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IRINA_888 [86]
4 years ago
15

Databases offer many security features including passwords to provide authentication, access levels to determine who can access

the data, and access controls to determine what type of access they have to the information.1. True2. False
Computers and Technology
1 answer:
Murljashka [212]4 years ago
7 0

Answer:

True

Explanation:

When using databases in a project, not everyone has the same access level, e.g the database admin may have the highest level of access (access to data on live mode), the software testers have their own level of access (access to data on test mode) and so on.  

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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
Computer design replaced ______________
bazaltina [42]
<span>Computer design replaced (B) models draw or created by hand. Technology nowadays has a big contribution in terms of planning and designing a building or a structure without burning your eyebrow facing that paper and handling your pen. Computer designs are used to make the design more accurate and more precise compared to traditional hand drawn designs.</span>
8 0
3 years ago
When a router forwards incoming packets closer to their destination hosts, this is called?
il63 [147K]
Port forwarding, I think.
3 0
3 years ago
How do you use a iPad when it has a password ?
BaLLatris [955]
Take it to the Apple Store and they will help you or you can go on their site and press "Forgot Apple ID." If you choose the second option you have to enter the first and last name, and the email you used for the iCloud on the iPad. 
6 0
3 years ago
Read 2 more answers
What statement would you use to assign the value 32 to the variable cheeses?
bezimeni [28]

Answer:

Statement to assign 32 to variable cheeses.

cheeses=32;

Explanation:

To assign any value to a variable in any programming language, we use "="operator.First declare a variable "cheeses"of type "int" in this case.Then assign 32 to variable "cheeses" with the help of "=" operator.After assigning 32 , variable will store 32 in it.

Implementation in c++.

#include <bits/stdc++.h>

using namespace std;

int main()

{

int cheeses;

cheeses=32;

return 0;

}

6 0
4 years ago
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