Math, asked by alvairis1995, 1 year ago

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

Answered by souhit12
4

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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Answered by amitnrw
1

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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