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0
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avatar+964 

 

(a) P(t) = 8500(0.995)^t (not sure if its correct) 

(b) i'm not sure 

 Jun 8, 2018
 #1
avatar+24862 
0

.955    not  .995

 

B  7000 = 8500 (.955)^t      solve for t (the number of years past 2010)

 Jun 8, 2018
edited by ElectricPavlov  Jun 8, 2018
 #2
avatar+111396 
+1

7000  = 8500(.955)^t      divide both sides by  8500

 

7000/8500  = .955^t

 

7000/8500  = .955^t      take the log of both sides

 

log (7000/8500)  =  log (.955)^t       and we can write

 

log (7000/8500)  = t * log (.955)  divide both sides by log(.955)

 

log (7000/8500)  / log (.955)  = t ≈ 4.2  years after 2010  =  2014

 

 

cool cool cool

 Jun 8, 2018
 #3
avatar+2421 
+1

I could be overthinking this, but wouldn't 4.2 years after 2010 be 2015?

 

4 years after 2010 would certainly land in 2014, but the extra .2 years would spill over into the year 2015. 

TheXSquaredFactor  Jun 9, 2018
 #4
avatar+111396 
0

Yeah, X^2... I think that's correct....my bad   !!!  2015 should be the answer

 

 

cool cool cool

CPhill  Jun 9, 2018

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