Proportional Power

Score: 4 Points Total

2

Ocean Research Submersible (Graph Interpretation) — Solutions

4 Points (1 pt each)

Submersible dives at steady rate. Plotted points: \((0,0), (3, 75), (6, 150), (9, 225)\).

Time (min) Depth (m) 0 3 6 9 0 75 150 225 (3, 75) (6, 150) (9, 225) (1, 25) Red point marks unit rate \((1, 25)\)

a) Identify ordered pair and compute \(k = \frac{y}{x}\):

Using point \((3, 75)\): \(k = \frac{75\text{ m}}{3\text{ min}} = \mathbf{25\text{ m/min}}\)

Also accept: \((6, 150) \implies \frac{150}{6} = 25\) or \((9, 225) \implies \frac{225}{9} = 25\).

Answer: \(k = 25\text{ m/min}\)

b) Formulate equation relating depth (\(y\)) to time (\(x\)):

Substitute \(k = 25\) into \(y = kx\):

\(\mathbf{y = 25x}\)

(\(y\) in meters, \(x\) in minutes)

c) Depth after 18.5 minutes of diving:

Substitution: \(y = 25(18.5)\)

Arithmetic: \(25 \times 18.5 = 462.5\)

Calculated Depth: \(462.5\text{ meters}\)

d) Real-world meaning of the point \((1, k) = (1, 25)\):

Key Explanation Requirement:

The point \((1, 25)\) represents the unit rate of descent: In exactly 1 minute, the research submersible descends a depth of 25 meters below sea level (or dives at a speed of \(25\text{ m/min}\)).

Accept any equivalent explanation that explicitly connects 1 minute to 25 meters.

Scoring Guide: 1 pt for ordered pair & correct \(k\) with units; 1 pt for \(y = 25x\); 1 pt for \(462.5\text{ m}\); 1 pt for complete conceptual interpretation of \((1, 25)\).

Constant Ratio Lab • Teacher Answer Key Page 2 of 6

QUESTION 3 KEY Electric Delivery Van Charging (Table Analysis)

Score: 4 Points Total

3

Electric Delivery Van Charging (Table Analysis) — Solutions

4 Points (1 pt each)

Logistics DC fast charger delivering energy (\(y\), in kWh) over charging time (\(x\), in minutes). Data pairs: \((5, 18), (10, 36), (15, 54), (20, 72)\).

a) Find constant of proportionality \(k = \frac{y}{x}\) with units:

Formula: \(k = \frac{18\text{ kWh}}{5\text{ min}} = \mathbf{3.6\text{ kWh/min}}\)

Answer: \(k = 3.6\text{ kWh/min}\) (or \(\frac{18}{5}\) kWh/min)

b) Write the equation in \(y = kx\) form:

Substitute \(k = 3.6\) into \(y = kx\):

\(\mathbf{y = 3.6x}\)

(also accept \(y = \frac{18}{5}x\))

c) Energy (\(y\)) delivered in 45 minutes:

Step 1: Given input: \(x = 45\text{ min}\).

Step 2: Substitute: \(y = 3.6(45)\).

Step 3: Compute: \(3.6 \times 45 = 162\).

Total Energy: \(162\text{ kWh}\)

d) Time (\(x\)) for 108 kWh full charge:

Step 1: Given output: \(y = 108\text{ kWh}\).

Step 2: Set up: \(108 = 3.6x\).

Step 3: Solve for \(x\): \(x = \frac{108}{3.6} = 30\).

Charging Time: \(30\text{ minutes}\)

Scoring Guide: 1 pt for \(k = 3.6\text{ kWh/min}\); 1 pt for \(y = 3.6x\); 1 pt for \(y = 162\text{ kWh}\); 1 pt for \(x = 30\text{ min}\).

Constant Ratio Lab • Teacher Answer Key Page 3 of 6

QUESTION 4 KEY Offshore Wind Turbine Generation (Graph Interpretation)

Score: 4 Points Total

4

Offshore Wind Turbine Generation (Graph Interpretation) — Solutions

4 Points (1 pt each)

Wind turbine energy (\(y\), in MWh) over operating time (\(x\), in hours). Points: \((0,0), (2, 8), (4, 16), (6, 24)\).

