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3 years ago

What information is required for a complete citation of a website source?

Computers and Technology

2 answers:

Llana [10]3 years ago

7 0

Answer:

The correct answer is C) the responsible person or organization, date accessed, and URL.

Explanation:

Reference making is always dependant on the rules and patterns the institutions, journals, magazines, among others, are having as their main source pattern. However, most of them have in common the pattern explicit in letter C, for they all expect to have sources with detailed informations about the responsible person or organization for the website, the date of the last access to the website, and the URL for the website. All other options are lacking in some way or are not specifically about website source references.

Vlada [557]3 years ago

5 0

The correct answer is D!

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Write an application that displays every perfect number from 1 through 1,000. A perfect number is one that equals the sum of all

matrenka [14]

Answer:

Written in Python

for num in range(1,1001):

sum=0

for j in range(1,num+1):

if num%j==0:

sum = sum + j

if num == sum:

print(str(num)+" is a perfect number")

Explanation:

This line gets the range from 1 to 1000

for num in range(1,1001):

This line initializes sum to 0 for each number 1 to 1000

sum=0

This line gets the divisor for each number 1 to 1000

for j in range(1,num+1):

This following if condition line checks for divisor of each number

The following if condition checks for perfect number

if num == sum:

print(str(num)+" is a perfect number")

5 0

3 years ago

Place a check next to each of the choices that describe an example of proper ergonomics. (more than 1 answer)

alexandr1967 [171]

I think it’s c and d not 100% sure though

4 0

2 years ago

Read 2 more answers

Write a recursive program that requests an answer to the question "Are we there yet?" using an input statement and terminates if

inna [77]

Answer:

def recursive_func():

x = input("Are we there yet?")

if x.casefold() == 'Yes'.casefold():

return

else:

recursive_func()

Explanation:

We define the required function as recursive_func().

The first line takes user input. The user input is stored in variable x.

The next line compares the user input to a string yes. The function executes the else block if the condition isn't met, that is a recursive call is executed.

IF condition returns the function. The string in variable X is compared to a string 'Yes'. the casefold() is a string function that ignores the upper/lower cases when comparing two strings. (This is important because a string 'yes' is not the same yes a string 'Yes' or 'YES'. Two equal strings means their cases and length should match).

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3 years ago

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Assoli18 [71]

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3 years ago

Let's implement a classic algorithm: binary search on an array. Implement a class named BinarySearcher that provides one static

yKpoI14uk [10]

Answer:

Hope this helped you, and if it did , do consider giving brainliest.

Explanation:

import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

// main method

public static void main(String[] args) {

// create a list of Comparable type

List<Comparable> list = new ArrayList<>();

// add elements

list.add(1);

list.add(2);

list.add(3);

list.add(4);

list.add(5);

list.add(6);

list.add(7);

// print list

System.out.println("\nList : "+list);

// test search method

Comparable a = 7;

System.out.println("\nSearch for 7 : "+search(list,a));

Comparable b = 3;

System.out.println("\nSearch for 3 : "+search(list,b));

Comparable c = 9;

System.out.println("\nSearch for 9 : "+search(list,c));

Comparable d = 1;

System.out.println("\nSearch for 1 : "+search(list,d));

Comparable e = 12;

System.out.println("\nSearch for 12 : "+search(list,e));

Comparable f = 0;

System.out.println("\nSearch for 0 : "+search(list,f));

}

// static method named search takes arguments Comparable list and Comparable parameter

public static boolean search(List<Comparable> list, Comparable par) {

// if list is empty or parameter is null the throw IllegalArgumentException

if(list.isEmpty() || par == null ) {

throw new IllegalArgumentException();

}

// binary search

// declare variables

int start=0;

int end =list.size()-1;

// using while loop

while(start<=end) {

// mid element

int mid =(start+end)/2;

// if par equal to mid element then return

if(list.get(mid).equals(par) )

{

return true ;

}

// if mid is less than parameter

else if (list.get(mid).compareTo(par) < 0 ) {

start=mid+1;

}

// if mid is greater than parameter

else {

end=mid-1;

}

// if not found then retuen false

return false;

}

}import java.util.ArrayList;

import java.util.List;

//classs named BinarySearcher

public class BinarySearcher {

// main method

public static void main(String[] args) {

// create a list of Comparable type

List<Comparable> list = new ArrayList<>();

// add elements

list.add(1);

list.add(2);

list.add(3);

list.add(4);

list.add(5);

list.add(6);

list.add(7);

// print list

System.out.println("\nList : "+list);

// test search method

Comparable a = 7;

System.out.println("\nSearch for 7 : "+search(list,a));

Comparable b = 3;

System.out.println("\nSearch for 3 : "+search(list,b));

Comparable c = 9;

System.out.println("\nSearch for 9 : "+search(list,c));

Comparable d = 1;

System.out.println("\nSearch for 1 : "+search(list,d));

Comparable e = 12;

System.out.println("\nSearch for 12 : "+search(list,e));

Comparable f = 0;

System.out.println("\nSearch for 0 : "+search(list,f));

}

// static method named search takes arguments Comparable list and Comparable parameter

public static boolean search(List<Comparable> list, Comparable par) {

// if list is empty or parameter is null the throw IllegalArgumentException

if(list.isEmpty() || par == null ) {

throw new IllegalArgumentException();

}

// binary search

// declare variables

int start=0;

int end =list.size()-1;

// using while loop

while(start<=end) {

// mid element

int mid =(start+end)/2;

// if par equal to mid element then return

if(list.get(mid).equals(par) )

{

return true ;

}

// if mid is less than parameter

else if (list.get(mid).compareTo(par) < 0 ) {

start=mid+1;

}

// if mid is greater than parameter

else {

end=mid-1;

}

// if not found then retuen false

return false;

}

7 0

3 years ago