The answer is 4 because all computers use variables to process something
Option c is answer because most of us are literature in English
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:
The owls population increased by 13 in one year
Explanation:
The given data on the owl population are as follows;
The number of new owls born = 20 (positive increase in population)
The number of owls that die = 10 (negative decrease in population)
The number of owls that enter the forest = 5 (positive increase in population)
The number of owls that left the forest = 2 (negative decrease in population)
Let 'w' represent the number of new owls born, let 'x' represent the number of owls that die, let 'y' represent the number of owls that enter the forest and let 'z' represent the number of owls that left the forest, we get;
The change in population, ΔP = w - x + y - z
By plugging in the values, we get;
ΔP = 20 - 10 + 5 - 2 = 13
The change in the population of owls in one year is an increase in 13 owls.
While the reader of a pie chart can clearly see which categories dominate, it can be difficult to see categories that occupy only a small percentage of the chart. This problem becomes worse if there are many small categories in the chart. Labeling the categories can also be difficult.
