Math 3 Quiz: Sensitivity Analysis
7 questions · exam conditions
0:00
Sensitivity AnalysisQuestion 1 of 7

A bridge's load capacity LL (in tons) varies with temperature TT (°F) according to L=20000.5T+0.001T2L = 2000 - 0.5T + 0.001T^2. During a heat wave, the temperature rises from 75°F to 105°F. If the bridge currently supports 80% of its capacity at 75°F, what safety margin remains at the higher temperature?

The safety margin decreases to approximately 12%
The safety margin decreases to approximately 18%
The safety margin decreases to approximately 25%
The safety margin increases to approximately 22%
← Back to quizzes

Math 3 Quiz

Math 3 Quiz: Sensitivity Analysis

Practice Sensitivity Analysis in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Sensitivity Analysis, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A bridge's load capacity LL (in tons) varies with temperature TT (°F) according to L=20000.5T+0.001T2L = 2000 - 0.5T + 0.001T^2. During a heat wave, the temperature rises from 75°F to 105°F. If the bridge currently supports 80% of its capacity at 75°F, what safety margin remains at the higher temperature?

  1. The safety margin decreases to approximately 12%
  2. The safety margin decreases to approximately 18% (correct answer)
  3. The safety margin decreases to approximately 25%
  4. The safety margin increases to approximately 22%
Explanation: At T = 75°F: L₁ = 2000 - 0.5(75) + 0.001(75)² = 2000 - 37.5 + 5.625 = 1968.125 tons. Current load = 0.8 × 1968.125 = 1574.5 tons. At T = 105°F: L₂ = 2000 - 0.5(105) + 0.001(105)² = 2000 - 52.5 + 11.025 = 1958.525 tons. New capacity utilization = 1574.5/1958.525 = 0.804 = 80.4%. Safety margin = 100% - 80.4% = 19.6% ≈ 18%. Choice A would require much higher utilization. Choice C underestimates the remaining capacity. Choice D incorrectly suggests improvement.

Question 2

A manufacturing process has a yield rate Y=0.950.002T1.5Y = 0.95 - 0.002T^{1.5}, where TT is the operating temperature in degrees above room temperature. If an equipment malfunction causes the temperature to fluctuate 8°C higher than the planned 25°C above room temperature, what is the approximate change in yield rate?

  1. A decrease of approximately 0.038 in yield rate
  2. A decrease of approximately 0.052 in yield rate (correct answer)
  3. A decrease of approximately 0.024 in yield rate
  4. A decrease of approximately 0.067 in yield rate
Explanation: Originally at T=25: Y = 0.95 - 0.002(25)^1.5 = 0.95 - 0.25 = 0.70. After malfunction at T=33: Y = 0.95 - 0.002(33)^1.5 = 0.95 - 0.3018 = 0.648. The change is 0.648 - 0.70 = -0.052. Choice A uses T^1 instead of T^1.5. Choice C uses an incorrect base calculation. Choice D compounds the exponent error with arithmetic mistakes.

Question 3

A company's quarterly profit PP (in thousands of dollars) is modeled by P=50x2x215P = 50x - 2x^2 - 15, where xx represents the number of products sold (in hundreds). If the company currently sells 800 products and is considering increasing sales by 10%, what is the approximate change in quarterly profit?

  1. An increase of $18,000
  2. An increase of $22,000
  3. An increase of $26,000
  4. An increase of $14,000 (correct answer)
Explanation: Current sales: 800 products = 8 hundreds, so x = 8. Current profit: P = 50(8) - 2(8)² - 15 = 400 - 128 - 15 = 257 thousand dollars. With 10% increase: 880 products = 8.8 hundreds. New profit: P = 50(8.8) - 2(8.8)² - 15 = 440 - 154.88 - 15 = 270.12 thousand dollars. Change = 270.12 - 257 = 13.12 ≈ 14 thousand dollars increase. Choice A uses incorrect derivative calculation, B assumes linear relationship, C miscalculates the quadratic term.

Question 4

An environmental scientist models the pH level of a lake as pH=7.20.8log(x)pH = 7.2 - 0.8\log(x) where xx is the concentration of pollutants in parts per million (ppm). If pollution increases from 10 ppm to 15 ppm, what is the approximate change in pH level?

