And, or, not are examples of boolean logic

[1] Please find all the candidate keys and the primary key (or composite primary key) Candidate Key: _______________________ Pri

AVprozaik [17]

Answer:

Check the explanation

Explanation:

1. The atomic attributes can't be a primary key because the values in the respective attributes should be unique.

So, the size of the primary key should be more than one.

In order to find the candidate key, let the functional dependencies be obtained.

The functional dependencies are :

Emp_ID -> Name, DeptID, Marketing, Salary

Name -> Emp_ID

DeptID -> Emp_ID

Marketing -> Emp_ID

Course_ID -> Course Name

Course_Name -> Course_ID

Date_Completed -> Course_Name

Closure of attribute { Emp_ID, Date_Completed } is { Emp_ID, Date_Completed , Name, DeptID, Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { Name , Date_Completed } is { Name, Date_Completed , Emp_ID , DeptID, Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { DeptID, Date_Completed } is { DeptID, Date_Completed , Emp_ID,, Name, , Marketing, Salary, Course_Name, Course_ID}

Closure of attribute { Marketing, Date_Completed } is { Marketing, Date_Completed , Emp_ID,, Name, DeptID , Salary, Course_Name, Course_ID}.

So, the candidate keys are :

{ Emp_ID, Date_Completed }

{ Name , Date_Completed }

{ DeptID, Date_Completed }

{ Marketing, Date_Completed }

Only one candidate key can be a primary key.

So, the primary key chosen be { Emp_ID, Date_Completed }..

2.

The functional dependencies are :

Emp_ID -> Name, DeptID, Marketing, Salary

Name -> Emp_ID

DeptID -> Emp_ID

Marketing -> Emp_ID

Course_ID -> Course Name

Course_Name -> Course_ID

Date_Completed -> Course_Name

3.

For a relation to be in 2NF, there should be no partial dependencies in the set of functional dependencies.

The first F.D. is

Emp_ID -> Name, DeptID, Marketing, Salary

Here, Emp_ID -> Salary ( decomposition rule ). So, a prime key determining a non-prime key is a partial dependency.

So, a separate table should be made for Emp_ID -> Salary.

The tables are R1(Emp_ID, Name, DeptID, Marketing, Course_ID, Course_Name, Date_Completed)

and R2( Emp_ID , Salary)

The following dependencies violate partial dependency as a prime attribute -> prime attribute :

Name -> Emp_ID

DeptID -> Emp_ID

Marketing -> Emp_ID

The following dependencies violate partial dependency as a non-prime attribute -> non-prime attribute :

Course_ID -> Course Name

Course_Name -> Course_ID

So, no separate tables should be made.

The functional dependency Date_Completed -> Course_Name has a partial dependency as a prime attribute determines a non-prime attribute.

So, a separate table is made.

The final relational schemas that follows 2NF are :

R1(Emp_ID, Name, DeptID, Marketing, Course_ID, Course_Name, Date_Completed)

R2( Emp_ID , Salary)

R3 (Date_Completed, Course_Name, Course_ID)

For a relation to be in 3NF, the functional dependencies should not have any transitive dependencies.

The functional dependencies in R1(Emp_ID, Name, DeptID, Marketing, Date_Completed) is :

Emp_ID -> Name, DeptID, Marketing

This violates the transitive property. So, no table is created.

The functional dependencies in R2 ( Emp_ID , Salary) is :

Emp_ID -> Salary

The functional dependencies in R3 (Date_Completed, Course_Name, Course_ID) are :

Date_Completed -> Course_Name

Course_Name -> Course_ID

Here there is a transitive dependency as a non- prime attribute ( Course_Name ) is determining a non-attribute ( Course_ID ).

So, a separate table is made with the concerned attributes.

The relational schemas which support 3NF re :

R1(Emp_ID, Name, DeptID, Course_ID, Marketing, Date_Completed) with candidate key as Emp_ID.

R2 ( Emp_ID , Salary) with candidate key Emp_ID.

R3 (Date_Completed, Course_Name ) with candidate key Date_Completed.

R4 ( Course_Name, Course_ID ). with candidate keys Course_Name and Course_ID.

6 0

3 years ago

Which option ensures that a page break is automatically inserted ahead of a specific paragraph or heading?

gulaghasi [49]

The correct answer is option D, the last one! <3

6 0

3 years ago

Read 2 more answers

How to shutdown a computer by step by step

Reptile [31]

Answer:

1) Press Ctrl + Alt + Del

2) Click the power button in the bottom-right corner of the screen.

3) From the desktop, press Alt + F4 to get the Shut Down Windows screen.

and that's how to shut down your computer

Please mark as brainliest if answer is right

Have a great day, be safe and healthy

Thank u

XD

7 0

2 years ago

Read 2 more answers

You accidentally find someone's password and use it to get into a system. this is hacking.

Umnica [9.8K]

I'd go with yes, but are you doing something malicious or just being nosey?

4 0

3 years ago

Create a lottery game application. Generate three random numbers (see Appendix D for help in doing so), each between 0 and 9. Al

Pachacha [2.7K]

The question is not complete! Here is the complete question and its answer!

Create a lottery game application. Generate three random numbers (see Appendix D for help in doing so), each between 0 and 9. Allow the user to guess three numbers. Compare each of the user’s guesses to the three random numbers and display a message that includes the user’s guess, the randomly determined three-digit number, and the amount of money the user has won as follows:

no matches: 0

any one matching: 10$

two matching: 1000$

three matching: 100000$

Code with Explanation:

#include <iostream>

using namespace std;

int main()

{

int matches=0;

int guess_1, guess_2, guess_3;

int num_1, num_2, num_3;

// get 3 digits from the user

cout<<"Enter first guess digit 0 to 9"<<endl;

cin>>guess_1;

cout<<"Enter second guess digit 0 to 9"<<endl;

cin>>guess_2;

cout<<"Enter third guess digit 0 to 9"<<endl;

cin>>guess_3;

// every time program runs srand() generates new random numbers and rand()%10 makes sure that number is single digit 0 to 9

srand(time(NULL));

num_1=rand()%10;

cout<<"First lottery digit: "<<num_1<<endl;

num_2=rand()%10;

cout<<"Second lottery digit: "<<num_2<<endl;

num_3=rand()%10;

cout<<"Third lottery digit: "<<num_3<<endl;

// store random generated numbers and guess numbers in arrays to compare them

int num[3]= {num_1,num_2,num_3};

int guess[3]={guess_1,guess_2,guess_3};

// compare the arrays to find out how many are matching

for(int i=0; i<3; i++)

{

for(int j=0; j<3; j++)

{

if(num[i]==guess[j])

{

matches = matches + 1;

}

cout << "Total Matches are: " <<matches << endl;

// display reward according to the number of matches

if (matches==0)

cout<<"you won: $0"<<endl;

if (matches==1)

cout<<"you won: $10"<<endl;

if (matches==2)

cout<<"you won: $1000"<<endl;

if (matches==3)

cout<<"you won: $100000"<<endl;

return 0;

}

Output:

Enter first guess digit 0 to 9

7

Enter second guess digit 0 to 9

1

Enter third guess digit 0 to 9

5

First lottery digit: 3

Second lottery digit: 7

Third lottery digit: 2

Total Matches are: 1

You won: $10

Enter first guess digit 0 to 9

7

Enter second guess digit 0 to 9

3

Enter third guess digit 0 to 9

4

First lottery digit: 5

Second lottery digit: 4

Third lottery digit: 7

Total Matches are: 2

You won: $1000

5 0

3 years ago