Respuesta :

[tex]\begin{gathered} a)\text{ 0.1085} \\ b)\text{ 10.85 \%} \\ c)\text{ \$7,300} \end{gathered}[/tex]

Firstly, we have to write the general depreciation fomula as follows;

[tex]V(t)=I(1-r)^t[/tex]

where V(t) represents the value of the commodity after a certain t years

I is the initial value of the item

r is the annual percentage change (rate of depreciation)

t is the number of years

a) Here, we want to calculate the annual rate of change

According to the data given in the question;

V(t) = $13,000

I = $41,000

r = ?

t = 2004-1994 = 10 years

Now, we proceed to substitute these values, to find the find of r as follows;

[tex]\begin{gathered} 13,000=41,000(1-r)^{10} \\ (1-r)^{10\text{ }}\text{ = }\frac{13,000}{41,000} \\ (1-r)^{10\text{ }}=\text{ }\frac{13}{41} \\ \\ (1-r)\text{ = }\sqrt[10]{\frac{13}{41}} \\ \\ 1-r\text{ = 0.8915} \\ r\text{ = 1-0.8915} \\ r\text{ = 0.1085} \end{gathered}[/tex]

b) To write r in percentage form, we have to multiply the answer in 'a' above by 100

We have this as ;

[tex]0.1085\times100\text{ = 10.85 \%}[/tex]

c) Here, we want to get the car value by year 2009

In that instance;

V(t) = ?

I = $41,000

r = 0.1085

t = 2009-1994 = 15

Substituting these values, we have;

[tex]\begin{gathered} V(15)=41,000(1-0.1085)^{15} \\ V(15)\text{ = \$7,321.56} \end{gathered}[/tex]

To the nearest 50 dollars, this is $7,300