Interpreting Linear Relationships in Context

70 questions

Question 1Question

A commercial agricultural irrigation system distributes liquid fertilizer from a storage tank. The volume of fertilizer remaining in the tank, VV, in liters, mm minutes after the system is turned on is modeled by the equation:

V=1,80015(m3c)V = 1,800 - 15(m - 3c)

where cc is the number of times the system's nozzles are cleaned during the irrigation process. During each cleaning cycle, the flow of fertilizer is completely paused, and no fertilizer is distributed. Based on the model, what is the duration, in minutes, of a single cleaning cycle?

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Answer: 3

Answer

The duration of a single cleaning cycle is 3 minutes.
In the model V=1,80015(m3c)V = 1,800 - 15(m - 3c), the term (m3c)(m - 3c) represents the total time, in minutes, that the irrigation system is actively distributing fertilizer. Since the total elapsed time is mm minutes and the system pauses during cleaning cycles, the term 3c3c represents the total paused time for cc cleanings. Therefore, the duration of a single cleaning cycle is 3cc=3\frac{3c}{c} = 3 minutes.

Step-by-Step Solution

1
Analyze the structure of the equation to identify the meaning of each term.
The coefficient 15 is the active flow rate in liters per minute, and (m3c)(m - 3c) is the active distribution time in minutes.
To understand how the time variables affect the volume of fertilizer remaining.
2
Relate the total elapsed time to the active time and the paused time.
The total elapsed time is mm minutes, and the active time is m3cm - 3c minutes, meaning the system is paused for a total of 3c3c minutes.
To find the expression for the total duration of all cleaning cycles.
3
Determine the duration of a single cleaning cycle.
Since cc cleaning cycles result in a total pause of 3c3c minutes, each cycle lasts 3cc=3\frac{3c}{c} = 3 minutes.
To calculate the duration of one individual cleaning cycle.

Key Concept

Interpreting Linear Relationships in Context
Question 2Question

A tutor charges a total fee, CC, in dollars, for a tutoring session of hh hours according to the equation C=25h+40C = 25h + 40. What is the best interpretation of 4040 in this context?

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Answer: The flat sign-up fee, in dollars, charged by the tutor

Answer

The flat sign-up fee, in dollars, charged by the tutor
The constant term in the linear equation represents the value of the dependent variable when the independent variable is zero. In this context, when the tutoring hours are zero, the total fee is 40 dollars, representing the flat sign-up fee.

Step-by-Step Solution

1
Identify the structure of the linear equation
The equation C=25h+40C = 25h + 40 matches the slope-intercept form y=mx+by = mx + b.
The linear model consists of a variable term representing the hourly rate and a constant term representing the flat fee.
2
Determine the value of the constant term when the variable is zero
When h=0h = 0, the equation simplifies to C=40C = 40.
Setting the hours of tutoring to zero isolates the starting value of the function.
3
Interpret the starting value in the context of the problem
A session of 00 hours corresponds to the cost before any tutoring begins, which represents a flat initial sign-up fee of 4040.
Connecting the mathematical y-intercept to the real-world scenario identifies the interpretation of the constant.

Key Concept

Interpreting Linear Relationships in Context
Question 3Question

A shipping company offers two types of delivery services: standard and express. The total shipping cost, in dollars, for a package sent via standard service is modeled by the function C(w)=1.25w+bsC(w) = 1.25w + b_s, where ww is the weight of the package, in pounds, and bsb_s is a constant representing the flat handling fee. The total shipping cost, in dollars, for a package sent via express service is modeled by the function E(k)=mek+beE(k) = m_e k + b_e, where kk is the weight of the package, in kilograms, and beb_e is a constant representing the flat handling fee.

The rate of change of the express shipping cost with respect to the package's weight, in dollars per kilogram, is 2.42.4 times the rate of change of the standard shipping cost with respect to the package's weight, in dollars per pound. The flat handling fee for the express service is 4.504.50 dollars more than the flat handling fee for the standard service. If it costs 58.5058.50 dollars to ship a package weighing 26.426.4 pounds using the express service, what is the cost, in dollars, to ship a package weighing 2020 pounds using the standard service? (Assume 1 kilogram=2.2 pounds1\text{ kilogram} = 2.2\text{ pounds}.)

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Answer: 43

Answer

The cost to ship a package weighing 20 pounds using the standard service is 43 dollars.
The correct answer is obtained by first calculating the rate of change of the express shipping cost (2.4×1.25=3.002.4 \times 1.25 = 3.00 dollars per kilogram). Next, the weight of the package is converted to kilograms (26.4/2.2=1226.4 / 2.2 = 12 kilograms) to match the express cost function's variable. Using the given cost of 58.5058.50 dollars for the express shipment, the express handling fee is determined to be 58.503(12)=22.5058.50 - 3(12) = 22.50 dollars. The standard handling fee is then found by subtracting 4.504.50 dollars from the express fee (22.504.50=18.0022.50 - 4.50 = 18.00 dollars). Finally, the cost of a 2020-pound standard package is computed as 1.25(20)+18.00=43.001.25(20) + 18.00 = 43.00 dollars.

