1. **State the problem:** We want to determine if the bacteria count for the tap water sample, modeled by $$f(t) = \frac{5t}{t^2 + 3t + 2}$$, will ever exceed the bacteria count for the pond water sample, modeled by $$g(t) = \frac{15t}{t^2 + 9}$$, for $$t > 0$$. 2. **Set up the inequality:** To find if $$f(t) > g(t)$$ for some $$t > 0$$, write: $$\frac{5t}{t^2 + 3t + 2} > \frac{15t}{t^2 + 9}$$ 3. **Simplify the inequality:** Since $$t > 0$$, we can divide both sides by $$t$$ without changing the inequality direction: $$\frac{5}{t^2 + 3t + 2} > \frac{15}{t^2 + 9}$$ 4. **Cross-multiply to clear denominators:** $$5(t^2 + 9) > 15(t^2 + 3t + 2)$$ 5. **Expand both sides:** $$5t^2 + 45 > 15t^2 + 45t + 30$$ 6. **Bring all terms to one side:** $$5t^2 + 45 - 15t^2 - 45t - 30 > 0$$ 7. **Simplify:** $$-10t^2 - 45t + 15 > 0$$ 8. **Multiply both sides by -1 to make leading coefficient positive, reversing inequality:** $$10t^2 + 45t - 15 < 0$$ 9. **Divide entire inequality by 5:** $$2t^2 + 9t - 3 < 0$$ 10. **Find roots of quadratic $$2t^2 + 9t - 3 = 0$$ using quadratic formula:** $$t = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} = \frac{-9 \pm \sqrt{81 + 24}}{4} = \frac{-9 \pm \sqrt{105}}{4}$$ 11. **Approximate roots:** $$\sqrt{105} \approx 10.247$$ So, $$t_1 = \frac{-9 - 10.247}{4} \approx -4.312$$ (negative, discard since $$t > 0$$) $$t_2 = \frac{-9 + 10.247}{4} \approx 0.312$$ 12. **Analyze inequality:** Since the parabola opens upward (coefficient of $$t^2$$ positive), $$2t^2 + 9t - 3 < 0$$ between the roots. For $$t > 0$$, the inequality holds for $$0 < t < 0.312$$. 13. **Conclusion:** For $$t$$ in the interval $$0 < t < 0.312$$ days, $$f(t) > g(t)$$, so the bacteria count for tap water exceeds that of pond water. For $$t > 0.312$$, pond water bacteria count is higher. **Final answer:** Yes, the bacteria count for the tap water sample exceeds the pond water sample for $$t$$ between 0 and approximately 0.312 days.