Answer:
B. The cost of tour t is at most twice the cost of the optimal tour.
Explanation:
You are using a polynomial time 2-approximation algorithm to find a tour t for the traveling salesman problem.
The cost of tour t is at most twice the cost of the optimal tour
The equation represented as Cost(t) <= 2 Cost(T)
Where
Cost (t) represents cost of tour t
Cost(T) represents cost of the optimal tour
Answer:
Explanation:
There are various safety features in place to prevent such scenarios from happening. For starters phones usually have a pin code, pattern code, or fingerprint scanner which prevents unauthorized individuals from entering into the phone's services. Assuming that these features have been disabled by the phone's owner, payment applications usually require 2FA verification which requires two forms of approval such as email approval and fingerprint scanner in order for any transactions to go through. Therefore, an unauthorized individual would not have access to such features and would not be able to complete any transactions.
Answer:
The upgrade is possible and it will yield a remarkable increase in performance
Explanation:
It is a newer product, there is tendency of having a better application compatibility/performance
It has much higher multi threaded performance which is around 522% higher. This allows for higher performance in professional applications like encoding and heavy multitasking compared to the previous.
When considering gaming, it has higher performance compared to the previous.
Answer:
// A optimized school method based C++ program to check
// if a number is composite.
#include <bits/stdc++.h>
using namespace std;
bool isComposite(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return false;
// This is checked so that we can skip
// middle five numbers in below loop
if (n%2 == 0 || n%3 == 0) return true;
for (int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return true;
return false;
}
// Driver Program to test above function
int main()
{
isComposite(11)? cout << " true\n": cout << " false\n";
isComposite(15)? cout << " true\n": cout << " false\n";
return 0;
}
Explanation:
