Is it called a, "server".
I will use 0b to refer to binary numbers: ex: 0b1 = 1, 0b10 = 2
0b100 = 4 and 0b11 = 3
4+3 = 7
convert 7 to binary and you get 111.
Hope this helps, and May the Force Be With You!
<span>-Jabba</span>
Answer:
b. 2.9!
Explanation:
There are is a mistake in the question.
Suppose the group consist of 10 kids and 2 adults, the number of ways in which they can form the line is:
= 2! 10!
= 2× 1× 10!
= 2.10!
But since that is not in the given option.
Let assume that the group consists of 9 kids and 2 adults, the number of ways in which they can form the line is:
No of ways the kids can be permutated = 9 ways
No of ways the adult can be permutated = two ways.
Thus; the number of ways in which they can form the line = 2! 9!
= 2 × 1× 9!
= 2.9!
Answer: Your history and you can check everything using a MAC or IP password
Answer:
while(userNum>=1){
System.out.print(userNum/2+" ");
userNum--;
}
Explanation:
This is implemented in Java programming language. Below is a complete code which prompts a user for the number, receives and stores this number in the variable userNum.
<em>import java.util.Scanner;</em>
<em>public class TestClock {</em>
<em> public static void main(String[] args) {</em>
<em> Scanner in = new Scanner (System.in);</em>
<em> System.out.println("Enter the number");</em>
<em> int userNum = in.nextInt();</em>
<em> while(userNum>=1){</em>
<em> System.out.print(userNum/2+" ");</em>
<em> userNum--;</em>
<em> }</em>
<em> }</em>
<em>}</em>
The condition for the while statement is userNum>=1 and after each iteration we subtract 1 from the value of userNum until reaching 1 (Hence userNum>=1)
