To determine the population of the threatened animal species after 30 years, we can use the formula for exponential decay, which is analogous to exponential growth but for a declining population:

$P(t) = P_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}}$

where:

• $P(t)$ is the population after time $t$.
• $P_0$ is the initial population.
• $t$ is the time elapsed.
• $T$ is the half-life.

Given:

• $P_0 = 1.3$ million
• $t = 30$ years
• $T = 20$ years

Substituting the values into the formula:

$P(30) = 1.3 \times \left(\frac{1}{2}\right)^{\frac{30}{20}} = 1.3 \times \left(\frac{1}{2}\right)^{1.5}$

Now, calculate $\left(\frac{1}{2}\right)^{1.5}$:

$\left(\frac{1}{2}\right)^{1.5} = \frac{1}{2^{1.5}} = \frac{1}{\sqrt{2^3}} = \frac{1}{\sqrt{8}} \approx 0.3536$

Thus,

$P(30) \approx 1.3 \times 0.3536 \approx 0.4597 \text{ million}$

So, after 30 years, approximately 0.46 million animals (or 460,000 animals) will be left.

Would you like more details or have any questions? Here are some related questions:

1. What will the population be after 50 years?
2. How does changing the half-life affect the population decay rate?
3. If the initial population was 2 million instead of 1.3 million, what would be the population after 30 years?
4. How does exponential decay differ from exponential growth in real-world scenarios?
5. Can exponential decay apply to non-population scenarios, such as radioactive decay?

Tip: In exponential decay, the population never truly reaches zero but continues to halve at regular intervals based on the half-life.