subject

The fractional_part function divides the numerator by the denominator, and returns just the fractional part (a number between 0 and 1). Complete the body of the function so that it returns the right number. Note: Since division by 0 produces an error, if the denominator is 0, the function should return O instead of attempting the division. 1 - def fractional_part(numerator, denominator): 2 # Operate with numerator and denominator to 3 # keep just the fractional part of the quotient 4 return 0 олиш 6 print(fractional_part(5, 5)) # Should be 0 7 print(fractional_part(5, 4)) # Should be 0.25 8 print(fractional_part(5, 3)) # Should be 0.66... 9 print(fractional_part(5, 2)) # Should be 0.5 10 print(fractional_part(5, 0)) # Should be a 11 print(fractional_part(0, 5)) # Should be 0 Run Reset

ansver
Answers: 3

Another question on Computers and Technology

question
Computers and Technology, 22.06.2019 07:00
Idon understand these and need some ! ?
Answers: 2
question
Computers and Technology, 22.06.2019 09:30
Is a string of code written to hurt others by damaging or destroying
Answers: 1
question
Computers and Technology, 22.06.2019 15:20
The north and south regions had very diferent economies in the 1800s.
Answers: 1
question
Computers and Technology, 22.06.2019 20:00
Amanda needs to create an informative print brochure for her local library’s fundraiser dinner. what critical detail must she have before she starts designing the brochure?
Answers: 1
You know the right answer?
The fractional_part function divides the numerator by the denominator, and returns just the fraction...
Questions
question
Social Studies, 15.01.2021 19:20
question
Mathematics, 15.01.2021 19:20
question
English, 15.01.2021 19:20
question
Mathematics, 15.01.2021 19:20
question
Mathematics, 15.01.2021 19:20
question
Biology, 15.01.2021 19:20
question
Mathematics, 15.01.2021 19:20
question
Mathematics, 15.01.2021 19:20
question
Geography, 15.01.2021 19:20
question
Mathematics, 15.01.2021 19:20
question
Mathematics, 15.01.2021 19:20
Questions on the website: 13722367