The answer to your question is,
True.
-Mabel <3
Answer:
Low CPU utilization and high I/O utilization
Explanation:
Answer:
if(revenue.cents - expenses.cents < 0){
profit.dollars = revenue.dollars - expenses.dollars - 1;
profit.cents = 1 - revenue.cents - expenses.cents;
}
else{
profit.dollars = revenue.dollars - expenses.dollars;
profit.cents = revenue.cents - expenses.cents;
}
Explanation:
We know that profit is given as: revenue - expenses from the question.
From the given expression above;
if(revenue.cents - expenses.cents < 0)
then profit.dollar will be revenue.dollars - expenses.dollars - 1; the 1 is to be carry over to the cent part. And the profit.cent will be 1 - revenue.cents - expenses.cents;
else the profit.dollars and the profit.cent is computed directly without needing to carry over:
profit.dollars = revenue.dollars - expenses.dollars;
profit.cents = revenue.cents - expenses.cents;
Answer:
Explanation:
The following code is written in Python. It creates a class that takes in one ArrayList parameter and loops through it and calls two functions that check if the numbers are Perfect, Odd, or Even. Then it goes counting each and printing the final results to the screen.
class NumberAnalyzer:
def __init__(self, myArray):
perfect = 0
odd = 0
even = 0
for element in myArray:
if self.isPerfect(element) == True:
perfect += 1
else:
if self.isEven(element) == True:
even += 1
else:
odd += 1
print("# of Perfect elements: " + str(perfect))
print("# of Even elements: " + str(even))
print("# of Odd elements: " + str(odd))
def isPerfect(self, number):
sum = 1
i = 2
while i * i <= number:
if number % i == 0:
sum = sum + i + number / i
i += 1
if number == sum:
return True
else:
return False
def isEven(self, number):
if (number % 2) == 0:
return True
else:
return False
Answer: Find answers in the attachments
Explanation:
