A university wants to schedule the classrooms for final exams. The attributes are given below:
course : course id, name, and department
section : section id, enrollment, and dependent as a weak entity set on course
room : room number, capacity, and building
exam: course id, section id, room number, and time.
1) Determine the relationships among these entity sets. Clearly specify the arity, the attributes, and the mapping cardinality of these relationships.
2) Construct an E-R diagram to for the scheduling system.

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:

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;

Answer:

check if its pluged in

Explanation:

Answer:

(a) the number of times the value is performs is up to four cycles. and as such the integer i is executed up to 5 times.  (b)The point version of the floating point can have CPE of 3.00, even when the multiplication operation required is either 4 or 5 clock.

Explanation:

Solution

The two floating point versions can have CPEs of 3.00, even though the multiplication operation demands either 4 or 5 clock cycles by the latency suggests the total number of clock cycles needed to work the actual operation, while issues time to specify the minimum number of cycles between operations.

Now,

sum = sum + udata[i] * vdata[i]

in this case, the value of i performs from 0 to 3.

Thus,

The value of sum is denoted as,

sum = ((((sum + udata[0] * vdata[0])+(udata[1] * vdata[1]))+( udata[2] * vdata[2]))+(udata[3] * vdata[3]))

Thus,

(A)The number of times the value is executed is up to 4 cycle. And the integer i performed up to 5 times.

Thus,

(B) The floating point version can have CPE of 3.00, even though the multiplication operation required either 4 or 5 clock.

Explanation:

See the attached image for The interface (.h file) of a class Counter

Answer: Provided in the explanation segment

Explanation:

Below is the code to run this program.

we have that the

Program

import java.util.Scanner;

public class CharMatching {

  public static void main(String[] args) {

      //Scanner object for keyboard read

      Scanner scnr=new Scanner(System.in);

      //Variable for user input string

      String userInput;

      //Variable for firstletter input

      char firstLetter;

      //Read user input string

      userInput=scnr.nextLine();

      //Read first letter from user

      firstLetter=scnr.nextLine().charAt(0);

      //Comparison without case sensititvity and result

      if(Character.toUpperCase(firstLetter)==Character.toUpperCase(userInput.charAt(0))) {

          System.out.println("Found match: "+firstLetter);

      }

      else {

          System.out.println("No match: "+firstLetter);

      }

  }

}

Output

Hello

h

Found match: h

cheers i hope this helped !!!

=======================================================

//Class header definition

public class DisArray {

   //First method with int array as parameter

   public static void convert2D(int[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the

       //length of  the one dimensional array,

       //then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       // number of rows

       int col = ((row * row) >= arraylength) ? row : row + 1;

       //3. Create a 2D int array with the number of rows and cols

       int [ ][ ] twoD = new int [row][col];

       //4. Place the elements in the one dimensional array into

       //the two dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       // two  dimensional array.

       for (int i = 0; i < row; i++) {

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

              }

              //Otherwise, just pad the array with zeros

               else {

                   twoD[i][j] = 0;

               }

           }

       }

       //You might want to create another pair of loop to print the elements

       //in the  populated two dimensional array as follows

       for (int i = 0; i < twoD.length; i++) {

           for (int j = 0; j < twoD[i].length; j++) {

               System.out.print(twoD[i][j] + " ");

           }

           System.out.println("");

       }

   }     //End of first method

   //Second method with String array as parameter

   public static void convert2D(String[] oneD) {

       //1. First calculate the number of columns

       //a. get the length of the one dimensional array

       int arraylength = oneD.length;

       //b. find the square root of the length and typecast it into a float

       float squareroot = (float) Math.sqrt(arraylength);

       //c. round off the result and save in a variable called row

       int row = Math.round(squareroot);

       //2. Secondly, calculate the number of columns

       //a. if the square of the number of rows is greater than or equal to the length of  

       //the one dimensional array, then to minimize padding, the number of

       //columns  is the same as the number of rows.

       //b. otherwise, the number of columns in one more than the

       //number of rows.

       int col = (row * row >= arraylength) ? row : row + 1;

       //3. Create a 2D String array with the number of rows and cols

       String[][] twoD = new String[row][col];

       //4. Place the elements in the one dimensional array into the two

       // dimensional array.

       //a. First create a variable counter to control the cycle through the one

       // dimensional array.

       int counter = 0;

       //b. Create two for loops to loop through the rows and columns of the

       //two  dimensional array.

       for (int i = 0; i < row; i++) {

           for (int j = 0; j < col; j++) {

               //if counter is less then the length of the one dimensional array,

               //then  copy the element at that position into the two dimensional

               //array.  And also increment counter by one.

               if (counter < oneD.length) {

                   twoD[i][j] = oneD[counter];

                   counter++;

               }

               //Otherwise, just pad the array with null values

               else {

                   twoD[i][j] = null;

               }

           }

       }

       //You might want to create another pair of loop to print the elements  

       //in the populated two dimensional array as follows:

       for (int i = 0; i < twoD.length; i++) {

           for (int j = 0; j < twoD[i].length; j++) {

               System.out.print(twoD[i][j] + " ");

