Math 1 Quiz: Interpreting Model Parameters
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Interpreting Model ParametersQuestion 1 of 20

A water tank contains 500 gallons initially and drains at a constant rate. The amount of water WW (in gallons) after tt hours is W=500−25tW = 500 - 25t. If the tank must maintain at least 50 gallons for safety reasons and cannot exceed its 500-gallon capacity, what is the practical range of this model?

50≤W≤50050 ≤ W ≤ 500
0≤W≤5000 ≤ W ≤ 500
50≤W≤47550 ≤ W ≤ 475
25≤W≤50025 ≤ W ≤ 500
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Math 1 Quiz

Math 1 Quiz: Interpreting Model Parameters

Practice Interpreting Model Parameters in Math 1 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 Interpreting Model Parameters, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 1.

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.

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Question 1

A water tank contains 500 gallons initially and drains at a constant rate. The amount of water WW (in gallons) after tt hours is W=500−25tW = 500 - 25t. If the tank must maintain at least 50 gallons for safety reasons and cannot exceed its 500-gallon capacity, what is the practical range of this model?

  1. 50≤W≤50050 ≤ W ≤ 500 (correct answer)
  2. 0≤W≤5000 ≤ W ≤ 500
  3. 50≤W≤47550 ≤ W ≤ 475
  4. 25≤W≤50025 ≤ W ≤ 500
Explanation: The practical range is determined by the constraints on the water amount. The minimum is 50 gallons (safety requirement), and the maximum is 500 gallons (initial amount and tank capacity). Since the tank only drains, WW decreases from 500 to 50, so the range is 50≤W≤50050 ≤ W ≤ 500. Choice B ignores the safety constraint. Choice C incorrectly suggests the maximum is less than the initial amount. Choice D uses an arbitrary minimum of 25 gallons.

Question 2

A research study models the relationship between study hours hh per week and test scores TT using T=45+3.2hT = 45 + 3.2h. The study is valid for students who study between 5 and 25 hours per week. If a student needs a score of at least 85 to pass, what does this requirement impose on the model's domain?

  1. Students must study at least 14 hours per week, providing a safety margin above minimum
  2. Students must study at least 13 hours per week, accounting for whole-hour study increments
  3. Students must study at least 11.5 hours per week, which is within the study's validity range
  4. Students must study at least 12.5 hours per week, which is within the study's validity range (correct answer)
Explanation: When you encounter a linear model with constraints, you need to work backwards from the requirement to find what input values satisfy it, then check if those values fall within the given domain restrictions. To find the minimum study hours needed for a passing score of 85, substitute T=85T = 85 into the equation T=45+3.2hT = 45 + 3.2h: 85=45+3.2h85 = 45 + 3.2h 40=3.2h40 = 3.2h h=403.2=12.5h = \frac{40}{3.2} = 12.5 So students must study at least 12.5 hours per week to achieve a score of 85. Since this falls within the study's validity range of 5-25 hours, this requirement is mathematically sound and practically achievable. Looking at the wrong answers: Choice A incorrectly states 14 hours, which would actually require a score higher than the minimum 85. Choice B gives 13 hours, which is close but represents an unnecessary rounding up that wasn't asked for in the problem. Choice C provides 11.5 hours, which would only yield a score of 45+3.2(11.5)=81.845 + 3.2(11.5) = 81.8—below the required 85. The correct answer is D because it gives the exact mathematical requirement of 12.5 hours per week and correctly notes this falls within the study's validity range. Study tip: When working with linear models and minimum requirements, always solve algebraically for the exact boundary value first, then check whether it falls within any given domain restrictions. Don't round unless the problem specifically asks for it.

Question 3

A small business's weekly revenue RR (in dollars) is modeled by R=1200+45nR = 1200 + 45n, where nn is the number of new customers acquired that week. Market research indicates that the business can realistically acquire between 5 and 30 new customers per week. If the business needs at least $2000 in weekly revenue to remain profitable, what constraint does this place on $nn $?

  1. n≥18n ≥ 18 customers per week (correct answer)
  2. n≥17.8n ≥ 17.8 customers per week
  3. n≥20n ≥ 20 customers per week
  4. n≥16n ≥ 16 customers per week
Explanation: Setting R≥2000R ≥ 2000: 1200+45n≥20001200 + 45n ≥ 2000, so 45n≥80045n ≥ 800, giving n≥17.78n ≥ 17.78. Since nn must be a whole number of customers, n≥18n ≥ 18. This is within the realistic range of 5-30 customers. Choice B gives the exact calculation but doesn't account for whole customers. Choice C uses an incorrect calculation. Choice D rounds down instead of up, which wouldn't meet the revenue requirement.

