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Python level up Python problem
The given code fragment for a video game checks if the player is ready for the next level. Only players with a score greater than 100 can move to the next level. Task Complete the code to display True when the score is greater than 100 and False otherwise Help me on this one
18 Answers
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Geani Epuran
Code:
Score=input() (to take the score)
If score > 100: (to check the score)
Print(True)
Else:
Print(False)
edit:this won't work in playground this will
score=int(input())
if score > 100:
print("True")
else:
print("False")
+ 7
ZIZO Abd alkawy ,
maybe you you want to check your code for errors? (8 issues in 5 lines of code)
+ 6
Geani Epuran ,
please show us your code attempt. put your code in playground, save it and create a link to it.
then post the link to your code here.
+ 3
Try it by yourself! I’m sure you can complete it
+ 2
Print(score>100) dosent work
^
p
Look at the uppercase
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Thanks guys for your help D: thanks to you, I stopped learning programming and started learning Spanish 🤗
+ 1
Well the answer is: print(score > 100)
If you still want it
+ 1
the answer is print(100 >= score) or
if score > 100:
print(True)
else:
print(False)
+ 1
If it requires input, add a score = int(input()) above the print
And please repeat the python course
+ 1
Lothar
I noticed now
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# The program takes the score as an input
score = int(input())
# Add the comparison operation inside the parentheses
print(score > 100)
0
Print(score>100) dosent work
0
the interesting thing is that nobody knows the answer and can't help me
0
Variable score is not defineted is saying
0
Jesus you are the god :) thanks man
0
No problem
0
Score =100
Print (Score)
0
score = int(input())
print(score > 100)