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umka21 [38]
3 years ago
6

Probability can be used to determine the likelihood of specific __________

Computers and Technology
2 answers:
kupik [55]3 years ago
6 0
B occurrences because you taking the probability of something <span />
Minchanka [31]3 years ago
4 0
The answer is b of course
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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.

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3 years ago
If an application written for an earlier version of Windows doesn't run correctly the operating system can emulate its own older
n200080 [17]

Answer:

The answer is b) Right-clicking the program icon, clicking Properties, and then applying the required settings on the Compatibility tab.

Explanation:

If you have to run an application on a specific operating system, you have to configure the compatibility by accessing in its Properties, then Compatibility tab, and configure the operating system properly.

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2 years ago
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