Answer:. True
Explanation:
yes it is true in my opinion
Using a linear search to find a value that is stored in the last element of an array of 20,000 elements, 20,000 element(s) must be compared.
Answer:
// here is code in C++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int n;
double average,sum=0,x;
cout<<"enter the Value of N:";
// read the value of N
cin>>n;
cout<<"enter "<<n<<" Numbers:";
// read n Numbers
for(int a=0;a<n;a++)
{
cin>>x;
// calculate total sum of all numbers
sum=sum+x;
}
// calculate average
average=sum/n;
// print average
cout<<"average of "<<n<<" Numbers is: "<<average<<endl;
return 0;
}
Explanation:
Read the total number from user i.e "n".Then read "n" numbers from user with for loop and sum them all.Find there average by dividing the sum with n.And print the average.
Output:
enter the Value of N:5
enter 5 Numbers:20.5 19.7 21.3 18.6 22.1
average of 5 Numbers is: 20.44
Answer:
Number of strings = (10, 2)×(8,3) = 2520
Explanation:
The number of possible combinations for taking two 0's is C(10, 2)
It remains 8 Positions
The number of possible combinations for taking three 1's is C(8,3)
So there remains 5 spots
