False, when you make a graph you can make many changes to it. It doesn't matter what kind of graph or where you make the graph on most applications you have to the option to make it exactly how you want it or change it after you have created it.
Answer:
<u> FlashCard.java</u>
- public class FlashCard {
- String Question;
- String Answer;
-
- public FlashCard(String q, String a){
- this.Question = q;
- this.Answer = a;
- }
-
- public String toString(){
- String output = "";
- output += "Question: " + this.Question + "\n";
- output += "Answer: " + this.Answer;
- return output;
- }
-
- public boolean equals(String response){
- if(this.Answer.equals(response)){
- return true;
- }
- else{
- return false;
- }
- }
- }
<u>Main.java</u>
- public class Main {
- public static void main(String[] args) {
- FlashCard card1 = new FlashCard("What is highest mountain?", "Everest");
- FlashCard card2 = new FlashCard("What is natural satelite of earth?", "Moon");
- FlashCard card3 = new FlashCard("Who is the first president of US?", "George Washington");
- FlashCard cards [] = {card1, card2, card3};
-
- for(int i=0; i < cards.length; i++){
- System.out.println(cards[i]);
- }
- }
- }
Explanation:
In FlashCard.java, we create a FlashCard class with two instance variable, Question and Answer (Line 2 - 3). There is a constructor that takes two input strings to initialize the Question and Answer instance variables (Line 5-8). There is also a toString method that will return the predefined output string of question answer (Line 10 - 15). And also another equals method that will take an input string and check against with the Answer using string equals method. If matched, return True (Line 17 -24).
In Main.java, create three FlashCard object (Line 3-5) and then put them into an array (Line 6). Use a for loop to print the object (Line 8-10). The sample output is as follows:
Question: What is highest mountain?
Answer: Everest
Question: What is natural satelite of earth?
Answer: Moon
Question: Who is the first president of US?
Answer: George Washington
Answer:
Explanation:
The following is written in Java. It continues asking the user for inputs until they enter a -1. Then it saves all the values into an array and calculates the number of values entered, the highest, and lowest, and prints all the variables to the screen. The code was tested and the output can be seen in the attached image below.
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Scanner;
class Brainly {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int count = 0;
int highest, lowest;
ArrayList<Integer> myArr = new ArrayList<>();
while (true) {
System.out.println("Enter a number [0-10] or -1 to exit");
int num = in.nextInt();
if (num != -1) {
if ((num >= 0) && (num <= 10)) {
count+= 1;
myArr.add(num);
} else {
System.out.println("Wrong Value");
}
} else {
break;
}
}
if (myArr.size() > 3) {
highest = myArr.get(0);
lowest = myArr.get(0);
for (int x: myArr) {
if (x > highest) {
highest = x;
}
if (x < lowest) {
lowest = x;
}
}
System.out.println("Number of Elements: " + count);
System.out.println("Highest: " + highest);
System.out.println("Lowest : " + lowest);
} else {
System.out.println("Number of Elements: " + count);
System.out.println("No Highest or Lowest Elements");
}
}
}
Answer:
class Main {
public static void main(String args[]) {
int a = 5;
int delta = 5;
for(int i=0; i<10; i++) {
System.out.printf("%d, ", a);
a += delta;
delta += 2;
}
}
}
Answer:
Communication path basically define the path in which the information and messages can be exchange by using the efficient communication path.
There are simple formula for calculating the total number of communication channel that is :
Where, n is the number of stack holder.
Now, the maximum number of communication paths for a team of twenty people can be calculated as:
n=20
=\frac{20(20-1)}{2} = 190
Therefore, 190 is the total number of communication path.
