Answer:
#include <iostream>
using namespace std;
void PrintPopcornTime (int bagOunces){
if (bagOunces < 2){
cout << "Too small"<<endl;
}
else if (bagOunces > 10){
cout << "Too large"<<endl;
}
else{
cout << bagOunces*6 <<" seconds"<<endl;
}
}
int main(){
PrintPopcornTime(7);
return 0;
}
Explanation:
Create a function called PrintPopcornTime that takes one parameter, bagOunces
Check the bagOunces using if-else structure. If it is smaller than 2, print "Too small". If it is greater than 10, print "Too large". Otherwise, calculate and print 6*bagOunces followed by " seconds".
Call the function in the main function with parameter 7. Since 7 is not smaller than 2 or not greater than 10, "42 seconds" will be printed.
True
yes computers are capable of serving network servers.
Answer:
#include <iostream>
#include <cstring>
using namespace std;
bool isAPalindrome(char* palindrome);
int main()
{
char palindrome[30];
bool palindrome_check;
cout << "Please enter an word or phrase.\n";
cin.getline(palindrome, 30);
palindrome_check = isAPalindrome(palindrome);
if (palindrome_check = true)
{
cout << "Input is a palindrome\n";
}
else
{
cout << "Inputis not a palindrome\n;";
}
system("pause");
return 0;
}
bool isAPalindrome(char* palindrome)
{
char* front;
char* rear;
front = palindrome;// starts at the left side of the c string
rear = (palindrome + strlen(palindrome)) - 1;//starts at the right side of the c-string. adds the c string plus the incriment value of s
while (front <= rear)
{
if (front = rear)
{
front++;
rear--;
}
else
{
return false;
}
}
return true;
}
Answer:
(10^6 + 9.9)
Explanation:
Given:
Total number of machine instructions = 1000
Number of page fault in 100 instructions = 1
Number of page faults in 1000 instructions = 10
Time to serve one page fault = 100 milliseconds
Time to serve ten page faults = 100*10 milliseconds = 1000 milliseconds = 10^6 Microseconds
Number of instructions without any page fault = 1000 - 10 = 990
Time required to run 1000 instructions = 10 Microseconds
So, time required to run 990 instructions = (10*(990/1000)) Microseconds = 9.9 Microseconds
So, the total time required to run the program = (10^6 + 9.9) Microseconds
