Explain what is meant by a limited data set and how this HIPAA rule may affect medical assistants
<span>If you sort a portion of an Excel sheet and you get an error message such as #DIV/0, the cause of the error message is (B) one or more cells containing absolute cell references. The possible reason of this error message includes: (1) e</span><span>ntering division formula that divided by zero (0), (2) and that is being used as a reference.</span>
Answer:
Time Complexity of Problem - O(n)
Explanation:
When n= 1024 time taken is t. on a particular computer.
When computer is 8 times faster in same time t , n can be equal to 8192. It means on increasing processing speed input grows linearly.
When computer is 8 times slow then with same time t , n will be 128 which is (1/8)th time 1024.
It means with increase in processing speed by x factor time taken will decrease by (1/x) factor. Or input size can be increased by x times. This signifies that time taken by program grows linearly with input size n. Therefore time complexity of problem will be O(n).
If we double the speed of original machine then we can solve problems of size 2n in time t.
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;
}
}