Time (hours) Energy (MWh) 0 2 4 6 8 0 8 16 24 (2, 8) (4, 16) (6, 24) (1, 4) Red point marks unit rate \((1, 4)\)

a) Identify ordered pair and compute \(k = \frac{y}{x}\):

Using point \((2, 8)\): \(k = \frac{8\text{ MWh}}{2\text{ hours}} = \mathbf{4\text{ MWh/hour}}\)

Also accept: \((4, 16) \implies \frac{16}{4} = 4\) or \((6, 24) \implies \frac{24}{6} = 4\).

Answer: \(k = 4\text{ MWh/hour}\)

b) Formulate equation relating energy (\(y\)) to time (\(x\)):

Substitute \(k = 4\) into \(y = kx\):

\(\mathbf{y = 4x}\)

(\(y\) in MWh, \(x\) in hours)

c) Energy produced after 26.5 operating hours:

Substitution: \(y = 4(26.5)\)

Calculation: \(4 \times 26.5 = 106\)

Calculated Energy: \(106\text{ MWh}\)

d) Real-world meaning of the point \((1, k) = (1, 4)\):

Key Explanation Requirement:

The point \((1, 4)\) represents the unit rate: In 1 operating hour, the wind turbine generates 4 megawatt-hours (MWh) of electricity (or produces power at an hourly rate of \(4\text{ MWh/h}\)).

Accept any equivalent explanation that explicitly connects 1 hour to 4 MWh.

Scoring Guide: 1 pt for ordered pair & correct \(k\) with units; 1 pt for \(y = 4x\); 1 pt for \(106\text{ MWh}\); 1 pt for complete conceptual interpretation of \((1, 4)\).

Constant Ratio Lab • Teacher Answer Key Page 4 of 6

QUESTION 5 KEY Additive Manufacturing (Verbal Description)

Score: 4 Points Total

5

Additive Manufacturing (Verbal Description) — Solutions

4 Points (1 pt each)

3D printing lab produces prosthetic limbs: uses exactly 150 grams of filament every 25 minutes. Unit rate: \(k = \frac{150}{25} = 6\text{ g/min}\).

a) Completed Proportional Table:

b) Determine unit rate \(k\) and write the \(y = kx\) equation:

Unit rate: \(k = \frac{150\text{ g}}{25\text{ min}} = \mathbf{6\text{ g/min}}\)

Linear Equation: \(\mathbf{y = 6x}\)

where \(y\) = filament consumed in grams, \(x\) = print time in minutes.

c) Time to use 450 g of filament:

Step 1: Given output: \(y = 450\text{ g}\).

Step 2: Set up equation: \(450 = 6x\).

Step 3: Solve for \(x\): \(x = \frac{450}{6} = 75\).

Total Minutes: \(75\text{ minutes}\) (1 hr 15 min)

d) Grams used in 210 minutes:

Step 1: Given input: \(x = 210\text{ minutes}\).

Step 2: Substitute: \(y = 6(210)\).

Step 3: Multiply: \(6 \times 210 = 1,260\).

Total Filament: \(1,260\text{ grams}\) (1.26 kg)

Scoring Guide: 1 pt for completing table rows (60, 240, 360); 1 pt for \(k = 6\) & \(y = 6x\); 1 pt for \(x = 75\text{ min}\); 1 pt for \(y = 1,260\text{ g}\).

Constant Ratio Lab • Teacher Answer Key Page 5 of 6

QUESTION 6 KEY Clean Energy Challenge (Comparing Rates)

Score: 4 Points Total

6

Clean Energy Challenge (Comparing Rates) — Solutions

4 Points (1 + 1 + 2 pts)

Comparing solar productivity: Array Alpha (36 kWh in 4 h) vs. Array Beta (45 kWh in 6 h).

Array Alpha

36 kWh in 4 hours

\(k_A = \frac{36}{4} = \mathbf{9\text{ kWh/h}}\)

Array Beta

45 kWh in 6 hours

\(k_B = \frac{45}{6} = \mathbf{7.5\text{ kWh/h}}\)

a) Equations in \(y = kx\) form:

Array Alpha: \(\mathbf{y = 9x}\)

Array Beta: \(\mathbf{y = 7.5x}\) (\(y = \frac{15}{2}x\))

b) Rate comparison:

Array Alpha produces more energy per hour.

