Answer:
ICMP.
Explanation:
Hackers might use ICMP for gain knowledge about current system addresses and specific procedures as part of a reconnaissance process.
ICMP is a failure informing system used by network protocols such as routers to produce warning to that of the source Address while network issues interrupt IP packet transmission.
Answer:
//class Name
class Name {
// attributes gave you 3
String first_name;
String last_name;
int length = 0;
//constructor with 2 parameters
Name(String first_name, String last_name){
}
//methods
public String find_printed_name(){
return first_name + " " + last_name;
}
public void find_sortable_name(){
return last_name + ", " first_name.charAt(0);
}
public static void main(String[] args){
// instantiate the class
Name test_name = new Name("David", "Joyner");
System.out.println(test_name.first_name);
System.out.println(test_name.last_name);
System.out.println(test_name.find_printed_name());
System.out.println(test_name.find_sortable_name());
}
Explanation:
You didn't specify a language so I did it in Java.
You should know how to declare methods in Python.
Answer:
#include <iostream>
using namespace std;
void PrintPopcornTime(int bagOunces) {
if(bagOunces < 2)
cout << "Too small" << endl;
else if(bagOunces > 10)
cout << "Too large" << endl;
else
cout << 6 * bagOunces <<" seconds" << endl;
}
int main()
{
int userOunces;
cin >> userOunces;
PrintPopcornTime(userOunces);
return 0;
}
Explanation:
Inside the function PrintPopcornTime, check the parameter bagOunces using if structure. Depending on the value of bagOunces, print the required information.
Inside the main, ask the user for the userOunces. Call the PrintPopcornTime function with the userOunces.
Answer:
Answer is B, see explanations.
Explanation:
Answer is (B) NP−complete∩P=ϕ
Since, P≠NP, there is at least one problem in NP, which is harder than all P problems. Lets take the hardest such problem, say X. Since, P≠NP,X∉P .
Now, by definition, NP−complete problems are the hardest problems in NP and so X problem is in NP−complete. And being in NP, X can be reduced to all problems in NP−complete, making any other NP−complete problem as hard as X. So, since X∉P, none of the other NP−complete problems also cannot be in P.