Answer:
Number of packets ≈ 5339
Explanation:
let
X = no of packets that is not erased.
P ( each packet getting erased ) = 0.8
P ( each packet not getting erased ) = 0.2
P ( X ≥ 1000 ) = 0.99
E(x) = n * 0.2
var ( x ) = n * 0.2 * 0.8
∴ Z = X - ( n * 0.2 ) / 
~ N ( 0.1 )
attached below is the remaining part of the solution
note : For the value of <em>n</em> take the positive number
import java.util.Scanner;
public class JavaApplication83 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Enter Strings: ");
String word1 = scan.nextLine();
String word2 = scan.nextLine();
String newWord = "";
if (word1.length() == word2.length()){
for (int i = 0; i < word1.length(); i++)
{
newWord += word1.charAt(i) +""+word2.charAt(i);
}
}
else{
newWord = "error";
}
System.out.println(newWord);
}
}
I hope this helps!
The word you are looking for is active.
Answer:
The company might not be able to satisfy their customers
