void main(String args[]) {

     //object of derived class created

     Square1 ob = new Square1();

     ob.setLength(1);

     ob.setWidth(4);

     ob.setSide(1);

     ob.computeArea();

     ob.computePerimeter();

   }

}

Explanation:

1. The class for rectangle is defined having variables to hold the dimensions of a rectangle.

2. The getters and setters are included for both the dimensions.

3. The method to compute the area of the rectangle is defined.

4. The class for square is defined having variables to hold the side of a square.

5. The getter and setter are included for the dimension of the square.

6. The method to compute the perimeter of the square is defined.

7. The class Square1 is the derived class which inherits the class Rectangle1 through the keyword extends.

8. All the variables, getters, setters and methods are declared as static.

9. Another class is defined having only the main() method.

10. Inside main(), the object of the derived class, Square1, is created.

11. The setters and the methods are called through this object.

12. The value of the dimensions are passed as arguments to the setters.

13. The output is attached in an image.

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?

Answers

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.

The relationship between the temperature of a fluid (t, in seconds), temperature (T, in degrees Celsius), is dependent upon the initial temperature of the liquid (T0, in degrees Celsius), the ambient temperature of the surroundings (TA, in degrees Celsius) and the cooling constant (k, in hertz); the relationship is given by: ???? ???? ???????? ???? ???????????? ???? ???????????? ???????????????? Ask the user the following questions:  From a menu, choose fluid ABC, FGH, or MNO.  Enter the initial fluid temperature, in units of degrees Celsius.  Enter the time, in units of minutes.  Enter the ambient air temperature, in units of degrees Celsius. Enter the following data into the program. The vector contains the cooling constant (k, units of hertz) corresponding to the menu entries. K Values = [0.01, 0.03, 0.02] Create a formatted output statement for the user in the Command Window similar to the following. The decimal places must match. ABC has temp 83.2 degrees Celsius after 3 minutes. In

Answers

Answer:

See explaination

Explanation:

clc;

clear all;

close all;

x=input(' choose abc or fgh or mno:','s');

to=input('enter intial fluid temperature in celcius:');

t1=input('enter time in minutes:');

ta=input('enter ambient temperature in celcius:');

abc=1;

fgh=2;

mno=3;

if x==1

k=0.01;

elseif x==2

k=0.03;

else

k=0.02;

end

t=ta+((to-ta)*exp((-k)*t1));

X = sprintf('%s has temp %f degrees celcius after %d minutes.',x,t,t1);

disp(X);

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?

Answers

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.

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

Answers

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

Use a vector to solve the following problem. Read in 20 numbers, each of which is between 10 and 100, inclusive. As each number is read, validate it and store it in the vector only if it isn't a duplicate of a number already read. After reading all the values, display only the unique values that the user entered. Begin with an empty vector and use its push_back function to add each unique value to the vector.

Answers

Answer:

The following are the program in the C++ Programming Language.

//header files

#include <iostream>

#include <vector>

//using namespace

using namespace std;

//define a function

int find(vector<int> &vec, int n)  

{

   //set the for loop

   for(int i = 0; i < vec.size(); ++i)  

   {

       //check the elements of v is in the num

       if(vec[i] == n)  

       {

           //then, return the values of i

          return i;

       }

   }

  return -1;

}

//define the main method

int main()  

{

   //declare a integer vector type variable

   vector<int> vec;

   //print message

   cout << "Enter numbers: ";

   //declare integer type variable

   int n;

   //set the for loop

   for(int i = 0; i < 20; ++i)  

   {

       //get input in the 'num' from the user

       cin >> n;

       //check the following conditions

       if(n >= 10 && n <= 100 && find(vec, n) == -1)  

       {

           //then, assign the values of 'num' in the vector

          vec.push_back(n);

       }

   }

   //print the message

   cout << "User entered: ";

   //set the for loop

   for(int i = 0; i < vec.size(); ++i)  

   {

       //print the output

       cout << vec[i] << " ";

  }

  cout << "\n";

  return 0;  

}

Output:

Enter numbers: 14 18 96 75 23 65 47 12 58 74 76 92 34 32 65 48 46 28 75 56

User entered: 14 18 96 75 23 65 47 12 58 74 76 92 34 32 48 46 28 56

Explanation:

The following are the description of the program:

Firstly, set the required header files and required namespace.Define an integer data type function 'find()' and pass two arguments that is integer vector type argument 'vec' and integer data type argument 'n'.Then, set the for loop that iterates according to the size of the vector variable 'vec' inside the function, after that set the if conditional statement to check the following condition and return the variable 'i'.Define the main method that gets the input from the user then, check that the following inputs are greater than 10 and smaller than 100 then assign these variables in the vector type variable 'vec' and print all the elements.

*Sometimes it is difficult to convince top management to commit funds to develop and implement a SIS why*

Answers

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.

Evaluati urmatoarele expresii


5+2*(x+4)/3, unde x are valoare 18


7/ 2*2+4*(5+7*3)>18


2<=x AND x<=7 , unde x are valoare 23


50 %10*5=


31250/ 5/5*2=

Answers

Answer:

A) 22 ⅓

B) 111>18

C) There is an error in the expression

D) 25

E) 62500

Question:

Evaluate the following expressions

A) 5 + 2 * (x + 4) / 3, where x has a value of 18

B) 7/2 * 2 + 4 * (5 + 7 * 3) & gt; 18

C) 2 <= x AND x<= 7, where x has value 23

D) 50% 10 * 5 =

F) 31250/5/5 * 2 =

Explanation:

A) 5 + 2 * (x + 4) / 3

x = 18

First we would insert the value of x

5 + 2 * (x + 4) / 3

5 + 2(18 + 8) / 3

Then we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.

= 5 + 2(26) / 3

= 5 + 52/3

= 5 + 17 ⅓

= 22 ⅓

B) 7/2 * 2 + 4 * (5 + 7 * 3) > 18

we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.

7/2 * 2 + 4 * (5 + 7 * 3) >18

= 7/2 × 2 + 4× (5 + 7 × 3)>18

= (7×2)/2 + 4× (5+21) >18

= 14/2 + 4(26) >18

= 7 + 104 >18

= 111>18

C) 2 <= x AND x<= 7, where x has value 23

There is an error in the expression

D) 50% of 10 * 5

we would evaluate the expression by applying BODMAS : This stands for Bracket, Of, Division, Multiplication, addition and subtraction.

The 'of' expression means multiplication

= 50% × 10×5

= 50% × 50

50% = 50/100

=50/100 × 50

= 1/2 × 50

= 25

F) 31250/5/5 * 2

The expression has no demarcation. Depending on how it is broken up, we would arrive at different answers. Let's consider:

31250/(5/5 × 2)

Apply BODMAS

= 31250/[5/(5 × 2)]

= 31250/(5/10)

= 31250/(1/2)

Multiply by the inverse of 1/2 = 2/1

= 31250 × (2/1)

= 62500

6. Why did he choose to install the window not totally plumb?

Answers

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

Who is your favorite smite god in Hi-Rez’s “Smite”

Answers

Answer:

Variety

Explanation:

Which of the following should be the first page of a report?
O Title page
Introduction
O Table of contents
Terms of reference

Answers

Answer:

Title page should be the first page of a report.

hope it helps!

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.

Answers

Answer:

I don't remember much on this stuff but I think it was B

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.​

Answers

Answer:

1. Logical

2.=

3.IF

Explanation:

just did the assignment

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.

Answers

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.

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