Step-by-Step Solution

1
Determine the rate of change for the express service (mem_e).
me=3.00m_e = 3.00 dollars per kilogram
The rate of change of the standard shipping cost is 1.251.25 dollars per pound. Since the rate of change for the express shipping cost is 2.42.4 times this rate, me=2.4×1.25=3.00m_e = 2.4 \times 1.25 = 3.00 dollars per kilogram.
2
Convert the weight of the express package from pounds to kilograms.
k=12k = 12 kilograms
The weight of the package is given as 26.426.4 pounds. Using the conversion 1 kilogram=2.2 pounds1\text{ kilogram} = 2.2\text{ pounds}, the weight in kilograms is 26.42.2=12\frac{26.4}{2.2} = 12 kilograms.
3
Find the express flat handling fee (beb_e) using the given cost of the express shipment.
be=22.50b_e = 22.50 dollars
We are given that the cost of shipping a 1212-kilogram package using the express service is 58.5058.50 dollars. Substituting these values into the express cost function: 3(12)+be=58.5036+be=58.50be=22.503(12) + b_e = 58.50 \Rightarrow 36 + b_e = 58.50 \Rightarrow b_e = 22.50 dollars.
4
Find the standard flat handling fee (bsb_s).
bs=18.00b_s = 18.00 dollars
The express handling fee is 4.504.50 dollars more than the standard handling fee: be=bs+4.5022.50=bs+4.50bs=18.00b_e = b_s + 4.50 \Rightarrow 22.50 = b_s + 4.50 \Rightarrow b_s = 18.00 dollars.
5
Calculate the cost to ship a 2020-pound package using the standard service.
43.0043.00 dollars
Using the standard cost function C(w)=1.25w+bsC(w) = 1.25w + b_s with w=20w = 20 and bs=18.00b_s = 18.00: C(20)=1.25(20)+18.00=25.00+18.00=43.00C(20) = 1.25(20) + 18.00 = 25.00 + 18.00 = 43.00 dollars.

Key Concept

Interpreting slope, y-intercept, and rates of change of linear functions in a real-world context with unit conversions.
Question 4Question

A worker at a distribution center packages boxes at a constant rate. The total number of boxes, BB, the worker has packaged hh hours after starting their shift can be modeled by the equation B=12h+15B = 12h + 15. According to the model, how many boxes were already packaged at the start of the worker's shift?

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Answer: 15

Answer

At the start of the worker's shift, 1515 boxes were already packaged.
In the linear model B=12h+15B = 12h + 15, the term 1515 is the constant term (y-intercept), which represents the value of BB when h=0h = 0. In this context, h=0h = 0 represents the start of the worker's shift. Therefore, 1515 boxes were already packaged at the start of the shift.

Step-by-Step Solution

1
Identify the value of hh that represents the start of the shift.
h=0h = 0
The variable hh represents the number of hours since the shift started, so the start of the shift corresponds to 00 hours.
2
Substitute h=0h = 0 into the given equation to find the value of BB.
B=15B = 15
Evaluating the equation at h=0h = 0 gives the initial number of packaged boxes, which is represented by the constant term of the linear equation.

Key Concept

Interpreting the y-intercept of a linear relationship in context
Question 5Question

A hiker begins a climb at an elevation of 1,2001,200 feet above sea level and climbs at a constant rate of 350350 feet per hour. The hiker's elevation, EE, in feet, tt hours after beginning the climb is given by the equation E=350t+1,200E = 350t + 1,200. Which of the following is the best interpretation of 350350 in this context?

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Answer: The increase in the hiker's elevation, in feet, for each hour of climbing

Answer

The increase in the hiker's elevation, in feet, for each hour of climbing
The coefficient of tt, which is 350350, represents the rate of change of the hiker's elevation with respect to time. Since elevation is measured in feet and time is measured in hours, this rate is 350350 feet per hour. Therefore, the value 350350 represents the increase in the hiker's elevation, in feet, for each hour of climbing.

Step-by-Step Solution

1
Identify the structure of the linear equation.
The equation E=350t+1,200E = 350t + 1,200 is in the slope-intercept form y=mx+by = mx + b, where m=350m = 350 is the slope and b=1,200b = 1,200 is the yy-intercept.
Linear equations in context have constant rates of change (slope) and starting values (yy-intercept).
2
Determine the meaning of the slope in context.
The slope 350350 represents the change in the dependent variable EE (elevation in feet) per unit change in the independent variable tt (time in hours).
The unit of the slope is the unit of the dependent variable divided by the unit of the independent variable, which is feet per hour.
3
Interpret the positive sign of the slope.
Since 350350 is positive, the elevation increases by 350350 feet for each hour of climbing.
A positive slope indicates a constant increase over time.

