Question 765749

you can do it this way:

In the problem you are given the rate of decrease of {{{30}}}% or {{{0.30}}}.

Each year, the boat is worth {{{100-30=70}}}% of its value the previous year, so after {{{3}}}years, its value will be

${{{15000* 0.30 =4500}}} will cost less than year before; so price firs year after is ${{{15000-4500=10500}}}

next year value  {{{10500}}} depreciates again for {{{30}}}%; so new value will be
{{{10500*0.30=3150}}}; so price second  year after is ${{{10500-3150=7350}}}

and finally, 

${{{7350*0.30=2205}}}; so price in three years is 

${{{7350-2205=5145}}}


or use exponential growth/decay formula:

Since the decay equation takes on the form {{{P=P[0](1-k)^t}}}, we can take note that {{{k=.30}}}, {{{P[0]=15000}}} and {{{t=3}}}. 
Plugging this into the equation, we have 

{{{P=15000(1-.3)^3}}}

{{{P=5145}}}

So after 3 years, the car will be worth {{{5145}}}