Answer:
1G
Explanation:
1 'G' as in first generation.
Hope this helped. :)
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:
A = 120
B = 40
C = 70
Solution:
As per the question:
Manufacturer forced to make 10 more type C clamps than the total of A and b:
10 + A + B = C (1)
Also, 3 times as many type B as type A clamps are:
A = 3B (2)
The total no. of clamps produced per day:
A + B + C = 330 (3)
The no. of each type manufactured per day:
Now, from eqn (1), and (3):
A + B + 10 + A + B = 330
2A + 2B = 320
A + B = 160 (4)
Now, from eqn (2) and (4):
3B + B = 160
B = 40
Since, A = 3B
A = 
A = 120
Put the values of A and C in eqn (3):
120 + 40 + C = 330
C = 70
Answer:
apt-get
Explanation:
apt-get can be used to manually install a package, without need to manually install all the dependencies for the package.
