Question

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?

Answer 1

pleaseee help!.... im doing an exam now... i dont get this one :/

Answer 2

@Luis_Rivera @KonradZuse

Answer 3

No cheating n exams... sorry...

Answer 4

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

Answer 5

I dont know which formula to use... if the compound interest function, or the continuous change function

Answer 6

You have some # 48k. you want 2.5% continously....

Answer 7

well: A = 48961e^(0.975*13) ?? is that right? (I dont know if the percent is?

Answer 8

then: A= 48961e^12.675
A= 48961*319655.78
A= 15650666514.54 (U see, this is soooo not right!) :/

Answer 9

where doyou get .975?

Answer 10

(1-0.025) because is depreciating right? and its 2.5%

Answer 11

I have no idea honestly....

Answer 12

:_( thanks though!

@AriPotta @Agent_Sniffles @abb0t @andriod09 @Alyx_da_Boss @bloodrain

Answer 13

@Dunce

Answer 14

stop

Answer 15

@dpaInc any idea on this one? @TuringTest

Answer 16

thanks

Answer 17

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.

Answer 18

^^ 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