Question

The average price of a two-bedroom apartment in the uptown area of a prominent American city during the real estate boom from 1994 to 2004 can be approximated by p(t) = 0.12e0.10t million dollars (0 ≤ t ≤ 10) where t is time in years (t = 0 represents 1994). What was the average price of a two-bedroom apartment in this uptown area in 2002, and how fast was the price increasing? (Round your answers to two significant digits.)

Answer 1

Answer:

Step-by-step explanation:

Average price of a two bedroom apartment can be approximated by the equation,

p(t) = $0.12e^{0.10t}$

Here, t represents the duration from year 1994.

Duration from year 1994 to year 2004 = 10 years

p(10) = $0.12e^{0.10\times 10}$

        = 0.12e

        = 0.3262

        ≈ 0.33 million

Cost of the apartment in 2004 was 0.33 million.

Let the equation to calculate the rate of increase in the price is,

p(t) = $0.12(1+\frac{r}{100})^t$

Cost of the apartment after 10 years = 0.33

0.33 = $0.12(1+\frac{r}{100})^{10}$

$\frac{0.33}{0.12}=(1+\frac{r}{100})^{10}$

2.75 = $(1+\frac{r}{100})^{10}$

$(1+\frac{r}{100})=(2.75)^{0.10}$

1 + $\frac{r}{100}=1.106$

$\frac{r}{100}=0.106$

r = 10.6%

Therefore, price of the apartment is increasing with 10.6%