Difference: \(9 - 7.5 = \mathbf{1.5\text{ kWh/hour}}\) faster rate of production.

c) Operating hours for 180 kWh demand:

Array Alpha Calculation:

\(180 = 9x \implies x = \frac{180}{9}\)

Answer: \(\mathbf{20\text{ hours}}\)

Array Beta Calculation:

\(180 = 7.5x \implies x = \frac{180}{7.5}\)

Answer: \(\mathbf{24\text{ hours}}\)

Unit Mastery Scoring Guide (24 Points Total across 6 Pages)

4 Points per Problem

Exceeds (22-24 pts) Consistently calculates \(k\) with units, sets up \(y = kx\), solves multi-step, interprets \((1, k)\).

Proficient (18-21 pts) Correctly calculates \(k\) and linear equations; minor arithmetic or missing unit in application.

Developing (<18 pts) Inverts ratio (\(x/y\)) or needs scaffolding when interpreting graphs and unit rates.

Constant Ratio Lab • Teacher Answer Key Page 6 of 6

b) Write the linear equation: \(y = 25x\)

c) Depth after 18.5 minutes (\(x = 18.5\)): \(y = 25(18.5) = \mathbf{462.5\text{ meters}}\)

d) Meaning of unit point \((1, 25)\): In 1 minute, the sub dives 25 meters.

Proportional Power • Constant Ratio Lab Slide 4 of 9

3

Electric Delivery Van Charging (Table Analysis)

Worksheet Problem 3

DC fast charger delivers cumulative energy (\(y\), in kWh) over time (\(x\), in minutes):

a) Find unit rate \(k = \frac{y}{x}\): \(k = \frac{18\text{ kWh}}{5\text{ min}} = \mathbf{3.6\text{ kWh/min}}\)

b) Write the linear equation: \(y = 3.6x\)

c) Energy delivered in 45 minutes (\(x = 45\)): \(y = 3.6(45) = \mathbf{162\text{ kWh}}\)

d) Time required for 108 kWh full charge (\(y = 108\)): \(108 = 3.6x \implies x = \frac{108}{3.6} = \mathbf{30\text{ minutes}}\)

Proportional Power • Constant Ratio Lab Slide 5 of 9

4

Offshore Wind Turbine Generation (Graph Interpretation)

Worksheet Problem 4

Commercial wind turbine generates electricity at a steady rate. Output (\(y\), MWh) over time (\(x\), hours):

Time (hours) Energy (MWh) 0 2 4 6 0 8 16 24 (2, 8) (4, 16) (6, 24) (1, 4)

a) Find unit rate \(k\) from point \((2, 8)\): \(k = \frac{8\text{ MWh}}{2\text{ hours}} = \mathbf{4\text{ MWh/hour}}\)

b) Write the linear equation: \(y = 4x\)

c) Energy after 26.5 operating hours (\(x = 26.5\)): \(y = 4(26.5) = \mathbf{106\text{ MWh}}\)

d) Meaning of unit point \((1, 4)\): In 1 hour, the turbine generates 4 MWh.

Proportional Power • Constant Ratio Lab Slide 6 of 9

5

Additive Manufacturing (Verbal Description)

Worksheet Problem 5

3D printer extrudes filament at a constant rate: uses exactly 150 grams every 25 minutes.

a) Proportional Table:

b) Unit rate & linear equation:

\(k = \frac{150\text{ g}}{25\text{ min}} = \mathbf{6\text{ g/min}}\)

Equation: \(y = 6x\)

c) Minutes for 450 g model (\(y = 450\)): \(450 = 6x \implies x = \frac{450}{6} = \mathbf{75\text{ minutes}}\)

d) Filament used in 210 minutes (\(x = 210\)): \(y = 6(210) = \mathbf{1,260\text{ grams}}\)

Proportional Power • Constant Ratio Lab Slide 7 of 9

6

Clean Energy Challenge (Comparing Rates)

Worksheet Problem 6

Array Alpha

36 kWh in 4 hours

\(k_A = 9\text{ kWh/h}\)

\(y = 9x\)

Array Beta

45 kWh in 6 hours

\(k_B = 7.5\text{ kWh/h}\)

\(y = 7.5x\)

Rate Comparison:

Which solar array produces more energy per hour, and by how much?

