Send me if else statement code .the tuffest one

Question

Please

Answer 1

the variable coursework is not defined, so you get a NameError.

Answer 2

Round Up kid..

in this code you have to define coursework...
suppose

coursework = "English"
score_theory = 53
score_practical = 35

if(coursework == "Science" or coursework == "science"):
if(score_theory > 50):
print("Please check the input score for 'Science: Theory'.")
elif(score_practical > 50):
print("Please check the input score for 'Science: Practical'.")
else:
print("Score validated for Science. Your total is: ",score_theory + score_practical)
elif(coursework == "English" or coursework == "english"):
if(score_theory > 60):
print("Please check the input score for 'English: Theory'.")
elif(score_practical > 40):
print("Please check the input score for 'English: Practical'.")
else:
print("Score validated for English. Your total is: ",score_theory + score_practical)
else: print("Coursework not recognised. Please enter score for either Science or English.")

now this will work..!!

Answer 3

No..!
u'll get a name error

Answer 4

Iron Man Indent it properly...
x=man
If x==man:
Print ('yes')
else:
Print('no')

This one works out....

Answer 5

Round Up Kid check out this one...
guessing_number =100
user_type = int(input("guess a number"))
if user_type == guessing_number:
print("you win")
else:
if user_type < guessing_number:
print("too low")
else :
print("too high")

Answer 6

It is working thank you

Answer 7

Yes
Now it says different things
I don't know it is working or not
But no error now

Answer 8

Thank yoi

Answer 9

For every bit of code, therer can be a more complex one.

Answer 10

Iron Man thank you

Answer 11

#plz correct my code
guessing_number =100
user_type = int(input("guess a number"))
if user_type == guessing_number:
print("you win")
else:
if user_type < guessing_number:
print("too low")
else :
print("too high")

#plz correct my code

Answer 12

Nguyễn Thành Trung I get error of intendention
I don't know why but I press tab button then also

But it's okay now every thing is fine

Answer 13

Thank you

Answer 14

Unfortunately it's not working
Maybe it's my pydriod
It's not working if else statement

Answer 15

No it's not running

Answer 16

Traceback error

Answer 17

score_theory = 53
score_practical = 35

if(coursework == "Science" or coursework == "science"):
if(score_theory > 50):
print("Please check the input score for 'Science: Theory'.")
elif(score_practical > 50):
print("Please check the input score for 'Science: Practical'.")
else:
print("Score validated for Science. Your total is: ",score_theory + score_practical)
elif(coursework == "English" or coursework == "english"):
if(score_theory > 60):
print("Please check the input score for 'English: Theory'.")
elif(score_practical > 40):
print("Please check the input score for 'English: Practical'.")
else:
print("Score validated for English. Your total is: ",score_theory + score_practical)
else: print("Coursework not recognised. Please enter score for either Science or English.")

# I tried pydriod 3 app I also try solo learn app nothing works
#maybe it's my personal app problem
#how can I fix this?

Answer 18

wht error exactly u are getting?

Answer 19

Traceback error

Answer 20

I have screenshot uploading in web
There you see what problem I'm facing
Wait