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Please help to build this Investment Calculator
*amount that they are depositing, stored as âPâ. *interest rate (as a percentage), stored as âiâ *number of years of the investment. stored as âtâ. *ask the user to input whether they want âsimpleâ or âcompoundâ interest, and store this in a variable called âinterestâ. *interest divided by 100 must be stored as 'r'. *Simple interest is A =P*(1+r*t) *Compound interest rate is calculated A = P* math.pow((1+r),t)
7 Antworten
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I'm struggling to get the if and elif statement correct for the simple and compound interests.
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import math
P = float(raw_input('amount depositing ?'))
i = float(raw_input('The interest rate in percentage?'))
t = float(raw_input('Number of years of investment?'))
interest = str(raw_input('\'simple\' or \'compound\' interest ?\n'))
r = int((i) / 100)
simple = float(P * (1 + r * t))
compound = float(P * math.pow((1 + r), t))
if interest == simple:
print (simple)
elif interest == compound:
print (compound)
I get confused with the if elif statements. It does not want to print. Sorry if this is a stupid question. I'm trying really hard to understand python. Want to become a software developer one day.
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I wanted to work on python 3 but the online course I'm doing requires me to use 2.7 :(
I will try that and get back to you. Thanks so much for helping me
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So find new course with python 3))
Use codecademy? Use udacity?))
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I payed for this course to get a micro degree. If the course is finished I will definitely go to codecademy.
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But... you have the instruction how to do it... what's wrong?
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Wait a minute...
Please install python 3 and higher, python 2 is not great)
About code:
you have unnessesary str() on input(composed or ...)
And when you try:
if interest == simple
You should use "simple", coz input will return str value