Array Alpha produces more energy.
Difference: \(9 - 7.5 = \mathbf{1.5\text{ kWh/hour}}\) faster.

Demand: 180 kWh

Operating hours needed to reach 180 kWh demand:

Array Alpha: \(180 = 9x \implies x = \mathbf{20\text{ hours}}\)

Array Beta: \(180 = 7.5x \implies x = \mathbf{24\text{ hours}}\)

Proportional Power • Constant Ratio Lab Slide 8 of 9

Key Takeaways & Wrap-Up

Review & Debrief

1

Always Check \(k = \frac{y}{x}\)

Ensure orientation is output over input. Units matter! (e.g., km/min, m/min, kWh/h).

Avoid the common \(\frac{x}{y}\) invert trap.

2

Unit Point \((1, k)\)

On any proportional graph, the coordinate \((1, k)\) reveals the exact unit rate of the relationship.

The line always originates at \((0,0)\).

3

Power of \(y = kx\)

Use the equation to solve in both directions: plug in \(x\) to find \(y\), or set equal to \(y\) and divide to find \(x\).

Algebraic modeling for real applications.

Proportional Power • Constant Ratio Lab Slide 9 of 9

\(k = ?\)

b) Formulate equation: Write in \(y = kx\) form

\(y = ?\)

c) Depth at 18.5 min: Calculate for \(x = 18.5\)

\(y = ?\)

d) Meaning of \((1, k)\): What does the point \((1, k)\) reveal?

Proportional Power • Constant Ratio Lab Slide 4 of 9

3

Electric Delivery Van Charging (Table Analysis)

Worksheet Problem 3

DC fast charger delivers cumulative energy (\(y\), in kWh) over charging time (\(x\), in minutes):

a) Find constant of proportionality \(k = \frac{y}{x}\): State with kWh and min units

\(k = ?\)

b) Write the linear equation: Relate energy (\(y\)) to time (\(x\))

\(y = \underline{\hspace{15px}} x\)

c) Energy delivered in 45 minutes: Calculate \(y\) when \(x = 45\)

\(y = ?\)

d) Time required for 108 kWh full charge: Set \(y = 108\) and solve for \(x\)

\(x = ?\)

Proportional Power • Constant Ratio Lab Slide 5 of 9

4

Offshore Wind Turbine Generation (Graph Interpretation)

Worksheet Problem 4

Commercial wind turbine generates electricity at a steady rate. Output (\(y\), MWh) over time (\(x\), hours):

Time (hours) Energy (MWh) 0 2 4 6 0 8 16 24 (2, 8) (4, 16) (6, 24)

a) Find unit rate \(k\): Calculate ratio \(\frac{y}{x}\)

\(k = ?\)

b) Formulate equation: Express in \(y = kx\) form

\(y = ?\)

c) Energy at 26.5 h: Calculate for \(x = 26.5\)

\(y = ?\)

d) Meaning of \((1, k)\): What does the point \((1, k)\) reveal?

Proportional Power • Constant Ratio Lab Slide 6 of 9

5

Additive Manufacturing (Verbal Description)

Worksheet Problem 5

3D printer extrudes filament at a constant rate: uses exactly 150 grams every 25 minutes.

a) Complete Proportional Table:

b) Unit rate & linear equation: Determine grams/min & equation

\(y = ?\)

c) Minutes for 450 g model: Solve for \(x\) when \(y = 450\)

\(x = ?\)

d) Filament in 210 minutes: Calculate \(y\) when \(x = 210\)

\(y = ?\)

Proportional Power • Constant Ratio Lab Slide 7 of 9

6

Clean Energy Challenge (Comparing Rates)

Worksheet Problem 6

Array Alpha

Generates 36 kWh in 4 hours

\(k_A = ?\)

Array Beta

Generates 45 kWh in 6 hours

\(k_B = ?\)

a & b) Equations & Rate Comparison:

Write the \(y = kx\) equation for each array. Which produces more energy per hour, and by how much?

