A car loses value each year you own it. A certain car only maintains 88 percenta
ID: 2903917 • Letter: A
Question
A car loses value each year you own it. A certain car only maintains 88 percentage of its original value each year. A man purchases a new car in 2009 that is worth dollar 28,600. Make a table of values to show the cars value for the first three years. What type of function docs the table represent? Show your work to prove it. What is the initial value? What is they back number? Write the function to model the data. What is the value of the car in 2015? ( There is more than one way to work this question) Make sure you show your work. What year will the car be valued at dollar 3,699.04? ( There is more than one way to work this question) Make sure to show your work. Is this data an exponential growth or decay? How do you know?Explanation / Answer
Initial value of car = $ 28,600
value of car after 1 year = 88 % of original cost
value of car after 1 year = 28600 *88/100 = $ 25,168
value of car after 2nd year = 25168 *88/100 = $ 22,147.84
value of car after 3 rd year = 22147.84*88/100 = $ 19,490.10
1
the function is exponential decay
formula for exponential decay => y(t) = a × ekt
here "a" initial value
y(t) = price of the car after "t" years
k = rate constant
t= time
after 1 year the price is $ 25,168
subtituiting this in the equation
25168 = 28600 x ek*1
dividing both sides by 28600
25168/28600 = ek
0.88 = ek
taking natural log
ln(0.88) = ln(ek) // ln(ex)=x,
k = -0.128
therefore
3)
y(t) = 28600 * e(-0.128*t)
4)
value of car in 2015
t =6 years (2015-2009)=6 years
y(2015) = 28600 * e (-0.128 * 6)
y(2015) = $ 13268.68
5)
y(t) =$ 3669.04
3669.04 = 28600 *e(-0.128 * t)
0.128 = e(-0.128 * t) // divided both sides by 28600
taking natural log
ln(0.128) = ln (e(-0.128 * t) )
-2.055 = -0.128 * t // ln(ex)=x
t = 16 years
6)
this is an exponential decay as the value is decreasing in a fixed period by a fixed rate
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