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Leno4ka [110]
4 years ago
12

Write HTML code for inserting an image "cricket.jpeg" in size 500 width and 400 height.

Computers and Technology
1 answer:
Rainbow [258]4 years ago
5 0

Answer:

<img src="cricket.jpeg" style="width: 500px; height:400px">

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A small company with 100 computers has hired you to install a local area network. All of the users perform functions like email,
charle [14.2K]

Answer:

Answer to the following question is as follows;

Explanation:

Windows 7 for 100 users and Windows 8.1 for 25 users are the best options since they both enable networking and active directory, and Windows 8.1 also offers Windows server capabilities.

Winxp would therefore work for $100, but it is unsupported and has no updates.

If you wish to go with open source, you can choose Ubuntu 16 or 18 or Linux.

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
HURRRYYYY PLZZ!!
vivado [14]

Answer: A. How much is this vehicle's resale value?

6 0
3 years ago
Read 2 more answers
The process of redefining the functionality of a built-in operator, such as , -, and *, to operate on programmer-defined objects
Ksenya-84 [330]

It should be noted that the process of redefining the functionality of a built-in operator to operate is known as <u>operator overloading</u>.

Operator overloading simply means polymorphism. It's a manner in which the operating system allows the same operator name to be used for different operations.

Operator overloading allows the operator symbols to be bound to more than one implementation. It's vital in redefining the functionality of a built-in operator to operate on programmer-defined objects.

Read related link on:

brainly.com/question/25487186

3 0
3 years ago
In a school 50% of the students are younger than 10, 1/20 are 10 years old and 1/10 are older than 10 but younger than 12, the r
Zielflug [23.3K]

Answer: 10 students

Explanation:

Students younger than 10 = 50%

Students that are 10years old = 1/20 = 1/20 × 100 = 5%

Students that are older than 10 but younger than 12 = 1/10= 1/10 × 100 = 10%

Students that are 12 years or older

= 100% - (50% + 5% + 10%)

= 100% - 65%

= 35%

This means that 35% of the students are 12 years or older and we've been given the number as 70.

Let's say the total number of students is x. Therefore,

35% of x = 70

0.35 × x = 70

0.35x = 70

x = 70/0.35

x = 200

The total number of students is 200.

Therefore, the number of students that are 10years will be:

= 1/20 × 200

= 10 students

Therefore, 10 students are 10 years.

3 0
3 years ago
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