Select the correct answer from each drop-down menu.The function t relates the age of a plant fossil, in years, to the percentage, c, of carbon-14 remaining in the fossil relative to when it was alive.𝑡⁡(𝑐) = −3,970.9⁢(ln⁡𝑐)Use function t to complete the statements.A fossil with 47% of its carbon-14 remaining is approximately years old.A fossil that is 7,000 years old will have approximately of its carbon-14 remaining.

To solve the given problem, we need to use the function $t(c) = -3970.9 \cdot \ln(c)$, where $t$ is the age of the fossil in years and $c$ is the percentage of carbon-14 remaining.

Part 1: Finding the age of a fossil with 47% of its carbon-14 remaining

1. Convert the percentage to a decimal: $c = 47\% = 0.47$

2. Substitute $c$ into the function: $t(0.47) = -3970.9 \cdot \ln(0.47)$

3. Calculate the natural logarithm of 0.47: $\ln(0.47) \approx -0.755$

4. Multiply by -3970.9: $t(0.47) = -3970.9 \cdot (-0.755) \approx 2997.4$

So, a fossil with 47% of its carbon-14 remaining is approximately 2997.4 years old.

Part 2: Finding the percentage of carbon-14 remaining in a fossil that is 7,000 years old

1. Set up the equation: $7000 = -3970.9 \cdot \ln(c)$

2. Solve for $\ln(c)$: $\ln(c) = \frac{7000}{-3970.9} \approx -1.763$

3. Exponentiate both sides to solve for $c$: $c = e^{-1.763} \approx 0.171$

4. Convert the decimal to a percentage: $c \approx 0.171 \times 100\% = 17.1\%$

So, a fossil that is 7,000 years old will have approximately 17.1% of its carbon-14 remaining.

Summary

• A fossil with 47% of its carbon-14 remaining is approximately 2997.4 years old.
• A fossil that is 7,000 years old will have approximately 17.1% of its carbon-14 remaining.