Having trouble implementing this part of the code.not sure if I should use if/else for range of days then print how many months that is; if i do that I would have to do the range up until 750 days. or is there another way to do it? def months (cell_digit):days143(cell_digit)**((cell_digit%2)+1)* (cell_digit)(days/30)144145MONTHS146147148 Description: This function gets the last digit of your cellphone and returns an integer thatbe months used for your "Title Movie", where months >= 1.First you need to determine a number of days with the following formula:days = *(cell_digit)(cell_digit)((cell_digit%2)+1)After, you need to transform the number of days to months. Assume a month has 30 days,return the value based on the range of days. Note that 1 is the minimum number of monthsRange of days0 - 3031 - 60Months1261 - 903...721 - 75025Parameters: cell_digit (int) is the last digit of your cellphone.Return value: An integer that represents a number of months, where months >= 1.OU MUST implement a math operation to determine the number of months, NO hard coding allovxamples:months(0)O days >> 1 month# 92 * 9 = 729 davs >> 25 months→ 1# 01 * 0months (9) → 25

Having trouble implementing this part of the code. not sure if I should use if/else for range of days then print how many months that is; if i do that I would have to do the range up until 750 days. or is there another way to do it?

Having trouble implementing this part of the code. not sure if I should use if/else for range of days then print how many months that is; if i do that I would have to do the range up until 750 days. or is there another way to do it?

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Question

Having trouble implementing this part of the code.

not sure if I should use if/else for range of days then print how many months that is; if i do that I would have to do the range up until 750 days.

or is there another way to do it?

def months (cell_digit):
days
143
(cell_digit)**((cell_digit%2)+1)* (cell_digit)
(days/30)
144
145
MONTHS
146
147
148

Description: This function gets the last digit of your cellphone and returns an integer that
be months used for your "Title Movie", where months >= 1.
First you need to determine a number of days with the following formula:
days = *(cell_digit)
(cell_digit)((cell_digit%2)+1)
After, you need to transform the number of days to months. Assume a month has 30 days,
return the value based on the range of days. Note that 1 is the minimum number of months
Range of days
0 - 30
31 - 60
Months
1
2
61 - 90
3
...
721 - 750
25
Parameters: cell_digit (int) is the last digit of your cellphone.
Return value: An integer that represents a number of months, where months >= 1.
OU MUST implement a math operation to determine the number of months, NO hard coding allov
xamples:
months(0)
O days >> 1 month
# 92 * 9 = 729 davs >> 25 months
→ 1
# 01 * 0
months (9) → 25

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