What is the last step in conducting url search

Using a windows computer, to expand a branch of the folder tree you would _____.

s344n2d4d5 [400]

A folder can include everything from documents and music, to applications [like Word or Excel or your favorite browser]. But computer folders can also contain other folders, lots of them, and those sub-folders can contain folders of their own. This is the called folder tree. The trunk of the tree is the Desktop. Then there are several folders like Recycle Bin, Control Panel,..The most important branch is the one called ‘Libraries’. Here are the Photos, Music, Videos,...
Using a windows computer, to expand a branch of the folder tree you would right click beside various folders and drives to expand the folder tree.

8 0

3 years ago

[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 is the recommended solution to configure this automated behavior? UC has a requirement that an opportunity should have a fi

Dmitry_Shevchenko [17]

The answer will need to be workflow.

8 0

3 years ago

3.4 lesson practice quiz edhesive

Marina86 [1]

3.4 lesson practice quiz edhesive :

Write a program to check if user inputs "yellow"

Answer:

In Python:

col = input("Enter Color: ")

if col == "yellow":

print("True")

else:

print("False")

Explanation:

This prompts the user for color

col = input("Enter Color: ")

This checks if color is yellow

if col == "yellow":

If true, this prints true

print("True")

If otherwise, this prints false

else:

print("False")

5 0

2 years ago

Difference between switch case and if else statement.

ser-zykov [4K]

Answer:

<h3>SWITCH CASE:</h3>

☆ STATEMENT WILL BE EXECUTED IS DECIDED BY USER.

☆ SWITCH STATEMENT EVALUATES ONLY CHARACTER 《OR 》INTEGER VALUE.

☆ IT USING SINGLE EXPRESSION FOR MULTIPLE CHOICES.

<h3>IF - ELSE STATEMENT. </h3>

☆ STATEMENT WILL BE EXECUTED DEPEND UPON THE OUTPUT OF THE EXPRESSION INSIDE.

☆ IF THE STATEMENT EVALUATES INTEGER, CHARACTER, POINTER《 OR 》FLOATING- POINT TYPE 《 OR 》BOOLEAN TYPE.

☆ IF USING MULTIPLE STATEMENT FOR MULTIPLE CHOICES.

HOPE IT HELP....

7 0

3 years ago