Array Alpha: \(y = \underline{\hspace{20px}} x\)

Array Beta: \(y = \underline{\hspace{20px}} x\)

c) Demand Target: 180 kWh

Operating hours needed to reach 180 kWh demand:

Array Alpha Hours: \(x = \underline{\hspace{25px}}\)

Array Beta Hours: \(x = \underline{\hspace{25px}}\)

Proportional Power • Constant Ratio Lab Slide 8 of 9

Key Takeaways & Reflection

Review & Debrief

1

Always Check \(k = \frac{y}{x}\)

Ensure orientation is output over input. Always include real-world units (e.g., km/min, m/min, kWh/h).

Avoid the common \(\frac{x}{y}\) inversion trap.

2

Unit Point \((1, k)\)

On any proportional graph, the point \((1, k)\) reveals the exact unit rate of the relationship.

The line always passes through \((0,0)\).

3

Power of \(y = kx\)

Solve in both directions: substitute \(x\) to find output \(y\), or set equal to \(y\) and divide by \(k\) to find input \(x\).

Direct algebraic modeling.

Proportional Power • Constant Ratio Lab Slide 9 of 9

Be specific about the numbers and their units.

Proportional Power • Constant Ratio Practice Worksheet Page 2 of 4

QUESTION 3 Solar Desalination Plant (Verbal Description)

Name:

| Date:

3

Solar Desalination Plant (Verbal Description)

4 Points

A coastal solar desalination facility purifies ocean water into clean drinking water at a constant rate, generating exactly 320 liters of fresh water every 8 hours of continuous sunlight operation.

a) Complete the proportional table:

b) Determine unit rate \(k\) and write the \(y = kx\) equation:

Find \(k = \frac{\text{liters}}{\text{hours}}\) and state the formula:

c) A community emergency station requires 920 liters of water. How many hours will this take?

Substitute \(y = 920\) into your equation and solve for \(x\):

Total Hours:

d) If the facility operates under sunlight for 38 hours, how many liters are produced?

Substitute \(x = 38\) and show your complete calculation:

Total Output:

Proportional Power • Constant Ratio Practice Worksheet Page 3 of 4

QUESTION 4 Hydroelectric Power Challenge (Comparing Rates)

Name:

| Date:

4

Hydroelectric Power Challenge (Comparing Rates)

4 Points

A civil engineering team analyzes the electricity generated by two hydroelectric dam turbine designs operating at maximum water flow:

Turbine Alpha

Generates 48 MWh of energy in 6 hours

Turbine Beta

Generates 63 MWh of energy in 7 hours

a) Write the \(y = kx\) equation for each turbine:

Turbine Alpha:

Turbine Beta:

b) Which turbine generates power at a greater unit rate, and by how much?

Compare the two unit rates and state the numerical difference:

c) A regional power grid requires 216 MWh. Calculate how many hours each turbine must run individually to supply this energy:

Turbine Alpha Work & Hours:

Turbine Beta Work & Hours:

Practice Completion Checklist: All 4 problems solved, units clearly labeled, work shown?

Ready to Submit

Proportional Power • Constant Ratio Practice Worksheet Page 4 of 4

b) Unit rate \(k\) & equation (\(y = kx\)):

c) Seconds (\(x\)) for 252 cm cut:

4

Geothermal Power Comparison (Comparing Rates)

5 Points

Engineers test two geothermal steam plants operating at peak thermal capacity:

Plant Alpha

Generates 84 GJ in 7 hours

Plant Beta

Generates 96 GJ in 6 hours

a) Equations: Alpha:

Beta:

b) Greater rate & diff:

Which is faster & by how much?

c) Hours for 360 GJ:

Solve for Beta (\(y = 360\)):

5

Rate Analysis & Modeling Critique (Conceptual Reasoning)

5 Points

A student analyzes drone rental charges: 2 h costs $65, 4 h costs $115, 6 h costs $165. The student claims: "This is proportional because both quantities increase by constant steps."

a) Test claim using \(k = \frac{y}{x}\):

Find \(\frac{y}{x}\) for all 3 pairs. Is it proportional?

b) Proportional model:

If rate is $28/h with no fee, write \(y = kx\) and find cost for 7 h:

Proficiency: 23-25: Exceeds • 19-22: Meets • 15-18: Developing • <15: Review

Teacher Signature:

Proportional Power • Constant Ratio Quiz Page 2 of 2

YES

NO

If yes, \(k =\) If no, why?

