The function \(\mathrm{A(t) = 500(0.80)^{(t/2)}}\) models the activity, in counts per minute, of a radioactive sample t hours after it...

1. TRANSLATE the function structure

• Given: \(\mathrm{A(t) = 500(0.80)^{(t/2)}}\)
• What this tells us: Every time \(\mathrm{t/2}\) increases by 1 unit, the activity multiplies by \(\mathrm{0.80}\)
• Since t is in hours, this means every 2 hours the activity becomes \(\mathrm{0.80}\) times what it was

2. TRANSLATE the question requirement

• "Activity decreases by \(\mathrm{n\%}\) every 6 hours"
• We need to find what happens over a 6-hour period
• A decrease of \(\mathrm{n\%}\) means the remaining activity is \(\mathrm{(100-n)\%}\) of the original

3. INFER the time interval relationship

• \(\mathrm{6 \text{ hours} ÷ 2 \text{ hours per interval} = 3 \text{ intervals}}\)
• Over 3 intervals, the multiplicative factor is \(\mathrm{(0.80)^3}\)

4. SIMPLIFY the calculation

• \(\mathrm{(0.80)^3 = 0.512}\) (use calculator)
• This means \(\mathrm{51.2\%}\) of the original activity remains

5. INFER the percentage decrease

• If \(\mathrm{51.2\%}\) remains, then the decrease is: \(\mathrm{100\% - 51.2\% = 48.8\%}\)
• Therefore \(\mathrm{n = 48.8}\)

Answer: B (48.8)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skills: Students misunderstand what the exponent \(\mathrm{t/2}\) means for time intervals. They might think the \(\mathrm{0.80}\) factor applies every hour instead of every 2 hours, leading them to calculate \(\mathrm{(0.80)^6 = 0.262}\) for 6 hours. This gives them a decrease of \(\mathrm{73.8\%}\), which doesn't match any answer choice, causing confusion and guessing.

Second Most Common Error:

Poor INFER reasoning: Students correctly identify that \(\mathrm{0.80}\) applies every 2 hours but incorrectly calculate the 6-hour effect. They might multiply \(\mathrm{0.80 × 6/2 = 2.4}\) instead of raising to a power, or calculate \(\mathrm{(0.80) × 3 = 2.4}\), leading to nonsensical results and random answer selection.

The Bottom Line:

This problem requires careful attention to how the function's structure defines time intervals. The key insight is recognizing that \(\mathrm{t/2}\) in the exponent creates 2-hour intervals, not 1-hour intervals, and then properly applying exponent rules for cumulative effects.