Python level up Python problem

Question

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

Answer 1

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")

Answer 2

ZIZO Abd alkawy ,

maybe you you want to check your code for errors? (8 issues in 5 lines of code)

Answer 3

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.

Answer 4

Try it by yourself! I’m sure you can complete it

Answer 5

Print(score>100) dosent work
^
p
Look at the uppercase

Answer 6

Thanks guys for your help D: thanks to you, I stopped learning programming and started learning Spanish 🤗

Answer 7

Well the answer is: print(score > 100)
If you still want it

Answer 8

Bro the answer is print(100 >= score) or

if score >= 100:
print(True)
else:
print(False)

Answer 9

If it requires input, add a score = int(input()) above the print
And please repeat the python course

Answer 10

Lothar
I noticed now

Answer 11

Print(score>100) dosent work

Answer 12

the interesting thing is that nobody knows the answer and can't help me

Answer 13

Variable score is not defineted is saying

Answer 14

Jesus you are the god :) thanks man

Answer 15

No problem

Answer 16

Score =100
Print (Score)

Answer 17

score = int(input())
print(score > 100)

Answer 18

# The program takes the score as an input
score = int(input())

# Add the comparison operation inside the parentheses
print(score > 100)