Proportional Power • RP2: Proportionality Assessment Page 1 of 4

QUESTION 5 Deep-Sea Trench Exploration (Graph Interpretation)

Name:

| Date:

5

Deep-Sea Trench Exploration (Graph Interpretation)

4 Points

An autonomous oceanographic submersible dives downward into an oceanic trench at a constant descent speed. The graph below displays depth below the surface (\(y\), in meters) over time (\(x\), in minutes).

Time (min) Depth (m) 0 4 8 12 16 0 50 100 150 200 (4, 50) (8, 100) (12, 150)

a) Identify one ordered pair \((x, y)\) on the line and compute unit rate \(k = \frac{y}{x}\):

b) Formulate the equation relating depth (\(y\)) to time (\(x\)) in \(y = kx\) form:

c) Calculate the vehicle's total depth after 26 minutes of diving:

Show your complete multiplication work and state appropriate units:

Calculated Depth:

d) Explain the real-world meaning of the point \((1, k)\) in this context:

What does the point \((1, 12.5)\) tell marine researchers?

Be specific about both quantities and their units.

Proportional Power • RP2: Proportionality Assessment Page 2 of 4

QUESTION 6 Vertical Hydroponic Farming (Verbal Description)

Name:

| Date:

6

Vertical Hydroponic Farming (Verbal Description)

4 Points

An automated vertical urban greenhouse dispenses liquid plant nutrient solution at a steady rate, utilizing exactly 210 liters of nutrient mixture every 14 hours of continuous automated cultivation.

a) Complete the proportional table:

b) Determine unit rate \(k\) and write the \(y = kx\) equation:

Find \(k = \frac{\text{liters}}{\text{hours}}\) and state the formula:

c) A nursery greenhouse section requires 480 L of solution. How many hours will this take?

Substitute \(y = 480\) into your equation and solve for \(x\):

Total Hours:

d) If the greenhouse runs continuously for 72 hours, how many liters are dispensed?

Substitute \(x = 72\) and show your complete calculation:

Total Solution:

Proportional Power • RP2: Proportionality Assessment Page 3 of 4

QUESTION 7 Electric Aviation Charging (Comparing Rates)

Name:

| Date:

7

Electric Aviation Charging (Comparing Rates)

4 Points

An electric airport terminal evaluates two fast-charging stations for regional cargo aircraft:

Charger Alpha

Delivers 75 kWh of energy in 5 minutes

Charger Beta

Delivers 108 kWh of energy in 6 minutes

a) Write the \(y = kx\) equation for each charger:

Charger Alpha:

Charger Beta:

b) Which charger supplies energy at a faster rate, and by how much?

Compare the two unit rates and state the numerical difference:

c) An electric airplane battery needs 270 kWh. Calculate how many minutes each charger takes:

Charger Alpha Work & Minutes:

Charger Beta Work & Minutes:

Assessment Completion Checklist: All problems solved, steps detailed, units labeled on answers?

Ready to Submit

Proportional Power • RP2: Proportionality Assessment Page 4 of 4

Constant ratio = 2.5 (or \(\frac{5}{2}\))!

Proportional? YES

Constant of proportionality: \(k = 2.5\) (\(y = 2.5x\))

4 Table D Solution NOT PROPORTIONAL

Calculated Ratios \(\frac{y}{x}\):

\(\frac{12}{3} = \mathbf{4}\); \(\frac{24}{6} = \mathbf{4}\); \(\frac{36}{9} = \mathbf{4}\)

\(\frac{45}{12} = \mathbf{3.75} \neq 4\)

The 4th pair breaks the constant ratio!

Proportional? NO

Why: Last pair has ratio 3.75, not 4. A table must maintain the constant ratio for ALL pairs.

Grading Page 1 (4 Pts): Award 1 pt per question (0.5 pt for correct ratio calculations, 0.5 pt for correct YES/NO decision with \(k\) or justification).

RP2: Proportionality Assessment • Teacher Answer Key Page 1 of 4

QUESTION 5 KEY Deep-Sea Trench Exploration (Graph Interpretation)

Score: 4 Points Total (1 pt each)

5

Deep-Sea Trench Exploration (Graph Interpretation) — Solutions

4 Points

Autonomous submersible diving downward at a constant speed. Plotted points on the line: \((0,0), (4, 50), (8, 100), (12, 150)\).