Key Concept

Interpreting slope in a linear context
Question 6Question

A logistics company operates two distribution centers, Center A and Center B. The daily operating cost, CAC_A, in dollars, at Center A when processing pp packages is given by the equation CA=1.75p+3,200C_A = 1.75p + 3,200. The daily operating cost, CBC_B, in dollars, at Center B when processing pp packages is given by the equation CB=2.25(p400)+2,800C_B = 2.25(p - 400) + 2,800, where p400p \geq 400. The difference in daily operating costs, DD, in dollars, between Center B and Center A is defined as D=CBCAD = C_B - C_A. For p400p \geq 400, this difference is modeled by the equation D=0.50p1,300D = 0.50p - 1,300. Which of the following is the best interpretation of the number 0.500.50 in this context?

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Answer: For each additional package processed, the daily operating cost at Center B increases by 0.500.50 dollars more than the daily operating cost at Center A.

Answer

For each additional package processed, the daily operating cost at Center B increases by 0.500.50 dollars more than the daily operating cost at Center A.
The equation D=0.50p1,300D = 0.50p - 1,300 models the difference in daily operating costs, DD, between Center B and Center A (CBCAC_B - C_A) as a function of the number of packages processed, pp. The coefficient of pp, which is 0.500.50, represents the slope of this linear relationship. This means that for each unit increase in pp (each additional package processed), the difference DD increases by 0.500.50 dollars. Since D=CBCAD = C_B - C_A, an increase in DD indicates that Center B's daily operating cost is increasing by 0.500.50 dollars more than Center A's daily operating cost for each additional package processed.

Step-by-Step Solution

1
Understand the meaning of the variables and the difference equation D=CBCAD = C_B - C_A.
The variable pp represents the number of packages processed, and DD represents the difference in daily operating costs between Center B and Center A.
Establishing the relationship between the independent variable and the dependent variable is necessary to interpret the slope.
2
Identify the slope in the linear equation D=0.50p1,300D = 0.50p - 1,300.
The equation is in slope-intercept form, y=mx+by = mx + b, where the slope mm is 0.500.50.
The slope represents the rate of change of the dependent variable DD with respect to the independent variable pp.
3
Interpret the rate of change in the context of the problem.
A slope of 0.500.50 means that for every increase of 11 in pp (each additional package processed), the value of DD increases by 0.500.50 dollars. Since D=CBCAD = C_B - C_A, an increase of 0.500.50 in DD means that CBC_B (Center B's cost) increases by 0.500.50 dollars more than CAC_A (Center A's cost).
This links the mathematical rate of change directly to the real-world difference in costs between the two centers.

Key Concept

Interpreting the slope of a combined linear relationship in context
Estimated Time:1m 30s
Question 7Question

An electric vehicle's battery is being charged. The charge of the battery, CC, as a percentage of its full capacity, tt minutes after the charging begins is modeled by the equation C=1.2t+18C = 1.2t + 18, where t60t \le 60. According to the model, what was the battery's charge percentage when the charging began?

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Answer: 18

Answer

The battery's charge percentage when the charging began was 18.
The linear relationship is given by the equation C=1.2t+18C = 1.2t + 18. The constant term in this linear model, 18, represents the y-intercept, which is the value of CC when t=0t = 0. In the context of this scenario, t=0t = 0 represents the time when the charging began, and CC represents the charge percentage. Substituting t=0t = 0 into the equation yields C=1.2(0)+18=18C = 1.2(0) + 18 = 18. Therefore, the battery's charge percentage was 18 when the charging began.

Step-by-Step Solution

1
Determine the value of the independent variable tt when charging began.
t=0t = 0
The initial state or the start of the charging process corresponds to a time of 0 minutes.
2
Substitute t=0t = 0 into the equation C=1.2t+18C = 1.2t + 18.
C=18C = 18
Evaluating the linear equation at t=0t = 0 yields the constant term, representing the initial charge percentage.

Key Concept

Interpreting the y-intercept of a linear relationship in context
Question 8Question

A custom t-shirt printing company charges a one-time design setup fee plus a fixed price for each t-shirt printed. The total cost, CC, in dollars, for an order of nn t-shirts is given by the equation C=15n+50C = 15n + 50. What does the number 5050 represent in this equation?

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Answer: The one-time design setup fee, in dollars

Answer

The one-time design setup fee, in dollars
The constant term 5050 in the linear equation C=15n+50C = 15n + 50 represents the initial value of the cost function when no t-shirts are ordered (n=0n = 0). In this context, this represents the one-time design setup fee.

Step-by-Step Solution

1
Identify the structure of the linear equation C=15n+50C = 15n + 50.
The equation is in the slope-intercept form y=mx+by = mx + b, where CC is the total cost, nn is the number of t-shirts, 1515 is the slope, and 5050 is the y-intercept.
Analyzing the linear equation structure helps map its parts to the contextual scenario.
2
Interpret the constant term 5050 within the context.
The constant term represents the value of CC when n=0n = 0, which is C=15(0)+50=50C = 15(0) + 50 = 50 dollars.
When no t-shirts are printed, the only cost incurred is the initial flat setup fee.

