A quantity with an initial value of 8100 grows exponentially at a rate of 0.35% every 4 decades. What is the value of the quantity after 47 years, to the nearest hundredth?
To solve this problem, we first need to find the growth factor for the quantity. The growth factor is given by the formula:
\[ growth\ factor = 1 + \frac{r}{100} \]
where r is the growth rate. In this case, the growth rate is 0.35%, so:
\[ growth\ factor = 1 + \frac{0.35}{100} = 1.0035 \]
Since the quantity grows every 4 decades, we need to find the number of growth periods in 47 years. Since 1 decade is equal to 10 years, there are \( \frac{47}{10} \) = 4.7 decades in 47 years. This means there have been 5 growth periods (4 decades) up to 40 years, and an additional 1.7 growth periods for the remaining 7 years.
Now, we can calculate the value of the quantity after 47 years using the formula for exponential growth:
\[ final\ value = initial\ value \times (growth\ factor)^{number\ of\ growth\ periods} \]
Plugging in the values, we get:
\[ final\ value = 8100 \times (1.0035)^5 \times (1.0035)^{0.7} \]
\[ final\ value ≈ 8100 \times 1.017675 \times 1.007676 \]
\[ final\ value ≈ 8100 \times 1.025692 \]
\[ final\ value ≈ 8312.30 \]
Therefore, the value of the quantity after 47 years is approximately \( $8312.30 \) to the nearest hundredth.