In the year 2009, Troy's boat had a value of $24,000. When he bought the boat in 2006 he paid $28,000. If the value of the boat depreciated linearly, what was the annual rate of change of the boat's value? Round your answer to the nearest hundredth if necessary.
Answers
In the year 2006 value of the boat was $28000 then in the year 2009 it's new value became $24000
thus decrease in price=$4000
time span=3 yrs
rate of change annually is=($4000÷3)=$1333.33 per annum.rounded up to nearest hundredth
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Given : In the year 2009, Troy's boat had a value of $24,000. When he bought the boat in 2006 he paid $28,000.
the value of the boat depreciated linearly,
To Find : annual rate of change of the boat's value to the nearest hundredth
Solution:
bought the boat in 2006 he paid $28,000
in Year 2009 Value = $ 24000
Value Depreciated = 28000 - 24000 = 4000
Number of years = 2009 - 2006 = 3
annual rate of change of the boat's value =4000/3
= 1333.3333
= 1333.33 nearest hundredth
annual rate of change of the boat's value = 1333.33 $
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