What is printed by the following code segment? doThus(3, 1); Given the method definitions: public static int doThat(int x, int y) { System.out.println("that " + x + " " + y); x = x + y; System.out.println("that " + x + " " + y); return x; } public static int doThus(int x, int y) { System.out.println("thus " + x + " " + y); y = doThat(y + x, x - y); System.out.println("thus " + x + " " + y); return y; }

Consider the following method: public static void method (int x, int y) { if (x < y) { method (x+1, y); System.out.print (x + " "); } } What will happen as the result of calling method (2,6)? (Select all correct answers) output will be 6 5 4 3 2 Program goes to an infinite loop because of errors in recursive calls 5 recursive calls will be made from method (3,6) to method (7,6) output will be 5 4 3 2 method (6, 6) will be called inside of the recursion 4 recursive calls will be made from method (3,6) to method (6,6) method (5, 6) will be called inside of the recursion

Define the method printValue() that takes two integer parameters and outputs the product of all integers starting with the first and ending with the second parameter, followed by a newline. The method does not return any value. Ex: If the input is 3 7, then the output is: 2520 Note: Assume the first integer parameter is less than the second. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16     import java.util.Scanner;   public class ValuePrinter {   /* Your code goes here */   publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); intnumber1; intnumber2;   number1=scnr.nextInt(); number2=scnr.nextInt(); printValue(number1, number2); } }

Consider the window size is 10, bandwidth is 1500 bps, transmission delay is 2 ms and propagation delay is 50 ms. What is the throughput using Go Back N protocol? a) 294.11 b) 140.23 c) 96 d) 154.21