Key Concept

Interpreting the y-intercept of a linear relationship in context
Question 9Question

An oceanographic research vessel measures hydrostatic pressure during deep-sea dives. During a dive in temperate waters, the pressure PP, in atmospheres (atm), at a depth of dd meters below the surface is modeled by the linear equation P=ρd+1.0P = \rho d + 1.0, where ρ\rho is a constant representing the rate of change of pressure with respect to depth. At a depth of 150150 meters, the pressure is 16.016.0 atm. During a second dive in a high-salinity region, the rate of change of pressure with respect to depth is 25%25\% greater than the rate in the temperate waters. According to the model for the second dive, at what depth, in meters, is the pressure 26.026.0 atm?

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Answer: 200

Answer

200
To find the depth on the second dive where the pressure is 26.026.0 atm, we first find the slope of the linear model for the first dive. Given P=ρd+1.0P = \rho d + 1.0, substituting the point (150,16.0)(150, 16.0) gives 16.0=150ρ+1.016.0 = 150\rho + 1.0, which yields ρ=0.1\rho = 0.1 atm/m. The slope for the second dive is 25%25\% greater, so the new slope is 0.1×1.25=0.1250.1 \times 1.25 = 0.125 atm/m. The equation for the second dive is P=0.125d+1.0P = 0.125d + 1.0. Substituting P=26.0P = 26.0 yields 26.0=0.125d+1.026.0 = 0.125d + 1.0. Solving for dd gives 25.0=0.125d25.0 = 0.125d, which results in a depth of 200200 meters.

Step-by-Step Solution

1
Find the rate of change of pressure with respect to depth in temperate waters.
ρ=0.1\rho = 0.1 atm/m
Substitute the depth d=150d = 150 and pressure P=16.0P = 16.0 into the linear equation P=ρd+1.0P = \rho d + 1.0 to get 16.0=150ρ+1.016.0 = 150\rho + 1.0. Solving for ρ\rho yields 15.0=150ρ15.0 = 150\rho, which means ρ=0.1\rho = 0.1.
2
Calculate the rate of change of pressure with respect to depth for the second dive.
ρnew=0.125\rho_{\text{new}} = 0.125 atm/m
The rate of change for the second dive is 25%25\% greater than the rate of 0.10.1 atm/m from the first dive. Thus, ρnew=0.1×1.25=0.125\rho_{\text{new}} = 0.1 \times 1.25 = 0.125.
3
Calculate the depth where the pressure is 26.026.0 atm using the new rate of change.
d=200d = 200 meters
Set up the equation for the second dive: P=0.125d+1.0P = 0.125d + 1.0. Substitute P=26.0P = 26.0 to get 26.0=0.125d+1.026.0 = 0.125d + 1.0. Subtracting 1.01.0 from both sides gives 25.0=0.125d25.0 = 0.125d. Solving for dd gives d=25.00.125=200d = \frac{25.0}{0.125} = 200.

Key Concept

Interpreting and manipulating slope in a linear relationship context
Question 10Question

A scientist is monitoring the volume of liquid nitrogen in a storage tank. The volume VV, in liters, of liquid nitrogen remaining in the tank tt hours after a cooling system malfunction is modeled by the equation V=2401.5tV = 240 - 1.5t. After how many hours of malfunction will there be exactly 180180 liters of liquid nitrogen remaining in the tank?

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Answer: 40

Answer

The cooling system malfunction must last 40 hours for there to be exactly 180 liters of liquid nitrogen remaining in the tank.
To find the number of hours of malfunction until 180180 liters of liquid nitrogen remain in the tank, substitute 180180 for VV in the given model equation, yielding 180=2401.5t180 = 240 - 1.5t. Subtracting 240240 from both sides of the equation results in 60=1.5t-60 = -1.5t. Dividing both sides by 1.5-1.5 gives the final value of t=40t = 40 hours.

Step-by-Step Solution

1
Substitute the target volume of liquid nitrogen into the linear equation.
180=2401.5t180 = 240 - 1.5t
The question asks for the time tt when the remaining volume VV is exactly 180180 liters.
2
Isolate the term containing the variable by subtracting the initial constant volume from both sides.
60=1.5t-60 = -1.5t
Subtracting 240240 from both sides of the equation begins the process of isolating the variable tt.
3
Divide both sides of the equation by the rate coefficient to solve for time.
t=40t = 40
Dividing by 1.5-1.5 isolates the variable tt and gives the solution in hours.

Key Concept

Interpreting values and solving equations in linear contexts
Estimated Time:45s
Question 11Question

A manufacturing company uses the equation below to model the total daily cost, KK, in thousands of dollars, of operating a factory when xx units are produced:

K=0.08(x150)+22K = 0.08(x - 150) + 22

where 150x800150 \le x \le 800. The company plans to transition to a new production setup that will increase the daily fixed operating cost by $3,000\$3,000 but will decrease the cost to produce each unit by $20\$20. Which of the following equations best models the new total daily cost, KnewK_{\text{new}}, in thousands of dollars, to produce xx units under the new setup?

