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grin007 [14]
3 years ago
15

Andrew has data in cell E14 and the cell should be blank. Andrew should _____.

Computers and Technology
2 answers:
GuDViN [60]3 years ago
8 0

Answer is B: use the mouse to click on cell E14 and press Delete

Moving from one cell to cell until Andrew gets to cell E14 will be tedious and time wasting. Pressing enter while Andrew is on cell E13 will only jump to cell E14 and will not delete or remove content in that cell. Clicking E14 and pressing either the delete key or the backspace will clear any content that was on that cell.

denpristay [2]3 years ago
4 0
Use the mouse to click on cell E14 and press delete
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A hardware compatibility list recommends striping with double parity as storage for an application. In a test environment, a tec
riadik2000 [5.3K]

Answer:

RAID level 5 can be used to compensate a limited number of available disks.

Explanation:

There are two type of RAID

  • Software RAID
  • Hardware RAID

Software RAID

deliver services form the host.

Hardware RAID

provides hardware services.

RAID has levels

0, 1, 5, 6, and 10

RAID 0, 1, and 5 work on both HDD and SSD media,

4 and 6 also work on both media.

RAID 0 :Striping

In this level minimum of two disks,RAID 0 split the file strip the data.Multiple hard drive are used to split the data.

RAID 1 : Mirroring

In this level Minimum two disk require and provide data tendency.

RAID 5 :Stripping with parity

Parity is a binary data.RAID system calculate the value which system used to recover the data.

Most RAID system with parity function store parity blocks.

RAID 5 combines the performance of RAID 0 with redundancy of RAID 1.

RAID 5 level should minimize the fault tolerance.

4 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
When using bits to represent fractions of a number, can you make all possible fractions?
saw5 [17]
This is your perfect answer

8 0
3 years ago
When data are entered into a form and saved, they are placed in the underlying database as knowledge?
luda_lava [24]
The answer would be and is true.
7 0
3 years ago
The explicit location make the query easier to understand and interpret
Step2247 [10]

I am assuming this is a true or false question? If so, the answer is true.

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