A sample of radioactive material is observed over time, and its mass is recorded at regular intervals. The scatterplot shows...

1. TRANSLATE the problem information

Looking at the graph and the model \(\mathrm{m = c(d)^t}\):

• We have data points we can read from the scatterplot
• We need to find the constant d (the decay factor)
• We know c and d are positive constants

Key insight from the equation: When \(\mathrm{t = 0}\), the exponent makes \(\mathrm{d^0 = 1}\), so \(\mathrm{m = c(1) = c}\)

2. TRANSLATE the y-intercept to find c

Reading from the graph at \(\mathrm{t = 0}\):

• \(\mathrm{m \approx 80}\) grams

Since \(\mathrm{m = c}\) when \(\mathrm{t = 0}\):

• \(\mathrm{c = 80}\)

Our model is now: \(\mathrm{m = 80(d)^t}\)

3. INFER the strategy to find d

Now that we know c, we need to find d. The strategic approach:

• Pick another data point from the graph
• Substitute the values into \(\mathrm{m = 80(d)^t}\)
• Solve for d

Why \(\mathrm{t = 1}\) is the best choice: At \(\mathrm{t = 1}\), the equation becomes \(\mathrm{m = 80d}\), which makes solving for d straightforward (no exponent rules needed).

4. TRANSLATE another data point

Reading from the graph at \(\mathrm{t = 1}\):

• \(\mathrm{m \approx 72}\) grams

5. SIMPLIFY to solve for d

Substitute into \(\mathrm{m = 80(d)^t}\):

\(\mathrm{72 = 80(d)^1}\)
\(\mathrm{72 = 80d}\)
\(\mathrm{d = \frac{72}{80}}\)
\(\mathrm{d = \frac{9}{10}}\)
\(\mathrm{d = 0.90}\)

6. Verify your answer (optional but recommended)

Check with \(\mathrm{t = 2}\) where \(\mathrm{m \approx 65}\) from the graph:

• \(\mathrm{m = 80(0.90)^2 = 80(0.81) = 64.8 \approx 65}\) ✓

Looking at the answer choices, \(\mathrm{d = 0.90}\) matches choice (A).

Answer: (A) 0.90

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize the strategic two-step approach (find c first, then use c to find d). Instead, they try to find d directly without determining c, or they attempt to use two arbitrary points without recognizing the efficiency of using \(\mathrm{t = 0}\).

This leads to confusion with setting up equations, getting stuck, and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when computing \(\mathrm{72 \div 80}\). Common mistakes include:

• Getting \(\mathrm{\frac{72}{80} = 0.80}\) (confusing the calculation)
• Getting \(\mathrm{\frac{8}{7} \approx 1.14}\) by flipping the fraction

If they calculate \(\mathrm{d = 0.80}\), they might choose a nearby answer or get confused since 0.80 isn't listed. If they flip to get \(\mathrm{\frac{80}{72} \approx 1.11}\), this may lead them to select Choice (B) (1.10).

Third Common Error:

Conceptual confusion about exponential decay: Students might not apply constraint reasoning about what d should represent. They might select:

• Choice (C) (8.00) thinking it relates to "80" from the initial mass
• Choice (D) (80.0) by confusing d with c

Students who understand that d must be less than 1 for decay (since \(\mathrm{0 \lt d \lt 1}\)) can eliminate choices (B), (C), and (D) immediately.

The Bottom Line:

This problem tests whether students can connect the abstract exponential model to concrete graph data systematically. The key is recognizing that \(\mathrm{t = 0}\) gives you c "for free," then using that to unlock d from the next data point. Students who jump in without a clear strategy, or who make arithmetic errors, will select incorrect choices.