Answer:
The answer is B hierchary protocal
Answer:
import java.util.Scanner;
public class TeenagerDetector {
public static void main (String [] args) {
Scanner scnr = new Scanner(System.in);
boolean isTeenager;
int kidAge;
kidAge = scnr.nextInt();
/* Your solution goes here */
isTeenager = (kidAge >= 13) && (kidAge <= 19);
if (isTeenager) {
System.out.println("Teen");
} else { System.out.println("Not teen"); } } }
Explanation:
A condition which check for the teenager age and return a boolean is assigned to isTeenager.
isTeenager = (kidAge >=13) && (kidAge <= 19);
So, if the kidAge is greater than/equal to 13 and less than/19, the boolean isTeenager will be true and the program will output "Teen" else "false" will be output.
The range of age for a teenager is 13 - 19.
Answer: So then every thing is not cluster together and you have everything where you can get to it faster and it just makes everything 100 times better then it not being organized :) hope this helped and if it didn't I'm so sorry
Explanation:
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Answer:
Pseudocode
////////////////////////////////////////////////////////////////////////////////////////////////////////////
Integer netElevation(list of elements of type elevation - type and number)
<em>function open</em>
Define running total = 0
for each element from list
<em>loop open</em>
elevation type = element[i].type
if (elevation type == Up)
running total = running total + element[i].number
else
running total = running total - element[i].number
<em>loop close</em>
return running total
<em>function close</em>