All categories

3 years ago

Are commonly used to control the number of times that a loop iterates?

1 answer:

5 0

Hmm... I feel like this query is much broader than it should be. However, I will start my initial answer, then another potential solution.

My initial answer to your query was: A condition controlled loop is used to control the number of times a loop iterates.

The potential answer, my secondary one, is: A count controlled loop iterates a specific number of times.

Two results, but my initial answer is the solution I opted when understanding this.

You might be interested in

Paraphin [41]

The answer is Document Inspector

5 0

2 years ago

In honor of Black History Month, write an essay about a black person who has inspired YOU in some way. This person does not have

babymother [125]

Answer:

I will be talking about Jackie Robinson. Robinson was known for playing baseball with the Brooklyn Dodgers from 1947 to 1956, I will share a few of his other accomplishment.

In 1935, Robinson graduated from Washington Junior High School and enrolled at John Muir High School Recognizing his athletic talents, Robinson's older brothers Mack (himself, an accomplished athlete and silver medalist, at the 1936 Summer Olympics) and Frank inspired Jackie to pursue his interest in sports.

At Muir high school , Robinson played several sports , football, basketball, track, baseball and tennis. He played shortstop and catcher on the baseball team, quarterback on the football team, and guard on the basketball team. With the track and field squad, he won awards for the broad jump.

In 1936, Robinson won the junior boys singles championship at the annual Pacific Coast Negro Tennis Tournament.

He earned a place on the annual baseball tournament all-star team,

In late January 1937, the News newspaper reported that Robinson "for two years had been the outstanding athlete at John Muir, starring in football, basketball, track, baseball and tennis.

As Jackie was encouraged to move forward by his brothers Mack and Frank. Today, his accomplishments inspires me to keep moving forward.

Explanation:

3 0

3 years ago

Excel solver does not allow for the inclusion of greater than or equal to

GuDViN [60]

Answer:

True

Explanation:

Excel Solver does not allow for the inclusion of greater than or equal to, less than or equal to or equal nomenclature as Excel will not recognize these.

6 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 }..

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

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

Print numbers 0, 1, 2, ..., userNum as shown, with each number indented by that number of spaces. For each printed line, print t

geniusboy [140]

Answer:

Please find the answer below

Explanation:

// Online C compiler to run C program online

#include <stdio.h>

int main() {

// Write C code here

//printf("Hello world");

int userNum;

int i;

int j;

scanf("%d", &userNum);

/* Your solution goes here */

for(i = 0; i<=userNum; i++){

for(j = 0; j <= i; j++){

printf(" ");

}

printf("%d\n", i);

}

return 0;

}

8 0

2 years ago