Answer:
The greater the number of exchanges are there greater is the complexity of the algorithm. Thus it is a measure to check for the complexity of the program.
Answer:
3
Explanation:
because I just got a new phone I looked up the awaner it is 3
Answer:
I think b is correct answer.
Answer:
def count_over_100(number_list):
count = 0
for number in number_list:
if number > 100:
count += 1
return count
Explanation:
Create a function called count_over_100 that takes one parameter, number_list
Initialize the count as 0
Create a for loop that iterates through the number_list. Inside the loop, check each number using if-else structure. If a number is greater than 100, increment the count by 1.
When the loop is done, return the count
You can just look up "python ide online" on google and paste this code:
n = -1
count = 0
while n < 0:
n = int(input("We're checking to see if a number is prime or not! Enter a positive number: "))
if n % 2 == 0:
if n == 2:
print("{} is a prime number".format(n))
else:
print("{} is not a prime number".format(n))
else:
for x in range(n, 1, -1):
if n % x == 0:
count += 1
if count > 1 or n == 1:
print("{} is not a prime number".format(n))
else:
print("{} is a prime number".format(n))
I've written some code that checks to see if a number entered by the user is a prime number or not.
Sorry, but I'm not too good with pseudocode plans and all that. I hope this helps.