           }

           System.out.println("");

       }

   }    // End of the second method

   //Create the main method

   public static void main(String[] args) {

       //1. Create an arbitrary one dimensional int array

       int[] x = {23, 3, 4, 3, 2, 4, 3, 3, 5, 6, 5, 3, 5, 5, 6, 3};

       //2. Create an arbitrary two dimensional String array

       String[] names = {"abc", "john", "dow", "doe", "xyz"};

       //Call the respective methods

       convert2D(x);

       System.out.println("");

       convert2D(names);

   }         // End of the main method

}            // End of class definition

=========================================================

==========================================================

23 3 4 3  

2 4 3 3  

5 6 5 3  

5 5 6 3  

abc john dow  

doe xyz null

==========================================================

The above code has been written in Java and it contains comments explaining each line of the code. Please go through the comments. The actual executable lines of code are written in bold-face to distinguish them from the comments.

Answer:

Explanation:

Please find attachment for the Python program

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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.

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

Answer:

Check the explanation

Explanation:

#include <iostream>

using namespace std;

void CoordTransform(int x, int y, int& xValNew,int& yValNew){

  xValNew = (x+1)*2;

  yValNew = (y+1)*2;

}

int main() {

  int xValNew = 0;

  int yValNew = 0;

  CoordTransform(3, 4, xValNew, yValNew);

  cout << "(3, 4) becomes " << "(" << xValNew << ", " << yValNew << ")" << endl;

  return 0;

}

Answer:

See explaination

Explanation:

/* Heap.h */

#ifndef HEAP_H

#define HEAP_H

#include<iostream>

using namespace std;

template<class T>

class Heap

{

public:

//constructor

Heap()

{

arr = nullptr;

size = maxsize = 0;

}

//parameterized constructor

Heap(const T elements[], int arraySize)

{

maxsize = arraySize;

size = 0;

arr = new T[maxsize];

for(int i=0; i<maxsize; i++)

{

add(elements[i]);

}

}

//Remove the root from the heap and maintain the heap property

T remove() //code is altered

{

T item = arr[0];

arr[0] = arr[size-1];

size--;

T x = arr[0];

int parent = 0;

while(1)

{

int child = 2*parent + 1;

if(child>=size) break;

if(child+1 < size &&arr[child+1]>arr[child])

child++;

if(x>=arr[child]) break;

arr[parent] = arr[child];

parent = child;

}

arr[parent] = x;

return item;

}

//Insert element into the heap and maintain the heap property

void add(const T&element)

{

if(size==0)

{

arr[size] = element;

size++;

return;

}

T x = element;

int child = size;

int parent = (child-1)/2;

while(child>0 && x>arr[parent])

{

arr[child] = arr[parent];

child = parent;

parent = (child-1)/2;

}

arr[child] = x;

size++;

}

//Get the number of element in the heap

int getSize() const

{

return size;

}

private:

T *arr; //code is altered

int size, maxsize;

};

#endif // HEAP_H

///////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

/* main.cpp */

#include<iostream>

#include "Heap.h"

using namespace std;

template <typename T>

void heapSort(T list[], int arraySize)

{

Heap<T> heap(list, arraySize);

int n = heap.getSize();

for(int i=n-1; i>0; i--)

{

list[i] = heap.remove();

}

}

int main()

{

const int SIZE = 9;

int list[] = {1, 7, 3, 4, 9, 11, 3, 1, 2};

heapSort<int>(list, SIZE);

cout << "Sorted elements using heap: \n";

for (int i = 0; i < SIZE; i++)

cout << list[i] << " ";

cout <<endl;

system("pause");

return 0;

}

Answer:

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).  

Answer:

Term describes the order of arrangement of files and folders on a computer would be ORGANIZATION.

:) Hope this helps!

Answer:

The term that describes the order of arrangement of files and folders on a computer is organization.

Explanation:

Answer:

#include <stdio.h>

#include <iostream>

using namespace std;

int main()

{

   //varible to indicate size of the array

   int size=20;

   //arrays to hold grocery item names and their quantities

   std::string grocery_item[size];

   int item_stock[size];

   //variables to hold user input

   std::string itemName;

   int quantity;

   //variable to check if item is already present

   bool itemPresent=false;

   //variable holding full stock value

   int full_stock=100;

   for(int n=0; n<size; n++)

   {

       grocery_item[n]="";

       item_stock[n]=0;

   }

   do

   {

       std::cout << endl<<"Enter the grocery item to be added(enter q to quit): ";

       cin>>itemName;

       if(itemName=="q")

       {

               cout<<"Program ends..."<<endl; break;

       }

       else

       {

           std::cout << "Enter the grocery item stock to be added: ";

           cin>>quantity;

       }

   for(int n=0; n<size; n++)

   {

       if(grocery_item[n]==itemName)

       {

           itemPresent=true;

           item_stock[n]=item_stock[n]+quantity;

       }

   }

   if(itemPresent==false)