Question 4

The height hh (in feet) of water in a reservoir is given by h=45−0.8dh = 45 - 0.8d, where dd is the number of days since measurements began. The reservoir is considered at critical low levels when the height drops to 15 feet. What does the value 0.8 represent in the context of this problem?

  1. The daily rate of water level decrease in the reservoir, measured in feet per day (correct answer)
  2. The number of days required for the water level to drop by one foot
  3. The fraction of the reservoir's total capacity lost each day during the drought
  4. The minimum sustainable water level for the reservoir's ecosystem, measured in feet
Explanation: In h=45−0.8dh = 45 - 0.8d, the coefficient 0.8 represents the rate of change of height with respect to days. Since it's subtracted, it indicates the water level decreases by 0.8 feet per day. Choice B confuses rate with its reciprocal (1/0.8 = 1.25 days per foot). Choice C incorrectly interprets 0.8 as a percentage or fraction of capacity. Choice D confuses the slope with a water level measurement.

Question 5

A farmer's corn yield YY (in bushels per acre) is modeled by Y=120+15f−200Y = 120 + 15f - 200, where ff represents the pounds of fertilizer applied per acre. The model is valid for 2≤f≤122 ≤ f ≤ 12 pounds per acre. What is the practical interpretation of the constant term when this equation is simplified?

  1. The baseline corn yield of negative 80 bushels per acre, representing a theoretical loss condition
  2. The maximum possible yield of 280 bushels per acre when optimal fertilizer is applied
  3. The yield intercept of negative 80 bushels per acre, indicating the model's limitation without fertilizer (correct answer)
  4. The break-even point of negative 80 bushels per acre where costs equal revenue from sales
Explanation: Simplifying: Y=120+15f−200=15f−80Y = 120 + 15f - 200 = 15f - 80. The constant term is -80, which represents the y-intercept when f=0f = 0. Since negative yield is impossible, this indicates the model is invalid without fertilizer, which aligns with the domain restriction f≥2f ≥ 2. Choice A incorrectly calls it 'baseline yield.' Choice B confuses it with maximum yield. Choice D incorrectly relates it to economic break-even analysis.

Question 6

A solar panel installation company models their monthly energy output EE (in kilowatt-hours) using E=1200+75s−15cE = 1200 + 75s - 15c, where ss is the number of sunny days and cc is the number of cloudy days in the month. The company operates in a region where s+c=30s + c = 30 (total days per month) and weather patterns ensure 10≤s≤2510 ≤ s ≤ 25.

Given the constraint that s+c=30s + c = 30, what is the simplified form of the energy model, and what does the coefficient of ss represent in this new form?

  1. E=1650+90sE = 1650 + 90s; the coefficient 90 represents the net energy gain per additional sunny day
  2. E=750+90sE = 750 + 90s; the coefficient 90 represents the net energy gain per additional sunny day (correct answer)
  3. E=1650+60sE = 1650 + 60s; the coefficient 60 represents the baseline energy production per sunny day
  4. E=750+60sE = 750 + 60s; the coefficient 60 represents the baseline energy production per sunny day
Explanation: Since s+c=30s + c = 30, we have c=30−sc = 30 - s. Substituting: E=1200+75s−15(30−s)=1200+75s−450+15s=750+90sE = 1200 + 75s - 15(30 - s) = 1200 + 75s - 450 + 15s = 750 + 90s. The coefficient 90 represents the net effect of one additional sunny day (which means one fewer cloudy day): gain of 75 from the sunny day plus gain of 15 from avoiding the cloudy day. Choice A has the wrong constant term. Choices C and D have incorrect coefficients.

Question 7

The distance dd (in miles) of a delivery truck from its depot is given by d=∣120−40t∣d = |120 - 40t|, where tt is time in hours. A linear approximation of this model for 0≤t≤30 ≤ t ≤ 3 is d=120−40td = 120 - 40t. What does the coefficient -40 represent in this linear approximation?

