Which piece of evidence from the article MOST

Researching the history of computers is a good troubleshooting technique. True or False?

Verdich [7]

Yes very much so! You could learn previous mistakes/bugs/etc and find solutions to fixing them and avoiding getting them. As well as much more. History is always a fun subject for anything, really in my opinion.

4 0

3 years ago

Which statement is most likely to be true about a computer network?

emmasim [6.3K]

Which statement is most likely to be true about a computer network?

A network can have several client computers and only one server.

4 0

3 years ago

Read 2 more answers

Two machines can finish a job in StartFraction 20 Over 9 EndFraction hours. Working alone, one machine would take one hour long

Alborosie

Answer

5 hours

Explanation

The two working together can finish a job in

$\frac{20}{9} \: hours$

Also, working alone, one machine would take one hour longer than the other to complete the same job.

Let the slower machine working alone take x hours. Then the faster machine takes x-1 hours to complete the same task working alone.

Their combined rate in terms of x is

$\frac{1}{x} + \frac{1}{x - 1}$

This should be equal to 20/9 hours.

$\frac{1}{x} + \frac{1}{x - 1} = \frac{9}{20}$

Multiply through by;

$20x(x - 1) \times \frac{1}{x} +20x(x - 1) \times \frac{1}{x - 1} = 20x(x - 1) \times \frac{9}{0}$

$20(x - 1) +20x = 9x(x - 1)$

$20x - 20+20x = 9{x}^{2} - 9x$

$9{x}^{2} - 9x - 20x - 20x + 20= 0$

$9{x}^{2} - 49x + 20= 0$

Factor to get:

$(9x - 4)(x - 5) = 0$

$x = \frac{4}{9} \: or \: x = 5$

It is not feasible for the slower machine to complete the work alone in 4/9 hours if the two will finish in 20/9 hours.

Therefore the slower finish in 5 hours.

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

4 years ago

Answers please !!!!!!!!!!!!!!!!!!!!!!!!!!!

alina1380 [7]

Hard to see please add a better image. Actually just type out the question.

4 0

4 years ago