I have a question about my HP Laptop

What are the challenges of Cyber Law???? Needed ASAP

lilavasa [31]

Answer:

Explanation:

Few challenges that the technology space faces in cyber security are the following:

Digital Data Threat: Growing online transactions generate bigger incentives for cybercriminals. Besides, establishments looking to mine data—for instance, customer information, results of product surveys, and generic market information—create treasured intellectual property that is in itself an attractive target.

Supply Chain Inter-connection: The supply chains are increasingly interconnected. Companies are urging vendors and customers to join their networks. This makes a company’s security wall thin.

Hacking: This action is penetrating into someone’s system in unauthorized fashion to steal or destroy data, which has grown hundred folds in the past few years. The availability of information online makes it easier for even non-technical people to perform hacking.

Phishing: The easiest to execute and can produce the results with very little effort. It is the act of sending out Fake emails, text messages and create websites to look like they're from authentic companies.

6 0

3 years ago

In 3 to 5 sentences, describe whether or not files should be deleted from your computer. Explain you answer.

daser333 [38]

I mean if your trying to delete like important files I dont think they should be deleted but that is My opinion. But if you do delete them then u could always got to the recycle bin in ur computer to get it back. I hope this helps !!

7 0

3 years ago

Read 2 more answers

The ratio of boys and girls is 4:5 calculate the number of boys if the are 1080 learners in schools

inna [77]

Explanation:

The ratio of Boys = 4x

The ratio of Girls = 5x

The Number of Boys in the School = 4x + 5x = 1080

= 9x = 1080

= x = 1080/9

= x = 120

The ratio of Boys = 4x = 4 x 120 = 480

The ratio of Girls = 5x = 5 x 120 = 600

Answer :- The ratio of Boys =480

8 0

2 years ago

Convert one billion byte into one storage unit

notka56 [123]

Answer:

Explanation:

To convert larger units to smaller units (i.e. take a number of gigabytes and convert it down in to megabytes, kilobytes, or bytes) you simply multiply the original number by 1,024 for each unit size along the way to the final desired unit

7 0

2 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