  1. The truck's acceleration of 40 mph per hour as it approaches the depot
  2. The truck's average speed of 40 mph regardless of direction during the time period
  3. The rate at which the truck's fuel efficiency decreases at 40 miles per gallon per hour
  4. The truck's constant velocity of 40 mph in the direction toward the depot (correct answer)
Explanation: When interpreting coefficients in linear models, you need to understand what each part of the equation represents in the real-world context. Here, the linear approximation d=120−40td = 120 - 40t describes distance from the depot over time. The coefficient -40 tells us how the distance changes with respect to time. Since distance is measured in miles and time in hours, this coefficient has units of miles per hour. The negative sign is crucial—it means the distance from the depot is decreasing at a rate of 40 miles per hour. In other words, the truck is moving toward the depot at a constant speed of 40 mph. Looking at the wrong answers: Choice A mentions acceleration, but acceleration would require a t2t^2 term in the equation, and the units would be mph per hour, not the simple rate we see here. Choice B incorrectly suggests this represents average speed "regardless of direction," but the negative sign specifically indicates direction—toward the depot, not away from it. Choice C introduces fuel efficiency, which isn't related to this distance-time relationship at all and would have completely different units (miles per gallon per hour makes no physical sense in this context). Choice D correctly identifies that -40 represents the truck's constant velocity of 40 mph toward the depot. The negative coefficient indicates the direction of movement (approaching rather than departing), while the magnitude gives the speed. Study tip: In linear models, always check the units of coefficients and pay attention to signs—they often indicate direction or whether a quantity is increasing or decreasing.

Question 8

An online retailer's shipping cost SS (in dollars) is modeled by S=3.50+0.75wS = 3.50 + 0.75w, where ww is the package weight in pounds. Due to carrier restrictions, packages cannot exceed 50 pounds, and the retailer offers free shipping when costs would exceed $25. What is the practical domain for this shipping cost model?

  1. 0<w≤28.670 < w ≤ 28.67 pounds, excluding packages with zero additional weight
  2. 0≤w≤500 ≤ w ≤ 50 pounds, based solely on the carrier's weight restriction
  3. 0≤w≤28.670 ≤ w ≤ 28.67 pounds, where both weight and cost constraints apply (correct answer)
  4. 0≤w<250 ≤ w < 25 pounds, ensuring shipping costs remain under the free shipping threshold
Explanation: Two constraints limit the domain: carrier weight limit (≤ 50 pounds) and free shipping policy (when S>25S > 25). Setting S=25S = 25: 3.50+0.75w=253.50 + 0.75w = 25, so w=28.67w = 28.67 pounds. Since 28.67 < 50, the cost constraint is more restrictive. The model applies for 0≤w≤28.670 ≤ w ≤ 28.67. Choice A incorrectly excludes w=0w = 0. Choice B ignores the free shipping policy. Choice D uses the dollar threshold instead of the weight threshold.

Question 9

A laboratory's chemical concentration CC (in mg/L) decreases according to C=250e−0.1tC = 250e^{-0.1t}, but for the first 10 hours, it's approximated linearly as C=250−22.75tC = 250 - 22.75t. If safety protocols require stopping the experiment when concentration falls below 50 mg/L, what does this constraint mean for the linear model's domain?

  1. The domain is unrestricted since both constraints can be satisfied independently
  2. The domain is restricted to 0≤t≤100 ≤ t ≤ 10 hours due to the linear approximation limit only
  3. The domain is restricted to 0≤t≤100 ≤ t ≤ 10 hours since the safety limit isn't reached within this timeframe
  4. The domain is restricted to 0≤t≤8.790 ≤ t ≤ 8.79 hours due to the safety constraint being more restrictive (correct answer)
Explanation: When you encounter a problem involving domain restrictions with multiple constraints, you need to identify which constraint is most limiting and determines the actual usable domain. The linear model C=250−22.75tC = 250 - 22.75t is only valid for the first 10 hours, giving us one potential restriction: 0≤t≤100 ≤ t ≤ 10. However, safety protocols add another constraint - the experiment must stop when concentration drops below 50 mg/L. To find when this occurs, solve: 50=250−22.75t50 = 250 - 22.75t. Subtracting 250 from both sides gives −200=−22.75t-200 = -22.75t, so t=20022.75≈8.79t = \frac{200}{22.75} ≈ 8.79 hours. Since the safety limit is reached at 8.79 hours (before the 10-hour linear approximation limit), the safety constraint is more restrictive and determines the domain: 0≤t≤8.790 ≤ t ≤ 8.79 hours. Choice A incorrectly assumes constraints can be satisfied independently - in reality, whichever is more restrictive governs the domain. Choice B only considers the linear approximation limit while ignoring the safety requirement. Choice C correctly identifies that the domain is within the 10-hour window but fails to recognize that the safety limit creates a more restrictive boundary at 8.79 hours, not 10 hours. The correct answer is D because the safety constraint (8.79 hours) is reached before the linear model's validity limit (10 hours), making it the determining factor. Study tip: When multiple constraints apply to a domain, always check which one is most restrictive - that's your actual domain limit.