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Answer: Knew=0.06x+13K_{\text{new}} = 0.06x + 13

Answer

The equation that best models the new total daily cost is Knew=0.06x+13K_{\text{new}} = 0.06x + 13.
The correct equation is determined by first expressing the original cost equation in slope-intercept form: K=0.08x+10K = 0.08x + 10. This reveals that the original variable cost is 0.080.08 thousand dollars ($80\$80) per unit and the original fixed cost is 1010 thousand dollars ($10,000\$10,000). Decreasing the variable cost by $20\$20 (0.020.02 thousand dollars) results in a new slope of 0.060.06. Increasing the fixed cost by $3,000\$3,000 (33 thousand dollars) yields a new y-intercept of 1313. Thus, the new cost model is Knew=0.06x+13K_{\text{new}} = 0.06x + 13.

Step-by-Step Solution

1
Expand the original equation to identify the baseline fixed and variable costs.
K=0.08x12+22    K=0.08x+10K = 0.08x - 12 + 22 \implies K = 0.08x + 10
Converting the equation to slope-intercept form, y=mx+by = mx + b, allows direct identification of the variable cost per unit (slope, mm) and the fixed cost (y-intercept, bb).
2
Convert the baseline parameters to their corresponding contextual values in dollars.
Variable cost: 0.080.08 thousand dollars = $80\$80 per unit. Fixed cost: 1010 thousand dollars = $10,000\$10,000 daily.
Since the cost KK is in thousands of dollars, a coefficient of 0.080.08 represents 0.08×1,000=$800.08 \times 1,000 = \$80 per unit, and a constant of 1010 represents 10×1,000=$10,00010 \times 1,000 = \$10,000.
3
Apply the described operational changes to determine the new variable and fixed costs.
New variable cost: $80$20=$60\$80 - \$20 = \$60 per unit, or 0.060.06 thousand dollars. New fixed cost: $10,000+$3,000=$13,000\$10,000 + \$3,000 = \$13,000 daily, or 1313 thousand dollars.
The unit cost decreases by $20\$20, lowering it from $80\$80 to $60\$60. The fixed cost increases by $3,000\$3,000, raising it from $10,000\$10,000 to $13,000\$13,000.
4
Assemble the new linear equation using the updated parameters.
Knew=0.06x+13K_{\text{new}} = 0.06x + 13
Combining the new variable cost per unit (0.060.06) and the new fixed cost (1313) yields the new daily cost equation.

Key Concept

Converting and modifying linear equations in a real-world context by identifying and altering slopes and y-intercepts.
Estimated Time:3m 0s
Question 12Question

An architect is tracking the height of a new skyscraper under construction. The total height HH, in feet, of the building ww weeks after construction of the main frames began is modeled by the equation H=120+15wH = 120 + 15w. According to the model, by how many feet does the height of the building increase each week?

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Answer: 15

Answer

The height of the building increases by 15 feet each week.
In the linear model H=120+15wH = 120 + 15w, the variable HH represents the total height of the building in feet and the variable ww represents the number of weeks since construction began. The coefficient of the variable ww, which is 1515, is the slope of the equation and represents the constant rate of change of the height per week. Therefore, the height of the building increases by 1515 feet each week.

Step-by-Step Solution

1
Analyze the linear model equation.
The equation H=120+15wH = 120 + 15w represents the total height HH as a function of the number of weeks ww, where 1515 is the coefficient of the variable ww.
To determine what each part of the linear equation represents in the given context.
2
Interpret the coefficient of the independent variable ww.
The coefficient of ww is the slope of the linear equation, which is 1515.
The slope represents the constant rate of change, which is the weekly increase in height.

Key Concept

Interpreting the slope of a linear equation in context
Question 13Question

A botanist tracks the growth of a bamboo plant. The height hh, in inches, of the plant dd days after the tracking began can be modeled by the equation h=4.5d+12h = 4.5d + 12. Which of the following is the best interpretation of 1212 in this context?

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Answer: The initial height of the bamboo plant, in inches, when the tracking began.

Answer

The initial height of the bamboo plant, in inches, when the tracking began.
In the linear equation h=4.5d+12h = 4.5d + 12, the constant term 1212 represents the y-intercept. This corresponds to the value of the function hh when the independent variable dd is equal to 00. Since dd is the number of days since tracking began, d=0d = 0 is the start of tracking, meaning 1212 represents the initial height of the bamboo plant, in inches.

Step-by-Step Solution

1
Identify the component of the linear equation h=4.5d+12h = 4.5d + 12 to be interpreted.
The number to interpret is the constant term 1212.
We must determine the contextual meaning of the constant value in the linear model.
2
Find the mathematical meaning of the constant in the slope-intercept form y=mx+by = mx + b.
The constant term 1212 represents the y-intercept of the linear equation, which occurs when the independent variable d=0d = 0.
In a linear function, the constant represents the value of the dependent variable when the independent variable is zero.
3
Translate the mathematical definition to the real-world context of the problem.
Since dd represents the number of days after tracking began, d=0d = 0 represents the start of tracking. At d=0d = 0, h=12h = 12, which is the initial height of the bamboo plant in inches.
Relating the y-intercept to the context yields the initial state of the measured quantity.

