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zysi [14]
3 years ago
15

Implement a Java program that creates math flashcards for elementary grade students. User will enter his/her name, the type (+,

-, *, /), the range of the factors to be used in the problems, and the number of problems to work. The system will provide problems, evaluate user responses to the problems, score the problems, and provide statistics about the session at the end. Functional Requirements
• User enters name at the beginning of a session. • System covers four math operations – addition, subtraction, multiplication, and division – with the user choosing which of the four operations to do in the session. Only 1 type of problem can be done in each session. • User will enter additional session parameters from prompts – number of problems to work and the range of values desired in the problems, e.g., addition with factors ranging from 0 to 12. • System will present problems to the user. • User will respond to problems with an answer and the system will provide im
Computers and Technology
1 answer:
jeka57 [31]3 years ago
8 0

import java.util.Scanner;

public class JavaApplication44 {

   public static void main(String[] args) {

       Scanner scan = new Scanner(System.in);

       System.out.print("Enter you name: ");

       String name = scan.nextLine();

       System.out.print("Enter your operation (+,-,*,/): ");

       String operator = scan.nextLine();

       System.out.print("Enter the range of the problems: ");

       int ran = scan.nextInt();

       System.out.print("Enter number of problems: ");

       int problems = scan.nextInt();

       int score = 0;

       for (int i = 1; i <= ran; i++){

           int first = (int)(Math.random()*ran);

           int sec = (int)(Math.random()*ran);

           System.out.print("Problem #"+i+": "+first + " "+operator+" "+sec+" = ");

           int ans = scan.nextInt();

           if (operator.equals("+")){

               if (ans == first + sec){

                   System.out.println("Correct!");

                   score++;

               }

               else{

                   System.out.println("Wrong. The correct answer is "+(first+sec));

               }

           }

           else if(operator.equals("-")){

               if (ans == first - sec){

                   System.out.println("Correct");

                   score++;

               }

               else{

                   System.out.println("Wrong. The correct answer is "+(first - sec));

               }

           }

           else if (operator.equals("*")){

               if (ans == first * sec){

                   System.out.println("Correct");

                   score++;

               }

               else{

                   System.out.println("Wrong. The correct answer is "+(first * sec));

               }

           }

           else if (operator.equals("/")){

               if (ans == first / sec){

                   System.out.println("Correct");

                   score++;

               }

               else{

                   System.out.println("Wrong. The correct answer is "+(first / sec));

               }

           }

           

       }

       System.out.println(name+", you answered "+score+" questions correctly.");

   }

   

}

I hope this helps!

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- Consider the relation R = {A, B, C, D, E, F, G, H, I, J} and the set of functional dependencies F = { {B, C} -&gt; {D}, {B} -&
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The key of R is {A, B}

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First thing we do is to identify partial dependencies that violate 2NF. These are attributes that are

functionally dependent on either parts of the key, {A} or {B}, alone.

We can calculate

the closures {A}+ and {B}+ to determine partially dependent attributes:

{A}+ = {A, D, E, I, J}. Hence {A} -> {D, E, I, J} ({A} -> {A} is a trivial dependency)

{B}+ = {B, F, G, H}, hence {A} -> {F, G, H} ({B} -> {B} is a trivial dependency)

To normalize into 2NF, we remove the attributes that are functionally dependent on

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along with the part of the key they depend on (A or B), which are copied into each of

these relations but also remains in the original relation, which we call R3 below:

R1 = {A, D, E, I, J}, R2 = {B, F, G, H}, R3 = {A, B, C}

The new keys for R1, R2, R3 are underlined. Next, we look for transitive

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dependency {B} -> {F} -> {G, H}:

R2 = {F, G, H}, R2 = {B, F}

The final set of relations in 3NF are {R11, R12, R21, R22, R3}

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