The recursive function divBy3And5 is defined in Python and is found in the attached image.
In the base case, the function divBy3And5 tests if the input list is empty. If so, the tuple returned is 
. This means no numbers are divisible by three and no numbers are divisible by five.
The recursive step gets the first element of the list and
- If divisible by 3, it sets <em>count_of_3</em> to 1, else it leaves it as 0
- If divisible by 5, it sets <em>count_of_5</em> to 1, else it leaves it as 0
It then makes a recursive call on the remaining elements, and stores it in a variable as follows
<em>divBy3And5_for_remaining_elem</em> = divBy3And5(remaining_elements)
Then, it returns the tuple
(<em>divBy3And5_for_remaining_elem</em>[0] + <em>count_of_3</em>,
<em>divBy3And5_for_remaining_elem</em>[1] + <em>count_of_5)</em>
Learn more about recursion in Python: brainly.com/question/19295093
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Number 1 and 4 are the same? I agree with your answers.
Answer: Linked cell
Explanation: I just did a test
Answer:
C++ code explained below
Explanation:
#include<bits/stdc++.h>
#include <iostream>
using namespace std;
int FiboNR(int n)
{
int max=n+1;
int F[max];
F[0]=0;F[1]=1;
for(int i=2;i<=n;i++)
{
F[i]=F[i-1]+F[i-2];
}
return (F[n]);
}
int FiboR(int n)
{
if(n==0||n==1)
return n;
else
return (FiboR(n-1)+FiboR(n-2));
}
int main()
{
long long int i,f;
double t1,t2;
int n[]={1,5,10,15,20,25,30,35,40,45,50,55,60,65,70,75};
cout<<"Fibonacci time analysis ( recursive vs. non-recursive "<<endl;
cout<<"Integer FiboR(seconds) FiboNR(seconds) Fibo-value"<<endl;
for(i=0;i<16;i++)
{
clock_t begin = clock();
f=FiboR(n[i]);
clock_t end = clock();
t1=double(end-begin); // elapsed time in milli secons
begin = clock();
f=FiboNR(n[i]);
end = clock();
t2=double(end-begin);
cout<<n[i]<<" "<<t1*1.0/CLOCKS_PER_SEC <<" "<<t2*1.0/CLOCKS_PER_SEC <<" "<<f<<endl; //elapsed time in seconds
}
return 0;
}