Question 10

A city's population PP (in thousands) is projected by P=85+2.3tP = 85 + 2.3t, where tt is years since 2020. City planners note that infrastructure can support a maximum of 150,000 people, and the model assumes current growth trends continue. Based on these constraints, when will the city need to address infrastructure limitations?

  1. In the year 2049, which is approximately 29 years after the baseline
  2. In the year 2048, which is approximately 28 years after the baseline (correct answer)
  3. In the year 2047, which is approximately 27 years after the baseline
  4. In the year 2050, which is approximately 30 years after the baseline
Explanation: When you encounter a linear growth model like this one, you need to set up an equation where the population reaches the given limit, then solve for the time variable. The city starts with 85,000 people in 2020 and grows by 2,300 people per year. To find when the population hits the infrastructure limit of 150,000 people, set up the equation: 150=85+2.3t150 = 85 + 2.3t. Since the population is given in thousands, 150 represents 150,000 people. Solving for tt: subtract 85 from both sides to get 65=2.3t65 = 2.3t, then divide by 2.3 to find t=28.26t = 28.26 years. Since tt represents years since 2020, this means the infrastructure limit will be reached in 2020+28.26=2048.262020 + 28.26 = 2048.26, or during the year 2048. Choice A incorrectly uses 29 years instead of the calculated 28.26, leading to 2049. This might come from rounding 28.26 up to 29, but the question asks when the city will need to address limitations, which happens during 2048. Choice C uses 27 years, which would be rounding down and missing the actual timeline. Choice D uses 30 years, which significantly overshoots the calculation and would delay necessary infrastructure planning. The correct answer is B: 2048, approximately 28 years after baseline. Study tip: In linear growth problems, always check your units carefully and remember that reaching a limit partway through a year means you need to plan for that entire year, not round to the next one.

Question 11

The value VV of a car (in thousands of dollars) depreciates according to V=28−2.5tV = 28 - 2.5t, where tt is the age of the car in years. Insurance companies consider a car 'totaled' if its value drops below $3,000. In this model, what is the significance of the ratio $282.5\frac{28}{2.5} $?

  1. The average annual percentage of value lost due to depreciation over the car's lifetime
  2. The number of years until the car's value reaches zero, representing its complete depreciation timeline (correct answer)
  3. The number of years until the car is considered totaled by insurance companies
  4. The ratio of initial value to annual depreciation, indicating the car's value retention factor
Explanation: When you encounter a linear depreciation model like V=28−2.5tV = 28 - 2.5t, you're looking at how value changes over time at a constant rate. The key insight is recognizing what happens when you set the equation equal to zero to find when the car becomes completely worthless. To find when the car's value reaches zero, set V=0V = 0: 0=28−2.5t0 = 28 - 2.5t 2.5t=282.5t = 28 t=282.5=11.2 yearst = \frac{28}{2.5} = 11.2 \text{ years} This ratio 282.5\frac{28}{2.5} represents exactly this calculation—the time until complete depreciation. The numerator (28) is the initial value in thousands, and the denominator (2.5) is the annual depreciation rate in thousands per year. Answer A is incorrect because this ratio gives you a time period (years), not a percentage. The annual percentage would require comparing the depreciation rate to the current value each year. Answer C confuses two different thresholds. The car is "totaled" when its value drops below $3,000, which happens when $28−2.5t<328 - 2.5t < 3 ,givingus, giving us t>10t > 10 $ years. This is different from reaching zero value. Answer D misinterprets what the ratio represents. While it does involve initial value and depreciation rate, the result isn't a "retention factor"—it's a time measurement showing when value hits zero. Study tip: In linear depreciation problems, always identify what \frac{\text{initial value}}{\text{rate of change}} represents—it's the time to reach zero, not percentages or other ratios.

