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RoseWind [281]
1 year ago
9

Nicole wants to create a database to collect information about videos in her video rental store. She would like to use the datab

ase to look up movies by rating, actors, producer, and title. She would also like to retrieve the movies that are in stock. What fields should she have in her database? title, rating, actor, and producer rating, director, and actor title, rating, actor 1, actor 2, actor 2, actor 4, producer, and in stock product number, price, and rating
Computers and Technology
1 answer:
Verizon [17]1 year ago
3 0

Answer:

All the field which she has mention has to created by Nichole in her database.

Explanation:

The following fields have to be created in the database

title, rating, actor, and producer rating, director, and actor title, rating, actor 1, actor 2, actor 2, actor 4, producer, and in the stock product number, price, and rating

 

And additional field also has to create movie-length also as integer field which should carry the length of the movie  

Date of release and movie certificate also can be captured so the search engine will be very useful

You might be interested in
Recall the problem of finding the number of inversions. As in the text, we are given a sequence of n numbers a1, . . . , an, whi
Kay [80]

Answer:

The algorithm is very similar to the algorithm of counting inversions. The only change is that here we separate the counting of significant inversions from the merge-sort process.

Algorithm:

Let A = (a1, a2, . . . , an).

Function CountSigInv(A[1...n])

if n = 1 return 0; //base case

Let L := A[1...floor(n/2)]; // Get the first half of A

Let R := A[floor(n/2)+1...n]; // Get the second half of A

//Recurse on L. Return B, the sorted L,

//and x, the number of significant inversions in $L$

Let B, x := CountSigInv(L);

Let C, y := CountSigInv(R); //Do the counting of significant split inversions

Let i := 1;

Let j := 1;

Let z := 0;

// to count the number of significant split inversions while(i <= length(B) and j <= length(C)) if(B[i] > 2*C[j]) z += length(B)-i+1; j += 1; else i += 1;

//the normal merge-sort process i := 1; j := 1;

//the sorted A to be output Let D[1...n] be an array of length n, and every entry is initialized with 0; for k = 1 to n if B[i] < C[j] D[k] = B[i]; i += 1; else D[k] = C[j]; j += 1; return D, (x + y + z);

Runtime Analysis: At each level, both the counting of significant split inversions and the normal merge-sort process take O(n) time, because we take a linear scan in both cases. Also, at each level, we break the problem into two subproblems and the size of each subproblem is n/2. Hence, the recurrence relation is T(n) = 2T(n/2) + O(n). So in total, the time complexity is O(n log n).

Explanation:

5 0
1 year ago
Design an application for Bob's E-Z Loans. The application accepts a client's loan amount and monthly payment amount. Output the
zmey [24]

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.

4 0
1 year ago
Given an n-element array X, algorithm D calls algorithm E on each element X[i]. Algorithm E runs in O(i) time when it is called
krek1111 [17]

Answer:

O(n^2)

Explanation:

The number of elements in the array X is proportional to the algorithm E runs time:

For one element (i=1) -> O(1)

For two elements (i=2) -> O(2)

.

.

.

For n elements (i=n) -> O(n)

If the array has n elements the algorithm D will call the algorithm E n times, so we have a maximum time of n times n, therefore the worst-case running time of D is O(n^2)  

5 0
2 years ago
An author is preparing to send their book to a publisher as an email attachment. The file on their computer is 1000 bytes. When
Yakvenalex [24]

Answer:

See explanation

Explanation:

My explanation to the author is that:

The reduction in size of the attachment doesn't mean that some parts of the book (i.e. the attachment) has been deleted.

I'll also made him understand that the book will retain its original size and contents after downloading by the recipient of the mail.

6 0
2 years ago
Write a function that implements another stack function, peek. Peek returns the value of the first element on the stack without
devlian [24]

Answer:

See explaination

Explanation:

StackExample.java

public class StackExample<T> {

private final static int DEFAULT_CAPACITY = 100;

private int top;

private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);

/**

* Returns a reference to the element at the top of this stack.

* The element is not removed from the stack.

* atreturn element on top of stack

* atthrows EmptyCollectionException if stack is empty

*/

public T peek() throws EmptyCollectionException

{

if (isEmpty())

throw new EmptyCollectionException("stack");

return stack[top-1];

}

/**

* Returns true if this stack is empty and false otherwise.

* atreturn true if this stack is empty

*/

public boolean isEmpty()

{

return top < 0;

}

}

//please replace "at" with the at symbol

Note:

peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.

3 0
2 years ago
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