A is wrong.
B is correct.
C is true but magnetic monopoles have been predicted in theory, but never found in practice.
Answer:
(11n-5) / 12 is correct answer.
Explanation:
The Probability that key will match to nth term = 1/2
The Probability that key will match to n-1th term = 1/3
As all other probabilities are equal
The Total Probability that key matches to any of 1 to n-2 index = 1 - 1/2 - 1/3 = 1/6
The Probability that key matches to any of 1 to n-2 index = (1/6) / n-2 = (1/6)* (n-2))
Let P(i) = Probability that key matches to ith index.
The Average time complexity = 22 i=1 P(i) * i
The Average time complexity = 1/(6(n-2) * ( sum of 1 to n-2 ) + (n-1) / 3 + n/2
The Average time complexity = 1/(6(n-2) * ( n-2)*(n-1) / 2 + ( n-1) / 3 + n/2
The Average time complexity = 1/6 * (n-1)/2 + (n-1)/3 + n/2
The Average time complexity = (n-1)/12 + (n-1)/3 + n/2
The Average time complexity = (n-1 + 4 * n - 4 * 1 +6 * n)/12
The Average time complexity = 11n-5 / 12
so (11n-5) / 12 is correct answer.
Answer:
the answer is priming read
Explanation:
hope it helps u
Answer:
// here is code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n,no_open=0;
cout<<"enter the number of lockers:";
// read the number of lockers
cin>>n;
// initialize all lockers with 0, 0 for locked and 1 for open
int lock[n]={};
// toggle the locks
// in each pass toggle every ith lock
// if open close it and vice versa
for(int i=1;i<=n;i++)
{
for(int a=0;a<n;a++)
{
if((a+1)%i==0)
{
if(lock[a]==0)
lock[a]=1;
else if(lock[a]==1)
lock[a]=0;
}
}
}
cout<<"After last pass status of all locks:"<<endl;
// print the status of all locks
for(int x=0;x<n;x++)
{
if(lock[x]==0)
{
cout<<"lock "<<x+1<<" is close."<<endl;
}
else if(lock[x]==1)
{
cout<<"lock "<<x+1<<" is open."<<endl;
// count the open locks
no_open++;
}
}
// print the open locks
cout<<"total open locks are :"<<no_open<<endl;
return 0;
}
Explanation:
First read the number of lockers from user.Create an array of size n, and make all the locks closed.Then run a for loop to toggle locks.In pass i, toggle every ith lock.If lock is open then close it and vice versa.After the last pass print the status of each lock and print count of open locks.
Output:
enter the number of lockers:9
After last pass status of all locks:
lock 1 is open.
lock 2 is close.
lock 3 is close.
lock 4 is open.
lock 5 is close.
lock 6 is close.
lock 7 is close.
lock 8 is close.
lock 9 is open.
total open locks are :3