Question 12

A taxi company charges according to the linear model C(d)=3.50+1.25dC(d) = 3.50 + 1.25d, where CC is the total cost in dollars and dd is the distance traveled in miles. What does the coefficient 1.25 represent in the context of this problem?

  1. The total cost for traveling one mile from the starting point
  2. The additional cost per mile traveled beyond the base fare (correct answer)
  3. The average cost per mile for any trip length
  4. The minimum cost for any taxi ride regardless of distance
Explanation: In the linear model C(d) = 3.50 + 1.25d, the coefficient 1.25 represents the rate of change, which is the additional cost per mile traveled. The constant term 3.50 is the base fare. Choice A is incorrect because it describes the total cost, not the incremental cost. Choice C confuses the rate with average cost, which would vary by trip length. Choice D describes the constant term (3.50), not the coefficient of d.

Question 13

The linear model P(x)=−0.02x+85P(x) = -0.02x + 85 represents the percentage of battery remaining in a laptop computer, where xx is the number of minutes since the laptop was unplugged. Based on this model, after how many hours will the battery be completely drained, and what assumption about battery discharge does this model make?

  1. 70.8 hours; assumes battery discharge rate increases over time
  2. 70.8 hours; assumes battery discharge rate remains constant over time (correct answer)
  3. 42.5 hours; assumes battery discharge rate remains constant over time
  4. 42.5 hours; assumes battery discharge rate decreases over time
Explanation: The battery is completely drained when P(x) = 0, so -0.02x + 85 = 0, giving x = 4250 minutes = 70.8 hours. Since this is a linear model with a constant slope, it assumes the discharge rate remains constant over time. Choice A has the correct time but wrong assumption about the discharge pattern. Choices C and D incorrectly calculate the time, possibly confusing minutes with hours or making arithmetic errors.

Question 14

The temperature of a cooling object follows the linear model T(m)=180−3.2mT(m) = 180 - 3.2m, where TT is temperature in degrees Fahrenheit and mm is time in minutes after cooling begins. If room temperature is 68°F and the object cannot cool below this temperature, what is the practical range for this model?

  1. 68≤T≤11268 ≤ T ≤ 112
  2. 0≤T≤1800 ≤ T ≤ 180
  3. 0≤T≤1120 ≤ T ≤ 112
  4. 68≤T≤18068 ≤ T ≤ 180 (correct answer)
Explanation: When working with mathematical models in real-world contexts, you need to identify the practical domain and range — the values that actually make sense given the physical constraints of the situation. Let's analyze this cooling model step by step. The equation T(m)=180−3.2mT(m) = 180 - 3.2m tells us the object starts at 180°F when m=0m = 0. As time passes, the temperature decreases by 3.2 degrees per minute. However, the object cannot cool below room temperature (68°F), so we need to find when it reaches this limit. Setting T(m)=68T(m) = 68: 68=180−3.2m68 = 180 - 3.2m, which gives us 3.2m=1123.2m = 112, so m=35m = 35 minutes. After 35 minutes, the object reaches room temperature and stops cooling further. Therefore, the practical range spans from the minimum temperature (68°F) to the maximum temperature (180°F): 68≤T≤18068 ≤ T ≤ 180. Looking at the wrong answers: Choice A (68≤T≤11268 ≤ T ≤ 112) incorrectly uses 112 as the maximum temperature — this number appears in our calculation but represents the total temperature drop, not the maximum temperature. Choice B (0≤T≤1800 ≤ T ≤ 180) ignores the physical constraint that the object cannot cool below room temperature. Choice C (0≤T≤1120 ≤ T ≤ 112) combines both errors, using neither the correct minimum nor maximum. Study tip: In real-world modeling problems, always identify the physical constraints that limit your mathematical model. The equation alone doesn't tell the whole story — context determines the practical boundaries.

Question 15

A spring's length follows the linear model L(w)=12+0.8wL(w) = 12 + 0.8w, where LL is length in centimeters and ww is the weight attached in grams. If the spring breaks when it reaches 28 cm and has a natural length limit, which parameter would change if a stiffer spring were used, and how would it change?

