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;
}
}
"Key logger" This could be software or hardware that does this.
1. The current is the same everywhere in the circuit. This means that wherever I try to measure
the current, I will obtain the same reading.
2. Each component has an individual Ohm's law Voltage Drop. This means that I can calculate
the voltage using Ohm's Law if I know the current through the component and the resistance.
3. Kirchoff's Voltage Law Applies. This means that the sum of all the voltage sources is equal to
the sum of all the voltage drops or
VS = V1 + V2 + V3 + . . . + VN
4. The total resistance in the circuit is equal to the sum of the individual resistances.
RT = R1 + R2 + R3 + . . . + RN
5. The sum of the power supplied by the source is equal to the sum of the power dissipated in
the components.
<span>PT = P1 + P2 + P3 + . . . + PN</span>
Answer:
False
Explanation:
Be safe and make good choices!!
Answer:
u can see the answer in this picture
