PLEASE HELP POWER POINT:

Veronika [31]

Answer: workbook

Explanation:
A workbook is the file that holds several worksheets.
The worksheets within a workbook can communicate with each other within the workbook or with other workbooks and their worksheets.
A worksheet within a workbook holds cells that are defined by rows and columns. Worksheets within a workbook can removed and new worksheets can be added.

8 0

2 years ago

Which method do software testers use to isolate new code from the rest of the network during the test stage of the software deve

disa [49]

Answer:

Sandboxing.

Explanation:

SDLC or software development life cycle is a systematic approach to software development. It marks the birth and death of an application.

The application development starts from the research of the properties of the application to the design and development or implementation of the application.

After the implementation of the software, it is tested, comparing the design and the codes in the development process. The sandboxing testing process helps to isolate and test new lines of code added during the testing phase.

3 0

2 years ago

The energy used by an appliance which operates at 240 volts at 15 amp for 4 hr. is A. 0.92 kwhr. B. 3.45 kwhr. C. 14.4 kwhr. D.

Tju [1.3M]

Answer:

C. 14.4 kwhr.

Explanation:

To calculate the energy used by appliance first calculate the power. The formulae for calculating power(<em>P</em>) is $P= I \times V$ , where V stands for voltage and I stands for current.

In the given problem, V = 240 volts and I = 15 amp. Therefore,

$P=240 \times 15 =3600\ watt(w) = 3.6\ kw$

Now, formulae for calculating energy used i.e. work(W) is $W = P\times t$ , where <em>t</em> is the time.

In the given question time is 4 hr. Therefore energy used will be,

Energy used (W) = $3.6\times4 =14.4\ kwhr$ .

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

2 years ago

To exclude members of a group from the basic permissions for a folder, which type of permission would you assign? Deny Allow Mod

oksano4ka [1.4K]

Answer:

Deny

Explanation:

Under the New Technology System, the Deny permission is applied when the administrator wants to overrule the permission given to members of a group. The Allow and Deny options help in regulating access to the components of the system. Only authorized groups are granted access to the files.

Most times, the Deny permission is considered before the Allow permission. If the user is denied access to some files and granted access to some, the deny permission is most times considered first.

4 0

2 years ago