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3 years ago

Compared to a virtual image, a real image

Computers and Technology

2 answers:

olganol [36]3 years ago

7 0

Answer: The answer is D: can be viewed on a screen.

Compared to a virtual image, a real image can be viewed on a screen.

Explanation:

A real image refers to an image that is formed when there is a meeting point between rays of light coming from an object after reflection. An example is the image seen on the cinema screen.

A virtual image refers to an image that is formed when there is no actual meeting point between rays of light coming from an object but they appear to meet when produced backwards. An example is the image seen in the mirror when standing in front of the mirror.

Musya8 [376]3 years ago

6 0

D. can be viewed on a screen

100% sure

hope it helps!

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In this exercise, use the following variables : i,lo, hi, and result. Assume that lo and hi each are associated with an int and

Assoli18 [71]

Answer:

result = 0

i = lo

while i <= hi:

result = result + i

i += 1

Explanation:

Initialize the result as 0 to hold the summation value.

Since we are asked not to change the value of lo and hi, our loop control variable is i and initially it starts from lo.

Since we are asked to add the number from lo to hi, while loop condition checks it.

While the condition satisfies (during each iteration), the value of i is added to the result and the value of i is incremented by one.

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3 years ago

Why might people feel differently about their digital lives?

denpristay [2]

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3 years ago

Your traffic light changes to yellow as you approach an intersection. In most cases, what action should you take?

mamaluj [8]

Make every reasonable effort to stop.

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3 years ago

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Clive wants to write a query that will display the names of the students who scored less than 10 in their final exams. Which num

Mice21 [21]

I'd go for FLOOR(x)

The numerical function FLOOR(x) can be used to return the largest integer value that is less than or equal to the numerical expression provided. The numerical function CEILING (x) is the opposite of FLOOR(x) since it gives the smallest integer value that is greater or equal to the numerical expression.

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3 years ago

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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 }..

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

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