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raimelgarcia5
The number of customers visiting a local business is 48,961 and has been continuously declining at a rate of 2.5% each year. What is the approximate number of customers in 13 years?
pleaseee help!.... im doing an exam now... i dont get this one :/
@Luis_Rivera @KonradZuse
No cheating n exams... sorry...
I'm not cheating.. its not an exam... its a practice for exam.. and I dont get hwo to do that... come on.. please a lil help would be appreciated... like, explain.. dont do it... please
I dont know which formula to use... if the compound interest function, or the continuous change function
You have some # 48k. you want 2.5% continously....
well: A = 48961e^(0.975*13) ?? is that right? (I dont know if the percent is?
then: A= 48961e^12.675 A= 48961*319655.78 A= 15650666514.54 (U see, this is soooo not right!) :/
where doyou get .975?
(1-0.025) because is depreciating right? and its 2.5%
I have no idea honestly....
:_( thanks though! @AriPotta @Agent_Sniffles @abb0t @andriod09 @Alyx_da_Boss @bloodrain
@dpaInc any idea on this one? @TuringTest
I can think of a long way to do it. 48961- (0.025*48961)=A(customer base after 1 year) A-(0.025*A)=B(customer base after 2 years) B-(0.025*B)=C(3 years.....) and so on. There is probably an easier way to solve this with n! or something, but I don't know off hand.
^^ that's basically the way to do it Since it's continuously declining, you can also use: \(\large y=Ce^{kt} \) where C is your initial number of customers and k = -0.025. Plug in t=13 and you should get 35375.66