Answer:
The correct answer to the following question will be Error-detection.
Explanation:
Error-detection: The detection of errors caused during the transmission from the transmitter to the receiver by damage and other noises, known as Error-detection. This error-detection has the ability to resolute if something went wrong and if any error occurs in the program.
There are mainly three types of error-detection, these types can be followed:
- Automatic Repeat Request (ARQ)
- Forward Error Correction
- Hybrid Schemes
There are two methods for error-detection, such as:
- Single parity check
- Two-dimensional parity check
Answer:
The answer is A
Explanation:
Basically, Shortest job first (SJF) is a scheduling policy that selects jobs on queue for execution within a short execution time.
From the definition of SJF above, it means that there are a lot of process on queue and the (SJF) job is to receive processes on queue to execute within a short execution time.
Therefore, if all the jobs or process arrives at the SJF at the same time, the SJF will forfeit one of its major purpose which is scheduling of jobs.
That will therefore not make Non-preemptive Shortest Job First (SJF) not function at its optimal point.
The simulation, player 2 will always play according to the same strategy.
Method getPlayer2Move below is completed by assigning the correct value to result to be returned.
Explanation:
- You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.
#include <bits/stdc++.h>
using namespace std;
bool getplayer2move(int x, int y, int n)
{
int dp[n + 1];
dp[0] = false;
dp[1] = true;
for (int i = 2; i <= n; i++) {
if (i - 1 >= 0 and !dp[i - 1])
dp[i] = true;
else if (i - x >= 0 and !dp[i - x])
dp[i] = true;
else if (i - y >= 0 and !dp[i - y])
dp[i] = true;
else
dp[i] = false;
}
return dp[n];
}
int main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
cout << 'A';
else
cout << 'B';
return 0;
}
<span>Server virtualization in windows server 2012 r2 is based on a module called the</span> hypervisor.
Answer:
class Simple{
public static void main(String args[]){
System.out.println("Hello Daddy and Mum);
}
}
Explanation:
First, we create a class, then a method and then give the Integrated Data Environment (IDE) the command to give out an output that says Hello, Daddy and Mum”
