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anonymous
 5 years ago
in 1980, your car was worth $25,000, now in 2010 your car is worth $13,000. What is the annual decay rate?
anonymous
 5 years ago
in 1980, your car was worth $25,000, now in 2010 your car is worth $13,000. What is the annual decay rate?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know decay is y=a(1=r)^x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y=Ae^(kx) Assuming reducing balance depreciation , which is a reasonable assumption

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoops , should be y= Ae^(kt) , for t=time in yrs

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now when t=0 , y= 25000 , so A= 25000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when t=30 , y= 13000 so 13000 = 25000e^(30k) e^(30k) = 13/25 30k = ln(13/25 ) k = [ln(13/25)] / 30 = 0.0279175...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or something like that, might like to calculate k for yourself

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now k is defined at the decay rate as a decimal because if you differentiate , y= Ae^(kt) you get dy/dt = k y , ie minus sign means its decreasing , and the factor of k is the rate ( as a decimal )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0EDIT correct value for k = 0.0217975

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so this means that every year the car losses 2.17975% of its previous years value
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