  1. The coefficient 0.8 would increase, making the spring extend more per gram
  2. The constant 12 would decrease, making the natural length shorter
  3. The constant 12 would increase, making the natural length longer
  4. The coefficient 0.8 would decrease, making the spring extend less per gram (correct answer)
Explanation: When analyzing linear models for physical systems like springs, you need to understand what each parameter represents. In the equation L(w)=12+0.8wL(w) = 12 + 0.8w, the constant 12 represents the natural length of the spring (when no weight is attached), while the coefficient 0.8 represents how much the spring stretches per gram of weight added. A stiffer spring resists stretching more than a flexible spring. This means it would extend less for each gram of weight added. Therefore, the coefficient 0.8 would become smaller—perhaps 0.5 or 0.3—indicating less extension per gram. The natural length (12 cm) wouldn't change since that's determined by the spring's construction, not its stiffness. Answer D correctly identifies that the coefficient 0.8 would decrease, making the spring extend less per gram. Answer A gets the direction wrong—claiming the coefficient would increase, which would actually make the spring more flexible, not stiffer. Answer B incorrectly suggests the natural length would decrease; however, stiffness doesn't affect the unstretched length of the spring. Answer C also focuses on natural length but suggests it would increase, which is equally incorrect since natural length is independent of stiffness. For linear models in physics contexts, always identify what each parameter physically represents before determining how changes in the system would affect them. The slope coefficient typically represents a rate of change (like stretch per unit weight), while the constant often represents an initial or baseline value.

Question 16

The model V(t)=25000−3200tV(t) = 25000 - 3200t represents a car's value in dollars, where tt is age in years. The car owner plans to sell when the value drops to $8000. What does the slope tell us about the depreciation, and what assumption does this model make about the car's minimum value?

  1. Depreciates at 12.8% annually; assumes car value approaches zero asymptotically
  2. Depreciates $3200 annually; assumes car retains $8000 minimum value indefinitely
  3. Depreciates $3200 annually; assumes car value can become negative after 7.8 years (correct answer)
  4. Depreciates at 12.8% annually; assumes car value can become negative after 7.8 years
Explanation: When analyzing linear depreciation models like V(t)=25000−3200tV(t) = 25000 - 3200t, focus on what the slope and y-intercept tell you about the real-world situation, and consider what happens when you extend the model beyond reasonable bounds. The slope of −3200-3200 represents the rate of change in value per year. Since the units are dollars per year, this means the car loses exactly $3200 in value annually - this is linear depreciation, not percentage-based. To find when the value becomes problematic, set $V(t)=0V(t) = 0 :: 25000−3200t=025000 - 3200t = 0 ,whichgives, which gives t=7.8t = 7.8 $ years. After this point, the model predicts negative car values, which is unrealistic since cars can't have negative worth. Answer A incorrectly calculates percentage depreciation ( 3200 ÷ 25000 = 12.8% only applies to the first year) and wrongly suggests an asymptotic approach to zero, which would require an exponential model. Answer B correctly identifies the 3200annualdepreciationbutmisinterpretsthesellingpoint(3200 annual depreciation but misinterprets the selling point (8000) as a minimum value assumption - the owner's selling decision doesn't change what the model assumes. Answer D makes the same percentage error as A while correctly noting the negative value issue. Answer C correctly identifies both the $3200 annual depreciation and recognizes that this linear model unrealistically allows negative values after 7.8 years. Study tip: In linear depreciation problems, always check what happens when you extend the model - linear functions continue indefinitely in both directions, which often creates unrealistic predictions in real-world contexts.

Question 17

A water reservoir's depth is modeled by D(d)=45−0.75dD(d) = 45 - 0.75d, where DD is depth in feet and dd is days since the drought began. If the reservoir is considered critically low at 15 feet and completely dry at 0 feet, what are the practical domain and range, and what do they represent?

