<span>Disconnect the host from the network.
</span><u>if that doesnt work then try these options</u>
Check the host hard drive for errors and file system issues.
<span>Examine the Device Manager on the host for device conflicts
</span>Unseat and then reconnect the hard drive connectors on the host.
Answer:
C. an example of open-source software.
Explanation:
open-source software is the type of software in which anyone can access, it can also be shared And modified by anyone simply because ita accessible to the public.
Hence and open source software's source code can be
inspected, enhanced and modified by anyone. A typical example is Linux.
<span>The correct answer is higher for both blank spaces.
We all know the famous saying: "No risk, no reward". What is true is the higher your risk you also have a higher degree of reaping a higher rewards. But the opposite is also true, the more you risk the more you stand to lose. In stockbroker business this is best exemplified, as you can se brokers trying to predict the stock market in order to make greater profits. Gambling is also the good example of this. </span>
Answer:
Explanation:
#include<iostream>
#include<ctime>
#include<bits/stdc++.h>
using namespace std;
double calculate(double arr[], int l)
{
double avg=0.0;
int x;
for(x=0;x<l;x++)
{
avg+=arr[x];
}
avg/=l;
return avg;
}
int biggest(int arr[], int n)
{
int x,idx,big=-1;
for(x=0;x<n;x++)
{
if(arr[x]>big)
{
big=arr[x];
idx=x;
}
}
return idx;
}
int main()
{
vector<pair<int,double> >result;
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
int choice;
cin>>choice;
while(choice!=2)
{
int n,m;
cout<<"Enter N"<<endl;
cin>>n;
cout<<"Enter M"<<endl;
cin>>m;
int c=m;
double running_time[c];
while(c>0)
{
int arr[n];
int x;
for(x=0;x<n;x++)
{
arr[x] = rand();
}
clock_t start = clock();
int pos = biggest(arr,n);
clock_t t_end = clock();
c--;
running_time[c] = 1000.0*(t_end-start)/CLOCKS_PER_SEC;
}
double avg_running_time = calculate(running_time,m);
result.push_back(make_pair(n,avg_running_time));
cout<<"Enter 1 for iteration\nEnter 2 for exit\n";
cin>>choice;
}
for(int x=0;x<result.size();x++)
{
cout<<result[x].first<<" "<<result[x].second<<endl;
}
}
Answer:
Output:
15
20
25
Explanation:
In first iteration value of num is 10. Condition is checked 10 is less than 21 so value of num is incremented by 5 and value 15 is printed than again condition is checked 15<21 so value of num is incremented again and 20 is printed. Again condition is checked 20<21. So 25 is printed. Then 4th time when condition is checked vakue of num is 25 and while loop condition becomes false because 25 is not less than 21 and program is terminated here.