Key Concept

Interpreting the y-intercept of a linear model in context
Estimated Time:45s
Question 14Question

A deep space communications satellite transmits a telemetry data file to a ground station on Earth. The remaining size of the file to be received at the ground station, SS, in megabytes (MB), can be modeled by the equation S=8504.5(t18)S = 850 - 4.5(t - 18), where tt is the number of seconds since the satellite initiated its transmission sequence, and t18t \ge 18. Which of the following is the best interpretation of the number 1818 in this context?

Show answer & explanation

Answer: The number of seconds after the transmission sequence is initiated before the ground station begins receiving the file.

Answer

The number of seconds after the transmission sequence is initiated before the ground station begins receiving the file.
The model S=8504.5(t18)S = 850 - 4.5(t - 18) is valid for t18t \ge 18. Substituting t=18t = 18 into the equation yields S=850S = 850 megabytes, which is the total size of the file before any data has been received. As tt increases beyond 1818, the remaining file size decreases at a rate of 4.54.5 megabytes per second. Therefore, the first 1818 seconds after the transmission sequence is initiated represent the time delay before the ground station starts receiving the file.

Step-by-Step Solution

1
Analyze the structure of the linear equation S=8504.5(t18)S = 850 - 4.5(t - 18) in context.
The variable SS represents the remaining file size in megabytes, and tt represents the time in seconds since the sequence was initiated. The model is defined only for t18t \ge 18.
Understanding the variables and constraints is the first step to interpreting the components of the equation.
2
Evaluate the equation at the boundary value t=18t = 18.
When t=18t = 18, S=8504.5(1818)=850S = 850 - 4.5(18 - 18) = 850 megabytes.
This determines the starting state of the data reception modeled by the equation.
3
Analyze how SS changes as tt increases beyond 1818.
For every second tt increases beyond 1818, SS decreases by 4.54.5 megabytes.
This confirms that the transmission starts at t=18t = 18 seconds and proceeds at a rate of 4.54.5 megabytes per second, meaning the first 1818 seconds represent the delay before data reception starts.

Key Concept

Interpreting Linear Relationships in Context
Question 15Question

A landscaping service uses a water tank to irrigate lawns. The volume of water, VV, in gallons, remaining in the tank after nn lawns have been irrigated is modeled by the equation V=85025nV = 850 - 25n. According to the model, by how many gallons does the volume of water in the tank decrease for each lawn that is irrigated?

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Answer: 25

Answer

The correct answer is 25, which represents the decrease in the volume of water in the tank, in gallons, for each lawn irrigated.
In the linear equation V=85025nV = 850 - 25n, the coefficient of nn is 25-25. This coefficient represents the rate of change of the volume of water with respect to the number of lawns irrigated. The negative sign shows that the volume is decreasing, and the magnitude, 25, indicates that the volume decreases by 25 gallons for each lawn irrigated.

Step-by-Step Solution

1
Identify the coefficient of the variable nn in the equation V=85025nV = 850 - 25n.
The coefficient of nn is 25-25.
The coefficient of the independent variable in a linear equation represents the rate of change of the dependent variable.
2
Interpret the coefficient in terms of the real-world context.
The coefficient 25-25 means the volume of water decreases by 25 gallons for each lawn irrigated.
The negative sign indicates a decrease, and the magnitude represents the amount of change per unit.

Key Concept

Interpreting the slope of a linear relationship in context.
Question 16Question

An environmental cleanup crew is removing a contaminant from a soil site. The remaining mass of the contaminant, CC, in kilograms, after dd days of treatment is modeled by the equation C=400pdC = 400 - p d, where pp is the daily removal rate, in kilograms per day, under the original protocol. Under a new treatment protocol, the daily removal rate is increased by 25%25\%, and the treatment time required to completely remove the contaminant is reduced by 88 days. What was the daily removal rate, in kilograms per day, under the original protocol?

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Answer: 10

Answer

The daily removal rate under the original protocol was 10 kilograms per day.
Under the original protocol, complete removal of the 400400 kg contaminant occurs when C=0C = 0, giving a duration of d=400pd = \frac{400}{p} days. Under the new protocol, the removal rate increases by 25%25\% to 1.25p1.25p, and the duration is reduced by 88 days to 400p8\frac{400}{p} - 8. Because the total mass removed must still equal 400400 kg, we write the equation (1.25p)(400p8)=400(1.25p)\left(\frac{400}{p} - 8\right) = 400. Distributing 1.25p1.25p yields 50010p=400500 - 10p = 400. Solving for pp gives 10p=10010p = 100, which simplifies to p=10p = 10 kilograms per day.