  1. Domain: 0≤d≤600 ≤ d ≤ 60, Range: 15≤D≤4515 ≤ D ≤ 45; represents 60 days with depth constraints applied
  2. Domain: 0≤d≤400 ≤ d ≤ 40, Range: 15≤D≤4515 ≤ D ≤ 45; represents 40 days from full to critically low
  3. Domain: 0≤d≤600 ≤ d ≤ 60, Range: 0≤D≤450 ≤ D ≤ 45; represents 60 days from full to empty (correct answer)
  4. Domain: 0≤d≤400 ≤ d ≤ 40, Range: 0≤D≤450 ≤ D ≤ 45; represents 40 days from critically low to full
Explanation: When working with practical domain and range problems, you need to identify the realistic constraints of the real-world situation, not just the mathematical possibilities of the function. Let's analyze what happens at key points. The function D(d)=45−0.75dD(d) = 45 - 0.75d starts at 45 feet when d=0d = 0 (beginning of drought). To find when the reservoir is completely empty, set D(d)=0D(d) = 0: 0=45−0.75d0 = 45 - 0.75d, so d=60d = 60 days. This gives us the domain: 0≤d≤600 ≤ d ≤ 60. For the range, the depth starts at 45 feet and decreases to 0 feet, so the range is 0≤D≤450 ≤ D ≤ 45. The correct answer is C because it captures the complete realistic scenario from when the drought begins until the reservoir is completely dry. Option A incorrectly restricts the range to 15≤D≤4515 ≤ D ≤ 45, ignoring that the reservoir can drop below the "critically low" threshold and continue decreasing to empty. Option B makes two errors: it stops the domain at 40 days (which corresponds to the critically low point, not empty) and restricts the range above 15 feet. Option D has the wrong domain (40 instead of 60 days) and describes the scenario backward—the reservoir goes from full to critically low, not the reverse. Remember: practical domain and range problems require you to consider all realistic values the variables can take, even if certain thresholds (like "critically low") are mentioned. These thresholds provide context but don't necessarily limit the mathematical boundaries.

Question 18

A water tank is being filled at a constant rate. The linear model h(t)=2.5t+15h(t) = 2.5t + 15 represents the height of water in the tank, where hh is measured in feet and tt is measured in hours after filling begins. If the tank has a maximum capacity corresponding to a height of 40 feet, what is the practical domain for this model?

  1. 0≤t≤100 ≤ t ≤ 10 (correct answer)
  2. 0≤t≤160 ≤ t ≤ 16
  3. 15≤t≤4015 ≤ t ≤ 40
  4. 0≤t≤250 ≤ t ≤ 25
Explanation: The practical domain represents the valid time interval for the model. At t = 0, the height is h(0) = 2.5(0) + 15 = 15 feet. The tank reaches maximum capacity when h(t) = 40, so 2.5t + 15 = 40, which gives t = 10 hours. Therefore, the domain is 0 ≤ t ≤ 10. Choice B incorrectly uses 16 as the maximum time. Choice C confuses domain (time) with range (height). Choice D incorrectly calculates the time as (40-15) = 25 without dividing by the rate.

Question 19

A company's profit model is given by P(n)=45n−2800P(n) = 45n - 2800, where PP is the profit in dollars and nn is the number of units sold. For this model to be realistic, what constraint must be placed on the range, and what does this constraint mean in business terms?

  1. P≥−2800P ≥ -2800; the company cannot lose more than $2800 regardless of sales (correct answer)
  2. P≥0P ≥ 0; the company will always operate at a profit or break-even point
  3. n≥62.2n ≥ 62.2; the company must sell at least 63 units to avoid bankruptcy
  4. n≥0n ≥ 0; the company cannot sell a negative number of units
Explanation: The range constraint P ≥ -2800 makes sense because when n = 0 (no sales), P = -2800, representing fixed costs. The company cannot lose more than its fixed costs. Choice B is incorrect because the company operates at a loss when n < 62.2. Choice C places a constraint on the domain (n), not the range (P). Choice D also constrains the domain rather than the range.

Question 20

A linear model for a car's fuel efficiency is E(s)=35−0.15sE(s) = 35 - 0.15s, where EE is fuel efficiency in miles per gallon and ss is speed in miles per hour. According to this model, what is the significance of the y-intercept, and what limitation does this reveal about the model's applicability?

  1. 35 mpg represents maximum efficiency; model invalid for speeds below 35 mph
  2. 35 mpg represents theoretical efficiency at zero speed; model unrealistic at very low speeds (correct answer)
  3. 35 mpg represents average efficiency; model invalid for speeds above 233 mph
  4. 35 mpg represents minimum efficiency; model unrealistic at very high speeds
Explanation: The y-intercept 35 represents the theoretical fuel efficiency when s = 0 (zero speed), which is unrealistic since cars don't achieve fuel efficiency when not moving. This reveals that the linear model breaks down at very low speeds. Choice A incorrectly calls it maximum efficiency. Choice C incorrectly identifies it as average efficiency and gives an irrelevant speed limit. Choice D incorrectly calls it minimum efficiency.