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Katyanochek1 [597]
3 years ago
9

It would be at least two decades before some of the technologies he demonstrated saw widespread use, but all of them are used by

people today. Here is a list of some of Engelbart’s technologies. Which ones do you use today?
A. the computer mouse

B. video conferencing

C. hypertext links

D. word processing

F. collaborative real-time editing
Computers and Technology
1 answer:
Phoenix [80]3 years ago
7 0

Answer:

all of the above

Explanation:

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What is targets net worth in 2020 (i will see if you put in 2019)
Schach [20]

Answer:

62.6 billion

Explanation:

3 0
3 years ago
Match each method of communication with its intended purpose.
Scorpion4ik [409]

Answer:

1. a late breaking news story = to inform

2. a poetry reading = to entertain

3. an advertisement = to persuade

4. a small group assignment = to collaborate

Explanation:

1. A "breaking news" tells people of what is happening in the society. It <em>informs </em>them of the occurrence of an important event such as the plane crash of Kobe Bryant.

2. Poetry reading is meant to touch the attention of listeners. It tries to entertain them through the poem's interesting verses.

3. An advertisement is being shown/displayed in order to convince people to buy a particular product or service.

4. A group assignment allows the members of the group to contribute their ideas together. Such situation is known as "collaboration." They try to brainstorm together towards a common goal.

7 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
7. Write a program in C to display the string "ARRAY" in the following format : A AR ARR ARRA ARRAY
Art [367]
This program will the string array in the given format.

int main(void){
char arr[6];
int counter;
strcpy(arr[], "ARRAY");

for(counter=0; counter<6; counter++){
printf("%c", arr[counter]);
}

return 0;
)

Note: Do not forget to include all the necessary library that is needed to run this program.

6 0
3 years ago
You need to perform maintenance on a router and need to temporarily reroute traffic through another office. which would be the b
Whitepunk [10]
<span>In order to perform maintenance on a router and need to temporarily reroute traffic through another office  the best way to perform this action would be to configure a static route on the router.
</span>By doing this, the router will use the manually-<span>configured routing entry to send the packets.</span>
3 0
3 years ago
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