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Function Arguments
Fill in the blanks to define a function that prints "Yes", if its parameter is an even number, and "No" otherwise. ____ even(x): if x%2 == 0: ____ ("Yes") ________ print("No")
8 odpowiedzi
+ 5
Attunuru Baby Sri Laya
else
print("No")
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Attunuru Baby Sri Laya
I think you missed a colon
else:
print("No")
+ 2
There i guess it would be else
Read the question well you will get your answer
" If it is even print yes **else** print no"
+ 2
Attunuru Baby Sri Laya If you are talking about Quizzes in the lessons ,You can get the answers from the comments below in each quiz if u couldn't make it , Check that out !!!
+ 1
Attunuru Baby Sri Laya, show your attempt please, we are not doing homework in the Q&A fourm.
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@Dino Wun
def even(x):
if x%2==0:
print ("Yes")
_______
print("No")
##Actually i tried a lot to fill that place but i couldn't get it!!
If possible could u pls help me in solving it...!!!
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@Krish @Alphin K Sajan
Yeah i tried it in first attempt only.....but it's incorrect.
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@Krish
Yup.....Thanks