   {

       for(int n=0; n<size; n++)

       {

           if(grocery_item[n]=="")

           {

               itemPresent=true;

               grocery_item[n]=itemName;

               item_stock[n]=item_stock[n]+quantity;

           }

           if(item_stock[n]==full_stock)

           {

               item_stock[n]=item_stock[n]*2;

           }

       }

   }

   }while(itemName!="q");

   return 0;

}

OUTPUT

Enter the grocery item to be added(enter q to quit): rice

Enter the grocery item stock to be added: 23

Enter the grocery item to be added(enter q to quit): bread

Enter the grocery item stock to be added: 10

Enter the grocery item to be added(enter q to quit): bread

Enter the grocery item stock to be added: 12

Enter the grocery item to be added(enter q to quit): q

Program ends...

Explanation:

1. The length of the array and the level of full stock has been defined inside the program.

2. Program can be tested for varying values of length of array and full stock level.

3. The variables are declared outside the do-while loop.

4. Inside do-while loop, user input is taken. The loop runs until user opts to quit the program.

5. Inside do-while loop, the logic for adding the item to the array and adding the quantity to the stock has been included. For loops have been used for arrays.

6. The program only takes user input for the item and the respective quantity to be added.

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

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:

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:

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.

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)

Answer:

Hewo, Here are some ways in which apps earn money :-

hope it helps!

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!!!

Answer:

See explaination

Explanation:

n = int(input())

if 20 <= n <= 98:

while n % 11 != 0: // for all numbers in range (20-98) all the identical

digit numbers are divisible by 11,So all numbers that

​​​​​​ are not divisible by 11 are a part of count down, This

while loop stops when output digits get identical.

print(n, end=" ")

n = n - 1

print(n)

else:

print("Input must be 20-98")

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.

Answer:

see explaination

Explanation:

#include <iostream>

#include <string>

#include <vector>

using namespace std;

vector<string> split(string, string);

int main()

{

vector<string> splitedStr;

string data;

string delimiter;

cout << "Enter string to split:" << endl;

getline(cin,data);

cout << "Enter delimiter string:" << endl;

getline(cin,delimiter);

splitedStr = split(data,delimiter);

cout << "\n";

cout << "The substrings are: ";

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

cout << "\"" << splitedStr[i] << "\"" << ",";

cout << endl << endl;

cin >> data;

return 0;

}

vector<string> split(string target, string delimiter)

{

unsigned first = 0;

unsigned last;

vector<string> subStr;

while((last = target.find(delimiter, first)) != string::npos)

{

subStr.push_back(target.substr(first, last-first));

first = last + delimiter.length();

}

subStr.push_back(target.substr(first));

return subStr;

}

Answer:

Pay raise is an example of the rewards & consequences characteristic of an organization.

Here is the complete question.

In this lab, you use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

start input testScore,

classRank if testScore >= 90 then if classRank >= 25 then output "Accept"

else output "Reject" endif else if testScore >= 80

then if classRank >= 50 then output "Accept" else output "Reject" endif

else if testScore >= 70

then if classRank >= 75 then output "Accept"

else output "Reject"

endif else output "Reject"

endif

endif

endif

stop

Study the pseudocode in picture above. Write the interactive input statements to retrieve: A student’s test score (testScore) A student's class rank (classRank) The rest of the program is written for you. Execute the program by clicking "Run Code." Enter 87 for the test score and 60 for the class rank. Execute the program by entering 60 for the test score and 87 for the class rank.

[comment]: <> (3. Write the statements to convert the string representation of a student’s test score and class rank to the integer data type (testScore and classRank, respectively).)

Function: This program determines if a student will be admitted or rejected. Input: Interactive Output: Accept or Reject

*/ #include using namespace std; int main() { // Declare variables

// Prompt for and get user input

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl; }

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl; }

else cout << "Reject" << endl; } } } //End of main() function

Answer:

Explanation:

The objective here is to use the flowchart and pseudocode found in the figures below to add code to a partially created C++ program. When completed, college admissions officers should be able to use the C++ program to determine whether to accept or reject a student, based on his or her test score and class rank.

PROGRAM:

#include<iostream>

using namespace std;

int main(){

// Declare variables

int testScore, classRank;

// Prompt for and get user input

cout<<"Enter test score: ";

cin>>testScore;

cout<<"Enter class rank: ";

cin>>classRank;

// Test using admission requirements and print Accept or Reject

if(testScore >= 90)  

{ if(classRank >= 25)

{ cout << "Accept" << endl; }

else

cout << "Reject" << endl;

}

else { if(testScore >= 80)

{ if(classRank >= 50)

cout << "Accept" << endl;

else cout << "Reject" << endl; }

else { if(testScore >= 70)

{ if(classRank >=75) cout << "Accept" << endl;

else cout << "Reject" << endl;

}

else cout << "Reject" << endl; }

}

return 0;

} //End of main() function

OUTPUT:

See the attached file below:

Answer:

The software can't locate the device

Explanation:

The device wasn't properly setup in order for the software to locate it.