Carlos Should use a Security Camera to keep track of his physical hardware. Carlos can then use a computer to monitor the security cameras.
Answer:
import java.util.*;
public class Main {
public static void main(String[] args)
{
Scanner scan = new Scanner();
double budget=0, num=0, total=0;
System.out.println("Your budget for the month? ");
budget=scan.nextDouble();
System.out.println("enter all expense, and after that type -9999 to quit: ");
while(num != -9999)
{
total+=num;
num=scan.nextDouble();
}
if(total<=budget)
{
System.out.println("under budget by ");
System.out.println(budget-total);
}
else
{
System.out.println("over budget by ");
System.out.println(total-budget);
}
}
}
Explanation:
- Take the budget as an input from user and store it to the budget variable.
- Loop until user has entered all his expenses and keep on adding them to the total variable.
- Check If the total is less than or equal to budget or otherwise, and then print the relevant message accordingly.
I think there's a typo in the question, otherwise they probably meant a 'she', since the name is Maureen.
Also, the answer is Personal hygiene and grooming.
Answer:
The program to this question as follows:
Program:
def isEvenPositiveInt(x): #defining method isEvenPositiveInt
if x>0: #checking number is positive or not
if x%2==0: #check true condition
return True #return value True
else:
return False #return value False
return False #return value False
print(isEvenPositiveInt(24)) #calling method and print return value
print(isEvenPositiveInt(-24)) #calling method and print return value
print(isEvenPositiveInt(23)) #calling method and print return value
Output:
True
False
False
Explanation:
In the above Python program, a method "isEvenPositiveInt" is defined, that accepts a variable "x" as its parameter, inside the method a conditional statement is used, which can be described as follows:
- In the if block the condition "x>0" is passed, that check value is positive, if this condition is true, it will go to another if block.
- In another, if block is defined, that checks the inserted value is even or not, if the value is even it will return "true", otherwise it will go to the else part, that returns "false".
Big-O notation is a way to describe a function that represents the n amount of times a program/function needs to be executed.
(I'm assuming that := is a typo and you mean just =, by the way)
In your case, you have two loops, nested within each other, and both loop to n (inclusive, meaning, that you loop for when i or j is equal to n), and both loops iterate by 1 each loop.
This means that both loops will therefore execute an n amount of times. Now, if the loops were NOT nested, our big-O would be O(2n), because 2 loops would run an n amount of times.
HOWEVER, since the j-loop is nested within i-loop, the j-loop executes every time the i-loop <span>ITERATES.
</span>
As previously mentioned, for every i-loop, there would be an n amount of executions. So if the i-loop is called an n amount of times by the j loop (which executes n times), the big-O notation would be O(n*n), or O(n^2).
(tl;dr) In basic, it is O(n^2) because the loops are nested, meaning that the i-loop would be called n times, and for each iteration, it would call the j-loop n times, resulting in n*n runs.
A way to verify this is to write and test program the above. I sometimes find it easier to wrap my head around concepts after testing them myself.
