Eased the mobilization and transportation of troops
Answer:
The correct answer to the following question will be "Data-in-use".
Explanation:
- Data-in-use is an IT term referring to active information that is usually preserved in a semi-persistent physical state in RAM of computer, CPU registers or caches.
- It might be created, modified or changed, deleted or accessed via different endpoints of the interface. This is indeed a useful term for IT departments to pursue institutional defense.
Therefore, it's the right answer.
Answer:
Following are the response to the given question:
Explanation:
The glamorous objective is to examine the items (as being the most valuable and "cheapest" items are chosen) while no item is selectable - in other words, the loading can be reached.
Assume that such a strategy also isn't optimum, this is that there is the set of items not including one of the selfish strategy items (say, i-th item), but instead a heavy, less valuable item j, with j > i and is optimal.
As 
, the i-th item may be substituted by the j-th item, as well as the overall load is still sustainable. Moreover, because 
and this strategy is better, our total profit has dropped. Contradiction.
Please Help! Unit 6: Lesson 1 - Coding Activity 2
Instructions: Hemachandra numbers (more commonly known as Fibonacci numbers) are found by starting with two numbers then finding the next number by adding the previous two numbers together. The most common starting numbers are 0 and 1 giving the numbers 0, 1, 1, 2, 3, 5...
The main method from this class contains code which is intended to fill an array of length 10 with these Hemachandra numbers, then print the value of the number in the array at the index entered by the user. For example if the user inputs 3 then the program should output 2, while if the user inputs 6 then the program should output 8. Debug this code so it works as intended.
The Code Given:
import java.util.Scanner;
public class U6_L1_Activity_Two{
public static void main(String[] args){
int[h] = new int[10];
0 = h[0];
1 = h[1];
h[2] = h[0] + h[1];
h[3] = h[1] + h[2];
h[4] = h[2] + h[3];
h[5] = h[3] + h[4];
h[6] = h[4] + h[5];
h[7] = h[5] + h[6];
h[8] = h[6] + h[7]
h[9] = h[7] + h[8];
h[10] = h[8] + h[9];
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
if (i >= 0 && i < 10)
System.out.println(h(i));
}
}