Step-by-Step Solution

1
Set C=0C = 0 in the original equation to represent complete removal.
0=400pd    d=400p0 = 400 - p d \implies d = \frac{400}{p}
Complete removal of the contaminant means that the remaining mass CC is 00 kilograms.
2
Express the new daily removal rate and the new treatment duration using the given percentage increase and day reduction.
pnew=1.25pp_{\text{new}} = 1.25p and dnew=d8=400p8d_{\text{new}} = d - 8 = \frac{400}{p} - 8
The new protocol increases the daily removal rate by 25%25\% and reduces the total treatment time by 88 days.
3
Set up the equation for complete removal under the new protocol using the new rate and duration.
400(1.25p)(400p8)=0400 - (1.25p) \left(\frac{400}{p} - 8\right) = 0
The total initial contaminant mass of 400400 kilograms must be completely removed by the new daily rate over the new duration.
4
Solve the equation for pp.
1.25p(400p8)=400    50010p=400    10p=100    p=101.25p \left(\frac{400}{p} - 8\right) = 400 \implies 500 - 10p = 400 \implies 10p = 100 \implies p = 10
Distribute 1.25p1.25p into the parentheses to eliminate the fraction, then isolate the variable pp.

Key Concept

Interpreting the rate (slope) and intercepts of a linear relationship in context, and modeling variations of those parameters.
Question 17Question

An industrial oven is used in a bakery. The temperature of the oven chamber, CC, in degrees Fahrenheit (F^\circ\text{F}), tt minutes after the heating element is turned on is modeled by the linear equation:

C=18.5t+72C = 18.5t + 72

After a system upgrade, the starting temperature of the oven is 8F8^\circ\text{F} warmer, and the rate at which the oven heats up is 20%20\% faster. During a test of the upgraded oven, the heating element is turned on for 1515 minutes, after which the oven is turned off and cools down at a constant rate of 12F12^\circ\text{F} per minute. If the cooling process is also linear, which of the following functions models the temperature of the upgraded oven, UU, in degrees Fahrenheit, mm minutes after it is turned off?

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Answer: U(m)=12m+413U(m) = -12m + 413

Answer

The function U(m)=12m+413U(m) = -12m + 413 models the temperature of the upgraded oven, UU, in degrees Fahrenheit, mm minutes after it is turned off.
To find the temperature model during the cooling phase, we must first determine the state of the oven when the cooling begins. The upgraded oven has a starting temperature of 72+8=80F72 + 8 = 80^\circ\text{F} and a heating rate of 18.5×1.20=22.2F18.5 \times 1.20 = 22.2^\circ\text{F} per minute. After 1515 minutes of heating, the temperature reaches 22.2×15+80=413F22.2 \times 15 + 80 = 413^\circ\text{F}. When the oven is turned off at m=0m = 0 minutes, its temperature is 413F413^\circ\text{F}, which serves as the y-intercept of the cooling function. Since the temperature decreases at a constant rate of 12F12^\circ\text{F} per minute, the rate of change (slope) is 12-12. Therefore, the linear model is the function showing a rate of change of 12-12 and a starting value of 413413.

Step-by-Step Solution

1
Determine the upgraded starting temperature and heating rate of the oven.
The upgraded starting temperature is 80F80^\circ\text{F} and the upgraded heating rate is 22.2F22.2^\circ\text{F} per minute.
The original starting temperature of 72F72^\circ\text{F} is increased by 8F8^\circ\text{F} to get 72+8=80F72 + 8 = 80^\circ\text{F}. The original heating rate (slope) of 18.5F18.5^\circ\text{F} per minute is increased by 20%20\%, which is calculated as 18.5×1.20=22.2F18.5 \times 1.20 = 22.2^\circ\text{F} per minute.
2
Calculate the temperature of the upgraded oven at the moment it is turned off.
The temperature is 413F413^\circ\text{F} at t=15t = 15 minutes.
The heating phase is modeled by the linear relationship H(t)=22.2t+80H(t) = 22.2t + 80. Substituting t=15t = 15 yields H(15)=22.2(15)+80=333+80=413FH(15) = 22.2(15) + 80 = 333 + 80 = 413^\circ\text{F}.
3
Construct the linear function for the cooling phase.
U(m)=12m+413U(m) = -12m + 413
When the oven is turned off (m=0m = 0), its temperature is 413F413^\circ\text{F}. Since it cools down at a constant rate of 12F12^\circ\text{F} per minute, the slope of the cooling function is 12-12. Thus, the linear model is U(m)=12m+413U(m) = -12m + 413.

Key Concept

Interpreting and modifying parameters of linear models in multi-stage contextual scenarios.

Alternative Method

Instead of writing the heating function explicitly, you can calculate the total temperature increase directly: the temperature rises by 18.5×1.20=22.2F18.5 \times 1.20 = 22.2^\circ\text{F} per minute for 1515 minutes, which is a total increase of 22.2×15=333F22.2 \times 15 = 333^\circ\text{F}. Adding this increase to the upgraded starting temperature of 72+8=80F72 + 8 = 80^\circ\text{F} gives the peak temperature of 80+333=413F80 + 333 = 413^\circ\text{F}. Since the cooling phase is linear with a slope of 12-12 and a y-intercept of 413413, the function is immediately determined.
Estimated Time:3m 0s
Question 18Question

To rent an electric scooter, a rider pays a flat unlocking fee plus a fee for each minute of the ride. The total cost CC, in dollars, for a ride of mm minutes is given by the equation C=0.15m+1.20C = 0.15m + 1.20. Which of the following is the best interpretation of the number 1.201.20 in this context?

