Question

modeling real life the table shows the coyote populations ( y ) in a national park after ( t ) decades. use technology to find a function that fits the data.

decade, ( t )	0	1	2	3	4
population, ( y )	15	26	41	72	123

• ( y = 15(1.7)^x )
• ( y = 51(1.7)^x )
• ( y = 15(7.1)^x )
• ( y = 51(7.1)^x )

predict the coyote population after 60 years.

• about 153 coyotes
• about 263 coyotes
• about 362 coyotes
• about 1,921,500 coyotes

Explanation:

Part 1: Identifying the Correct Function

Step 1: Analyze the initial condition (t = 0)

When \( t = 0 \), the population \( y \) should be the initial value. For an exponential function of the form \( y = a(b)^t \), when \( t = 0 \), \( y = a \) (since \( b^0 = 1 \)). From the table, when \( t = 0 \), \( y = 15 \), so \( a = 15 \). This eliminates options with \( a = 51 \).

Step 2: Check the growth factor (b)

Now check the ratio of consecutive populations to estimate \( b \). For \( t = 0 \) to \( t = 1 \): \( \frac{26}{15} \approx 1.73 \), close to 1.7. For \( t = 1 \) to \( t = 2 \): \( \frac{41}{26} \approx 1.58 \), still around 1.7. For \( t = 2 \) to \( t = 3 \): \( \frac{72}{41} \approx 1.76 \), and \( t = 3 \) to \( t = 4 \): \( \frac{123}{72} \approx 1.71 \). So \( b \approx 1.7 \). Thus the function is \( y = 15(1.7)^t \).

Part 2: Predicting Population After 60 Years

Step 1: Convert 60 years to decades

60 years is \( \frac{60}{10} = 6 \) decades, so \( t = 6 \).

Step 2: Substitute t = 6 into the function \( y = 15(1.7)^t \)

Calculate \( (1.7)^6 \):
\( (1.7)^2 = 2.89 \)
\( (1.7)^3 = 1.7 \times 2.89 = 4.913 \)
\( (1.7)^4 = 1.7 \times 4.913 = 8.3521 \)
\( (1.7)^5 = 1.7 \times 8.3521 = 14.19857 \)
\( (1.7)^6 = 1.7 \times 14.19857 \approx 24.137569 \)

Then \( y = 15 \times 24.137569 \approx 362.0635 \), which is about 362.

Answer:

• Correct function: \( y = 15(1.7)^t \)
• Population after 60 years: About 362 coyotes