  1. pH decreases by approximately 0.14 units (correct answer)
  2. pH decreases by approximately 0.18 units
  3. pH decreases by approximately 0.22 units
  4. pH increases by approximately 0.16 units
Explanation: At x = 10 ppm: pH = 7.2 - 0.8log(10) = 7.2 - 0.8(1) = 6.4. At x = 15 ppm: pH = 7.2 - 0.8log(15) = 7.2 - 0.8(1.176) = 7.2 - 0.941 = 6.259. Change = 6.259 - 6.4 = -0.141 ≈ -0.14. The pH decreases by about 0.14 units. Choice D has wrong sign, B and C use incorrect logarithm values or arithmetic errors.

Question 5

A projectile's height hh (in feet) after tt seconds is given by h(t)=16t2+64t+80h(t) = -16t^2 + 64t + 80. If the initial velocity coefficient changes from 64 to 60 ft/s while keeping other parameters constant, how does this affect the maximum height achieved?

  1. Maximum height decreases by exactly 7 feet (correct answer)
  2. Maximum height decreases by exactly 9 feet
  3. Maximum height decreases by exactly 11 feet
  4. Maximum height decreases by exactly 13 feet
Explanation: For h(t) = -16t² + 64t + 80, maximum occurs at t = 64/(2×16) = 2 seconds. Max height = -16(4) + 64(2) + 80 = -64 + 128 + 80 = 144 feet. For h(t) = -16t² + 60t + 80, maximum at t = 60/32 = 1.875 seconds. Max height = -16(1.875)² + 60(1.875) + 80 = -56.25 + 112.5 + 80 = 136.25 feet. Decrease = 144 - 136.25 = 7.75 ≈ 7 feet. Other choices reflect common errors in vertex formula or arithmetic mistakes.

Question 6

A population growth model predicts that the number of bacteria NN after tt hours is N(t)=100020.3tN(t) = 1000 \cdot 2^{0.3t}. If the growth rate parameter changes from 0.3 to 0.35, what is the approximate percent increase in population size after 8 hours?

  1. Approximately 15% increase in final population
  2. Approximately 25% increase in final population
  3. Approximately 35% increase in final population (correct answer)
  4. Approximately 45% increase in final population
Explanation: Original population at t = 8: N = 1000 × 2^(0.3×8) = 1000 × 2^2.4 ≈ 1000 × 5.278 = 5,278. New population: N = 1000 × 2^(0.35×8) = 1000 × 2^2.8 ≈ 1000 × 6.964 = 6,964. Percent change = (6,964 - 5,278)/5,278 × 100% ≈ 32% ≈ 35%. Choice A underestimates exponential sensitivity, B uses linear approximation, D overestimates by confusing the parameter change with output change.

Question 7

A logistics company models delivery time TT (in hours) as a function of distance dd (in miles) and traffic density ff using T(d,f)=0.02d+3fT(d,f) = 0.02d + 3\sqrt{f}. Currently, a route has d=150d = 150 miles and traffic density f=16f = 16. If traffic density increases to f=25f = 25 while distance remains constant, how does this change affect delivery time?

  1. Delivery time increases by exactly 4 hours
  2. Delivery time increases by exactly 3 hours (correct answer)
  3. Delivery time increases by exactly 5 hours
  4. Delivery time increases by exactly 2 hours
Explanation: When you encounter multivariable functions like this delivery time model, you're being asked to analyze how changing one variable affects the output while keeping others constant. This tests your ability to evaluate functions and calculate differences. To find how the traffic density change affects delivery time, you need to calculate TT at both traffic densities and find the difference. With the distance fixed at d=150d = 150 miles: Initial conditions (f=16f = 16): T(150,16)=0.02(150)+316=3+3(4)=3+12=15T(150, 16) = 0.02(150) + 3\sqrt{16} = 3 + 3(4) = 3 + 12 = 15 hours New conditions (f=25f = 25): T(150,25)=0.02(150)+325=3+3(5)=3+15=18T(150, 25) = 0.02(150) + 3\sqrt{25} = 3 + 3(5) = 3 + 15 = 18 hours The change in delivery time is 1815=318 - 15 = 3 hours, making B correct. Choice A (4 hours) likely comes from incorrectly calculating 2516=54=1\sqrt{25} - \sqrt{16} = 5 - 4 = 1, then multiplying by 4 instead of 3. Choice C (5 hours) results from forgetting the coefficient 3 and just using 2516=1\sqrt{25} - \sqrt{16} = 1, but somehow getting 5. Choice D (2 hours) might come from calculation errors in the square roots or arithmetic mistakes. Remember that when analyzing multivariable functions, always substitute the specific values carefully and double-check your square root calculations. The key is methodically evaluating the function at both sets of conditions, then finding the difference.