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zalisa [80]
3 years ago
12

A Python function cannot under normal circumstances reference a module variable for its value.

Computers and Technology
2 answers:
user100 [1]3 years ago
8 0

Answer:

dunno

Explanation:

marishachu [46]3 years ago
6 0
This is false because
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[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
What do you think of my profile picture
Gnom [1K]

ayo it's pretty sweet lma.o.

8 0
3 years ago
Read 2 more answers
A tablet computer is a low-cost, centrally managed computer with no internal or external attached drives for data storage.
Norma-Jean [14]

Answer: False

Explanation: Tablet computer is a type of the smartphone which is portable and has computer functions. It can be used for the data accessing, videos, pictures etc. They also have a internal storage for the storing of the data.So, the statement given is false.

Thin client is the device that persist these features mentioned in the question.It is a small computer without fan and any hard-drive for the storage.It is device that has a connection along with a network,.

5 0
3 years ago
Assume there are 15 students registered for a history class. You will need to declare an array of the StudentType structure to c
andrezito [222]

Answer:

#include <iostream>

using namespace std;

struct StudentType{

   string studentName;

   int studentId;

}

int n;

char answer[20];

int main(){

   cout<< "Enter the size of the array: ";

   cin >> n;

   StudentType *student = new StudentType(n);

   for (int i = 0; i < n; i++){

       int name;

       int number;

       cin>> name;

       cin >> number;

       student[i].studentName = name;

       student[i].studentId = number;

   }

   for (int i = 0; i < 20; i++){

       cout<< "Enter answers: ";

       cin >> ans;

       answer[i] = ans;

   }

}

Explanation:

The C++ source code has three global variables namely, answer which is an array of character data type, StudentType which is a structure data type and the integer variable n. The main function declares and initializes the dynamic-spaced student array of the structure datatype with the n variable.

3 0
3 years ago
What is the first character for a file or directory names if they should not be displayed by commands such as ls unless specific
riadik2000 [5.3K]

Answer: The .(dot) character

Explanation: in Linux, the period (dot) is short hand for the bash built in source. It will read and execute commands from a file in the current environment and return the exit status of the last command executed.

The .(dot) character is the first character for a file or directory names if they should not be displayed by commands such as ls unless specifically requested

4 0
3 years ago
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