Time (min) Depth (m) 0 4 8 12 16 0 50 100 150 200 (4, 50) (8, 100) (12, 150) (1, 12.5) Red point marks unit rate \((1, 12.5)\)

a) Identify ordered pair and compute \(k = \frac{y}{x}\):

Using point \((4, 50)\): \(k = \frac{50\text{ m}}{4\text{ min}} = \mathbf{12.5\text{ m/min}}\)

Also accept: \((8, 100) \implies \frac{100}{8} = 12.5\) or \((12, 150) \implies \frac{150}{12} = 12.5\).

Answer: \(k = 12.5\text{ m/min}\) (or \(\frac{25}{2}\) m/min)

b) Formulate the equation relating depth (\(y\)) to time (\(x\)):

Substitute \(k = 12.5\) into \(y = kx\):

\(\mathbf{y = 12.5x}\)

(also accept \(y = \frac{25}{2}x\))

c) Total depth after 26 minutes of diving:

Substitution: \(y = 12.5(26)\)

Calculation: \(12.5 \times 26 = 325\)

Calculated Depth: \(325\text{ meters}\)

d) Real-world meaning of the point \((1, 12.5)\):

Key Explanation Requirement:

The point \((1, 12.5)\) represents the unit rate: In exactly 1 minute of dive time, the oceanographic submersible descends a depth of 12.5 meters below the surface (or dives at \(12.5\text{ m/min}\)).

Accept any equivalent explanation that explicitly connects 1 minute to 12.5 meters.

Scoring Guide: 1 pt for ordered pair & correct \(k\) with units; 1 pt for \(y = 12.5x\); 1 pt for \(325\text{ m}\); 1 pt for complete conceptual interpretation of \((1, 12.5)\).

RP2: Proportionality Assessment • Teacher Answer Key Page 2 of 4

QUESTION 6 KEY Vertical Hydroponic Farming (Verbal Description)

Score: 4 Points Total (1 pt each)

6

Vertical Hydroponic Farming (Verbal Description) — Solutions

4 Points

Greenhouse misting system delivers exactly 210 liters of nutrient solution every 14 hours. Unit rate: \(k = \frac{210}{14} = 15\text{ L/hour}\).

a) Completed Proportional Table:

b) Determine unit rate \(k\) and write the \(y = kx\) equation:

Unit rate: \(k = \frac{210\text{ L}}{14\text{ hours}} = \mathbf{15\text{ L/hour}}\)

Linear Equation: \(\mathbf{y = 15x}\)

where \(y\) = nutrient solution in liters, \(x\) = irrigation time in hours.

c) Time to dispense 480 liters of solution:

Step 1: Given output: \(y = 480\text{ L}\).

Step 2: Set up equation: \(480 = 15x\).

Step 3: Solve for \(x\): \(x = \frac{480}{15} = 32\).

Total Hours: \(32\text{ hours}\)

d) Liters dispensed in 72 hours:

Step 1: Given input: \(x = 72\text{ hours}\).

Step 2: Substitute: \(y = 15(72)\).

Step 3: Multiply: \(15 \times 72 = 1,080\).

Total Solution: \(1,080\text{ liters}\)

Scoring Guide: 1 pt for completing table rows (90, 330, 525); 1 pt for \(k = 15\) & \(y = 15x\); 1 pt for \(x = 32\text{ h}\); 1 pt for \(y = 1,080\text{ L}\).

RP2: Proportionality Assessment • Teacher Answer Key Page 3 of 4

QUESTION 7 KEY Electric Aviation Charging (Comparing Rates)

Score: 4 Points Total

7

Electric Aviation Charging (Comparing Rates) — Solutions

4 Points (1 + 1 + 2 pts)

Comparing charging speed: Charger Alpha (75 kWh in 5 min) vs. Charger Beta (108 kWh in 6 min).

