Choice A is the correct answer. Use semi-annual compounding.

1. Values: $P = 8000$, $r = 0.045$, $n = 2$ (semi-annually), $t = 6$.
2. Formula: $A = 8000(1 + \frac{0.045}{2})^{2 \times 6} = 8000(1.0225)^{12}$.
3. Calculate: $(1.0225)^{12} \approx 1.30673$, so $A \approx 10,453.86$.

💡 Strategic Tip: Semi-annual means twice per year, so $n = 2$.

Choice B is incorrect because this assumes annual compounding. Choice C is incorrect because the calculation doesn't match. Choice D is incorrect because this growth is too high.