3 criteria must be taken into account to identify leap years: The year is evenly divisible by 4; If the year can be evenly divided by 100, it is NOT a leap year, unless; The year is also evenly divisible by 400. Then it is a leap year. Write a function that takes a year as a parameter and returns True if the year is a leap year, False otherwise. Save & Run Original - 1 of 1 Show Feedback Hide Code 1 import calendar 2 def is_leap (year): 3 leap = False 4 if year $ 4 == 0: 5 if year $100 == 0: 6 if year $ 400 == 0: 7 leap = True 8 else: 9 leap = True 10 return leap 11 12 print (is_leap (1996)) 13 print (is_leap (1900)) 14 print (is_leap (2020)) 15 000 Activity: 7.10.29 ActiveCode (ex_6_12) Result Actual Value Expected Value Notes ERROR None None Error: name 'isLeap' is not defined You passed: 0.0% of the tests Error NotImplementedError: calendar is not yet implemented in Skulpt on line 1 Description