Unit Quiz: Real-World Modeling
Instructions
- Time: 45 minutes
- Calculator: Allowed for all problems
- Show all work for full credit
Questions
1. A population of deer is modeled by P(t) = 200e^(0.15t), where t is years since 2020. What will the population be in 2030? Round to the nearest whole number.
2. A company's profit P (in thousands) is modeled by P(x) = -2x² + 40x - 150, where x is the selling price in dollars. Find the price that maximizes profit and the maximum profit.
3. The half-life of a medication in the bloodstream is 4 hours. If a patient takes 500mg at 8am, how much remains at 4pm?
4. Data shows a linear relationship between study hours (x) and test scores (y). Given points (2, 65) and (6, 85), find the equation and predict the score for 8 hours of study.
5. A savings account earns 3.5% interest compounded continuously. How long will it take $5,000 to grow to $7,500? Use A = Pe^(rt).
6. The table shows data for a projectile. Determine whether a linear or quadratic model is more appropriate and explain why.
| Time (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Height (m) | 0 | 35 | 60 | 75 | 80 |
7. A city's population is 50,000 and growing at 2.5% per year. Another city has 80,000 people but is declining at 1% per year. In how many years will their populations be equal?
8. The cost C of producing x items is C(x) = 5000 + 15x. The revenue is R(x) = 40x. Find the break-even point and interpret its meaning.
9. A researcher fits the model y = 3.2x + 12 to data with r² = 0.94. Interpret the slope and the r² value in context if x represents advertising spending (in $1000s) and y represents sales (in $1000s).
10. Water flows from a tank at a rate proportional to the current volume. If the tank starts with 1000 gallons and has 600 gallons after 5 hours, find the decay constant k and predict how long until only 100 gallons remain.
Answer Key
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- 897 deer — P(10) = 200e^(1.5) ≈ 200(4.48) ≈ 897
- $10 price, $50,000 max profit — Vertex at x = -40/(2×-2) = 10; P(10) = -200 + 400 - 150 = 50
- 62.5mg — 8 hours = 2 half-lives; 500 × (1/2)² = 125mg... wait, 4pm - 8am = 8 hours = 2 half-lives; 500 × 0.25 = 125mg
- y = 5x + 55; score for 8 hours = 95 — Slope = (85-65)/(6-2) = 5; y - 65 = 5(x-2) → y = 5x + 55
- ≈ 11.6 years — 7500 = 5000e^(0.035t); ln(1.5) = 0.035t; t = ln(1.5)/0.035 ≈ 11.58
- Quadratic — The differences in height (35, 25, 15, 5) are decreasing by 10 each time, indicating constant second differences characteristic of quadratic functions. Also, projectile motion follows a parabolic path.
- ≈ 13.5 years — 50000(1.025)^t = 80000(0.99)^t; solving: t = ln(1.6)/ln(1.025/0.99) ≈ 13.5
- Break-even at 200 items — 5000 + 15x = 40x; 5000 = 25x; x = 200. This means the company must sell 200 items to cover all costs.
- Slope interpretation: Each additional $1000 in advertising is associated with $3200 increase in sales. r² interpretation: 94% of the variation in sales is explained by advertising spending.
- k ≈ 0.102; t ≈ 22.5 hours — 600 = 1000e^(-5k); k = -ln(0.6)/5 ≈ 0.102; For 100: 100 = 1000e^(-0.102t); t = ln(10)/0.102 ≈ 22.5 hours