Look it up on a best buy website ion know
The answer to this is B=MODE(A1:A65)
Answer:
def leap_year(y):
if y % 4 == 0:
return 1
else:
return 0
def number_of_days(m,y):
if m == 2:
return 28 + leap_year(y)
elif m == 1 or m == 3 or m == 5 or m == 7 or m == 8 or m ==10 or m == 12:
return 31
elif m == 4 or m == 6 or m == 9 or m == 11:
return 30
def days(m,d):
if m == 1:
return 0 + d
if m == 2:
return 31 + d
if m == 3:
return 59 + d
if m == 4:
return 90 + d
if m == 5:
return 120 + d
if m == 6:
return 151 + d
if m == 7:
return 181 + d
if m == 8:
return 212 + d
if m == 9:
return 243 + d
if m == 10:
return 273 + d
if m == 11:
return 304 + d
if m == 12:
return 334 + d
def days_left(d,m,y):
if days(m,d) <= 60:
return 365 - days(m,d) + leap_year(y)
else:
return 365 - days(m,d)
print("Please enter a date")
day=int(input("Day: "))
month=int(input("Month: "))
year=int(input("Year: "))
choice=int(input("Menu:\n1) Calculate the number of days in the given month.\n2) Calculate the number of days left in the given year.\n"))
if choice == 1:
print(number_of_days(month, year))
if choice == 2:
print(days_left(day,month,year))
Explanation:
Hoped this helped
Answer:
d. The trigger is fired more than once.
Explanation:
What would happen in this situation is that the trigger would be fired more than once. This is because the trigger will be fired when the user updates the record. It will also be fired when the process builder is run.
If the trigger fires more than once, this can be problematic for the developer. Therefore, it is better if the trigger fires just once, as this is the time when the present changes can be placed.