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Answer: The flat unlocking fee, in dollars, to rent the scooter

Answer

The flat unlocking fee, in dollars, to rent the scooter
In the linear model C=0.15m+1.20C = 0.15m + 1.20, the constant term 1.201.20 represents the value of CC when m=0m = 0. Since mm represents the number of minutes, m=0m = 0 corresponds to the beginning of the rental. Therefore, 1.201.20 represents the initial flat unlocking fee, in dollars, to rent the scooter.

Step-by-Step Solution

1
Identify the component of the linear equation C=0.15m+1.20C = 0.15m + 1.20 that corresponds to the number 1.201.20.
The number 1.201.20 is the constant term (y-intercept) of the equation.
A linear equation in slope-intercept form is y=mx+by = mx + b, where bb is the y-intercept (the value of yy when x=0x = 0).
2
Determine the physical meaning of m=0m = 0 in the given context.
m=0m = 0 represents a ride that lasts 00 minutes, meaning no time has elapsed yet.
Evaluating the relationship at the initial state helps identify the physical meaning of the y-intercept.
3
Substitute m=0m = 0 into the equation to find the corresponding cost.
C=0.15(0)+1.20=1.20C = 0.15(0) + 1.20 = 1.20 dollars.
This shows that the initial cost, or flat unlocking fee before starting the ride, is 1.201.20 dollars.

Key Concept

Interpreting the y-intercept of a linear relationship in context
Question 19Question

During a chemical reaction, the temperature TT, in degrees Celsius, of a solution ss seconds after the reaction begins is modeled by the equation T=0.04(s150)+92T = -0.04(s - 150) + 92, where 150s900150 \le s \le 900. According to the model, how many minutes does it take for the temperature of the solution to decrease by 1212 degrees Celsius?

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Answer: 5

Answer

5
To find the number of minutes it takes for the temperature to decrease by 12C12^\circ\text{C}, we first determine the rate of temperature change from the linear model. The equation is given in the form T=m(ss0)+T0T = m(s - s_0) + T_0, where the slope m=0.04m = -0.04 represents the rate of change of temperature in degrees Celsius per second. Thus, the temperature decreases at a rate of 0.04C0.04^\circ\text{C} per second. To achieve a total decrease of 12C12^\circ\text{C}, the time in seconds required is 120.04=300\frac{12}{0.04} = 300 seconds. Converting 300300 seconds to minutes gives 30060=5\frac{300}{60} = 5 minutes.

Step-by-Step Solution

1
Identify the rate of change from the linear equation.
The rate of temperature decrease is 0.04C0.04^\circ\text{C} per second.
The slope of the linear equation T=0.04(s150)+92T = -0.04(s - 150) + 92 is 0.04-0.04, which represents a change of 0.04C-0.04^\circ\text{C} for every 11 second increase in time.
2
Calculate the time in seconds for a decrease of 12C12^\circ\text{C}.
300300 seconds
Divide the target temperature change of 12C-12^\circ\text{C} by the rate of change of 0.04C-0.04^\circ\text{C} per second: 120.04=300\frac{-12}{-0.04} = 300 seconds.
3
Convert the time from seconds to minutes.
55 minutes
Since there are 6060 seconds in 11 minute, divide 300300 seconds by 6060: 30060=5\frac{300}{60} = 5 minutes.

Key Concept

Interpreting Linear Relationships in Context
Question 20Question

A custom t-shirt printing company uses the function C(n)=12n+45C(n) = 12n + 45 to determine the total cost, in dollars, for an order of nn t-shirts. What is the best interpretation of the value 4545 in this function?

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Answer: The setup fee, in dollars, charged for the entire order regardless of the number of t-shirts printed.

Answer

The setup fee, in dollars, charged for the entire order regardless of the number of t-shirts printed.
In the linear function C(n)=12n+45C(n) = 12n + 45, the term 4545 is a constant that does not depend on the variable nn (the number of t-shirts). Therefore, it represents the initial or flat setup cost in dollars for placing an order, which is charged regardless of the number of t-shirts printed.

Step-by-Step Solution

1
Identify the structure of the linear function C(n)=12n+45C(n) = 12n + 45.
The function is in the slope-intercept form y=mx+by = mx + b, where m=12m = 12 is the slope and b=45b = 45 is the y-intercept.
Understanding the components of a linear function helps determine which part of the equation corresponds to which real-world quantity.
2
Determine the meaning of the constant term (y-intercept) in the context of the problem.
The constant term 4545 represents the value of C(n)C(n) when n=0n = 0, which is the cost before any t-shirts are added, representing a flat setup fee.
Evaluating the function at n=0n = 0 isolates the base cost or initial charge.

Key Concept

Interpreting Linear Relationships in Context
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