Question

A new car is purchased for $17,000 and over time its value depreciates by one half every 3 years. What is the value of the car 19 years after it was purchased, to the nearest hundred dollars?

Answer 1

Answer:

200

Step-by-step explanation:

Halving Formula:

y=a\left(\frac{1}{2}\right)^{\frac{t}{h}}

y=a(

2

1

​

)

h

t

​

a=17000\hspace{40px}h=3\hspace{40px}t=19

a=17000h=3t=19

h is the halving time

\text{Plug in:}

Plug in:

y=17000\left(\frac{1}{2}\right)^{\frac{19}{3}}

y=17000(

2

1

​

)

3

19

​

y=210.826702215

y=210.826702215

y\approx 200

y≈200

Answer 2

Answer:

The answer is "$238".

Step-by-step explanation:

Current worth$= \ 17,000$

depreciates by $\frac{1}{2}$ in 3 years.

time=  19 years

depreciates rate=?

Using formula:

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$\to A_t=A_0(1-\frac{r}{100})^t$

calculates depreciate value in 3 year $= \frac{1}{2} \times 17,000$

                                                              $= 8,500$

so,

$A_t=8,500\\\\A_0=17,000\\\\t=3\ years$

$\to A_t=A_0(1-\frac{r}{100})^t\\\\\to 8,500= 17,000(1-\frac{r}{100})^3\\\\\to \frac{8,500}{17,000}= (1-\frac{r}{100})^3\\\\\to \frac{1}{2}= (1-\frac{r}{100})^3\\\\\to (\frac{1}{2})^{\frac{1}{3}}= (1-\frac{r}{100})\\\\\to 0.793700526 = (1-\frac{r}{100})\\\\\to \frac{r}{100} = (1-0.793700526)\\\\\to \frac{r}{100} = (1-0.8)\\\\\to r= 0.2 \times 100 \\\\\to r= 20 \%$

depreciates rate= 20%

$\to \text{Worth= Current worth}(1- \frac{\text{depreciates rate}}{100})^{time}$

$= \ 17,000 (1- \frac{20}{100})^{19}\\\\= \ 17,000 (1-0.2)^{19}\\\\= \ 17,000 (0.8)^{19}\\\\= \ 17,000 \times 0.014\\\\= \ 238$