A user complains that there are not enough USB ports available on a desktop. What can a technician do to solve the problem?

Interest on loan paid to sangam stores Rs5000(journal entry)

ratelena [41]

I’m confused... whats the question?

6 0

3 years ago

Suppose that you have the following declaration:

masha68 [24]

Answer:

Explanation:

The code is written in C++:

#include <iostream>

#include <stack>

#include <math.h>

#include <iomanip>

using namespace std;

/*

* Supporting method to print contents of a stack.

*/

void print(stack<double> &s)

{

if(s.empty())

{

cout << endl;

return;

}

double x= s.top();

s.pop();

print(s);

s.push(x);

cout << x << " ";

}

int main(){

// Declaration of the stack variable

stack<double> stack;

//rray with input values

double inputs[] = {25,64,-3,6.25,36,-4.5,86,14,-12,9};

/*

* For each element in the input, if it is positive push the square root into stack

* otherwise push the square into the stack

*/

for(int i=0;i<10;i++){

if(inputs[i]>=0){

stack.push(sqrt(inputs[i]));

}else{

stack.push(pow(inputs[i],2));

}

//Print thye content of the stack

print(stack);

}

OUTPUT:

5 8 9 2.5 6 20.25 9.27362 3.74166 144 3

-------------------------------------------------------------------------

Process exited after 0.01643 seconds with return value 0

Press any key to continue . . . -

6 0

3 years ago

Henri is working in a complex financial spreadsheet that has hundreds of columns of data. If he is at column AN and would like t

Alisiya [41]

Answer: Contro; + Home

This key combination will return you to the first row, first column of the current worksheet.

8 0

3 years ago

This program will convert a set of temperatures from Fahrenheit to Celsius and Kelvin. Your program will be reading in three dou

Naily [24]

Answer:

see explaination

Explanation:

#include<iostream>

#include<iomanip>

using namespace std;

int main()

{

double temp1,temp2,inc,cel;

int i=1;

while(i==1)

{

i=0;

cin>>temp1>>temp2>>inc;

if(temp2<temp1||inc<=0)

{

i=1;

cout<<"Starting temperature must be <= ending temperature and increment must be >0.0\n";

}

cout<<endl;

cout<<setw(15)<<"Fahrenheit"<<setw(15)<<"Celsius";

while(temp1<=temp2)

{

cel=(temp1-32)/1.8;

cout<<endl;

cout<<fixed<<setprecision(3)<<setw(15)<<temp1<<setw(15)<<cel;

temp1+=inc;

}

Please kindly check attachment for output.

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

3 years ago