Answer:
2
Explanation:
The output of the Java program is 2. The public Vehicle class is defined with the class variable 'counter'. When a Vehicle class object is instantiated, the counter variable increments by one.
In the program, the two instances of the class are created, incrementing the counter variable to two, the print statement outputs 2 as the result of the program.
Answer:
The flashdrive can hold 35389 400-pages-books
Explanation:
If 
of a page occupies 1 kB of memory, we can calculate how much memory a book will take
Now that we know that a book average file size is about 266,67 kB, we calculate how many of them can a 9 GB flash drive hold.
To do the calculation, we have to know how many kilobytes are in 9 gigabytes.
There is 1024 kilobytes in a megabyte, and 1024 megabytes in a gigabyte, so:
Finally, knowing the average file size of a book and how much memory in kilobytes the 9 GB flash drive holds, we calculate how many books can it hold.
The flashdrive can hold 35389 400-pages-books, or 14155776 pages of typical text.
Answer:
Written in Python
import math
principal = 8000
rate = 0.025
for i in range(1, 11):
amount = principal + principal * rate
principal = amount
print("Year "+str(i)+": "+str(round(amount,2)))
Explanation:
This line imports math library
import math
This line initializes principal amount to 8000
principal = 8000
This line initializes rate to 0.025
rate = 0.025
The following is an iteration from year 1 to 10
for i in range(1, 11):
This calculates the amount at the end of the year
amount = principal + principal * rate
This calculates the amount at the beginning of the next year
principal = amount
This prints the calculated amount
print("Year "+str(i)+": "+str(round(amount,2)))
Answer:
// code in C++
#include <bits/stdc++.h>
using namespace std;
// main function
int main()
{
// variables
int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;
int largest=INT_MIN;
int smallest=INT_MAX;
int n;
cout<<"Enter 10 Integers:";
// read 10 Integers
for(int a=0;a<10;a++)
{
cin>>n;
// find largest
if(n>largest)
largest=n;
// find smallest
if(n<smallest)
smallest=n;
// if input is even
if(n%2==0)
{
// sum of even
sum_even+=n;
// even count
eve_count++;
}
else
{
// sum of odd
sum_odd+=n;
// odd count
odd_count++;
}
}
// print sum of even
cout<<"Sum of all even numbers is: "<<sum_even<<endl;
// print sum of odd
cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;
// print largest
cout<<"largest Integer is: "<<largest<<endl;
// print smallest
cout<<"smallest Integer is: "<<smallest<<endl;
// print even count
cout<<"count of even number is: "<<eve_count<<endl;
// print odd cout
cout<<"count of odd number is: "<<odd_count<<endl;
return 0;
}
Explanation:
Read an integer from user.If the input is greater that largest then update the largest.If the input is smaller than smallest then update the smallest.Then check if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.
Output:
Enter 10 Integers:1 3 4 2 10 11 12 44 5 20
Sum of all even numbers is: 92
Sum of all odd numbers is: 20
largest Integer is: 44
smallest Integer is: 1
count of even number is: 6
count of odd number is: 4
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