Charger Alpha

75 kWh in 5 minutes

\(k_A = \frac{75}{5} = \mathbf{15\text{ kWh/min}}\)

Charger Beta

108 kWh in 6 minutes

\(k_B = \frac{108}{6} = \mathbf{18\text{ kWh/min}}\)

a) Equations in \(y = kx\) form:

Charger Alpha: \(\mathbf{y = 15x}\)

Charger Beta: \(\mathbf{y = 18x}\)

b) Rate comparison:

Charger Beta delivers energy at a faster rate.

Difference: \(18 - 15 = \mathbf{3\text{ kWh/minute}}\) faster charging speed.

c) Charging minutes for 270 kWh demand:

Charger Alpha Calculation:

\(270 = 15x \implies x = \frac{270}{15}\)

Answer: \(\mathbf{18\text{ minutes}}\)

Charger Beta Calculation:

\(270 = 18x \implies x = \frac{270}{18}\)

Answer: \(\mathbf{15\text{ minutes}}\)

Unit Mastery Scoring Guide (16 Points Total across 4 Pages)

4 Points per Page

Exceeds (15–16 pts) Calculates \(k\) with exact units, sets up \(y = kx\), tests all table rows, and thoroughly interprets \((1, k)\).

Proficient (13–14 pts) Correctly calculates \(k\) and equations; minor arithmetic flaw or omitted unit in real-world application.

Developing (<13 pts) Inverts ratio (\(x/y\)), checks only the first row in tables, or requires scaffolding for graph interpretation.

RP2: Proportionality Assessment • Teacher Answer Key Page 4 of 4

Proportional Power • RP2: Proportionality Assessment (Independent Form) Page 1 of 4

QUESTION 5 Deep-Sea Trench Exploration (Graph Interpretation)

Name:

| Date:

5

Deep-Sea Trench Exploration (Graph Interpretation)

4 Points

An autonomous oceanographic submersible dives downward into an oceanic trench at a constant descent speed. The graph below displays depth below the surface (\(y\), in meters) over time (\(x\), in minutes).

Time (min) Depth (m) 0 4 8 12 16 0 50 100 150 200 (4, 50) (8, 100) (12, 150)

a) Identify one ordered pair on the line and determine the unit rate (constant of proportionality):

b) Write an equation that represents depth (\(y\)) in terms of time (\(x\)):

c) Calculate the vehicle's total depth after 26 minutes of diving:

Show your complete mathematical work and state units:

Calculated Depth:

d) Explain the real-world meaning of the point \((1, 12.5)\) in this context:

What does this coordinate tell marine researchers?

Be specific about both quantities and their units.

Proportional Power • RP2: Proportionality Assessment (Independent Form) Page 2 of 4

QUESTION 6 Vertical Hydroponic Farming (Verbal Description)

Name:

| Date:

6

Vertical Hydroponic Farming (Verbal Description)

4 Points

An automated vertical urban greenhouse dispenses liquid plant nutrient solution at a steady rate, utilizing exactly 210 liters of nutrient mixture every 14 hours of continuous automated cultivation.

a) Complete the proportional table:

b) Determine the constant of proportionality and write an equation for this relationship:

Show your work and state your equation clearly:

c) A nursery greenhouse section requires 480 L of solution. How many hours will this take?

Show your algebraic setup and calculation steps:

Total Hours:

d) If the greenhouse runs continuously for 72 hours, how many liters are dispensed?

Show your complete calculation:

Total Solution:

Proportional Power • RP2: Proportionality Assessment (Independent Form) Page 3 of 4

QUESTION 7 Electric Aviation Charging (Comparing Rates)

Name:

| Date:

7

Electric Aviation Charging (Comparing Rates)

4 Points

An electric airport terminal evaluates two fast-charging stations for regional cargo aircraft:

Charger Alpha

Delivers 75 kWh of energy in 5 minutes

Charger Beta

Delivers 108 kWh of energy in 6 minutes

a) Write an equation for each charger:

Charger Alpha:

Charger Beta:

b) Which charger supplies energy at a faster rate, and by how much?

Compare the two unit rates and state the numerical difference:

c) An electric airplane battery needs 270 kWh. Calculate how many minutes each charger takes:

Charger Alpha Work & Minutes:

Charger Beta Work & Minutes:

Assessment Completion Checklist: All problems completed, calculations detailed, units clearly stated?

Ready to Submit

Proportional Power • RP2: Proportionality Assessment (Independent Form) Page 4 of 4