Data Representation
Tables, graphs, trends, interpolation, combined data sets, variables and claims.
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Data Representation
Tables, graphs, trends, interpolation, combined data sets, variables and claims.
Worked Examples
<details> <summary><b>Example 1: A running total</b></summary>A student dropped magnesium ribbon into acid and recorded the total volume of hydrogen gas collected.
| Time (min) | Total H₂ collected (mL) |
|---|---|
| 0 | 0 |
| 1 | 18 |
| 2 | 33 |
| 3 | 45 |
| 4 | 54 |
| 5 | 60 |
Question 1: How much gas was produced from minute 2 to minute 4?
The header says total, so subtract: 54 − 33 = 21 mL. Answering 54 mL would count the gas from minutes 0 to 2 as well.
Question 2: During which 1-minute interval was the least gas produced?
The per-minute amounts are 18, 15, 12, 9, and 6 mL. The least, 6 mL, came from minute 4 to minute 5, even though that row has the largest total.
</details> <details> <summary><b>Example 2: A reverse lookup with a unit conversion</b></summary>A student measured the current through a resistor at four voltages.
| Voltage (V) | Current (mA) |
|---|---|
| 1.5 | 12 |
| 3.0 | 24 |
| 4.5 | 36 |
| 6.0 | 48 |
Question: At what voltage was the current 0.024 A?
- Units first. The table is in mA, the question is in A. Convert: 0.024 A × 1,000 = 24 mA.
- Reverse lookup. Find 24 in the Current column and read left: 3.0 V.
A student who skips the conversion looks for "0.024" in the table, finds nothing, and guesses. Converting the question's value to the table's units is almost always faster than converting the whole table.
</details>Worked Examples
<details> <summary><b>Example 1: Slopes along a heating curve</b></summary>A beaker of crushed ice was heated steadily. Figure 1 is a line graph of temperature (°C, vertical axis) versus time (min, horizontal axis) through these points:
| Time (min) | 0 | 2 | 4 | 6 | 8 | 10 | 12 |
|---|---|---|---|---|---|---|---|
| Temperature (°C) | −20 | −10 | 0 | 0 | 0 | 15 | 30 |
Question 1: What is the slope from 0 to 4 min?
The slope is 5 °C/min: the temperature rose 5 degrees each minute.
Question 2: What does the graph show from 4 to 8 min?
The graph is flat (slope 0): temperature did not change even though heat was still being added. The description explains why (the ice was melting), but the data alone tell you the temperature stayed at 0°C.
Question 3: Which stretch is steepest?
From 8 to 12 min the slope is 30 ÷ 4 = 7.5 °C/min, steeper than the 5 °C/min at the start. The liquid water warmed faster than the ice did.
</details> <details> <summary><b>Example 2: Two curves, two axes</b></summary>Figure 2 shows an algae culture over 10 days. One curve is algae density (left axis, thousands of cells/mL); the other is dissolved nitrate (right axis, mg/L).
| Day | 0 | 2 | 4 | 6 | 8 | 10 |
|---|---|---|---|---|---|---|
| Algae (thousands of cells/mL) | 5 | 9 | 20 | 38 | 42 | 43 |
| Nitrate (mg/L) | 40 | 34 | 22 | 10 | 4 | 3 |
Question: On the day the nitrate level was 10 mg/L, what was the algae density?
- Read the nitrate curve against the right axis: 10 mg/L occurs on Day 6.
- Go straight up or down to the algae curve and read the left axis: 38 thousand cells/mL.
The trap answer is 10 thousand cells/mL, which reads the nitrate value off the wrong axis. Notice also the relationship: as algae rose, nitrate fell, and both curves flatten after Day 8.
</details>Worked Examples
<details> <summary><b>Example 1: Classify two trends with first differences</b></summary>Table A — Distance a ball has rolled down a long ramp
| Time (s) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| Distance (m) | 0 | 0.3 | 1.2 | 2.7 | 4.8 |
Differences: 0.3, 0.9, 1.5, 2.1. They grow each second, so distance increases at an accelerating rate; the ball is speeding up. "Linear increase" is wrong because the steps are not equal.
Table B — Oxygen output of a leaf at different light levels
| Light (lux) | 0 | 200 | 400 | 600 | 800 | 1,000 |
|---|---|---|---|---|---|---|
| Oxygen (units/h) | 0 | 14 | 22 | 26 | 27 | 27 |
Differences: 14, 8, 4, 1, 0. They shrink to zero, so oxygen output increases, then levels off. Adding light beyond about 800 lux produced no more oxygen.
</details> <details> <summary><b>Example 2: Is it inversely proportional?</b></summary>A gas was trapped in a syringe and squeezed at constant temperature.
| Pressure (kPa) | 100 | 150 | 200 | 300 |
|---|---|---|---|---|
| Volume (mL) | 60 | 40 | 30 | 20 |
Step 1: Direction. As pressure rises, volume falls, so the relationship is inverse.
Step 2: Check the product. 100 × 60 = 6,000; 150 × 40 = 6,000; 200 × 30 = 6,000; 300 × 20 = 6,000. The product is constant, so volume is inversely proportional to pressure.
Step 3: Use it. At 400 kPa, volume = 6,000 ÷ 400 = 15 mL. Doubling the pressure from 200 to 400 kPa halves the volume from 30 to 15 mL.
Why not linear? The volume drops by 20, then 10, then 10 mL over pressure steps of 50, 50, and 100 kPa. Per kPa, that is 0.4, 0.2, and 0.1 mL; the drop keeps slowing, so the graph curves.
</details>Worked Examples
<details> <summary><b>Example 1: Chaining two tables</b></summary>Table 1 — Light at different depths in a lake
| Depth (m) | 0 | 2 | 4 | 6 | 8 |
|---|---|---|---|---|---|
| Light (% of surface) | 100 | 60 | 36 | 22 | 13 |
Table 2 — Growth of an alga at different light levels
| Light (% of surface) | 13 | 22 | 36 | 60 | 100 |
|---|---|---|---|---|---|
| Growth (doublings/day) | 0.4 | 0.9 | 1.6 | 2.3 | 2.6 |
Question 1: What is the algae's growth rate at a depth of 4 m?
The bridge is light. Table 1: 4 m → 36%. Table 2: 36% → 1.6 doublings/day. Answering "36" reports the light level, not the growth rate.
Question 2: At about what depth would the alga grow at 0.9 doublings per day?
Work backward. Table 2: 0.9 → 22% light. Table 1: 22% → 6 m.
</details> <details> <summary><b>Example 2: Do the error bars overlap?</b></summary>Three groups of tomato plants were grown for 4 weeks. Each mean height is shown with its uncertainty.
| Group | Mean height (cm) | Uncertainty (cm) | Range (cm) |
|---|---|---|---|
| Control (no supplement) | 21.0 | ± 1.5 | 19.5 to 22.5 |
| Low dose | 22.5 | ± 1.0 | 21.5 to 23.5 |
| High dose | 26.0 | ± 1.2 | 24.8 to 27.2 |
Question: Which dose gives clear evidence of taller plants than the control?
- Low dose: 21.5 to 23.5 overlaps the control's 19.5 to 22.5. The 1.5 cm gap in the means is uncertain.
- High dose: 24.8 to 27.2 sits entirely above 22.5. That is clear evidence.
Answer: the high dose only. A choice saying "both doses clearly increased height" trusts the low-dose mean without checking its error bar.
</details>Worked Examples
<details> <summary><b>Example 1: Uneven spacing, forward and backward</b></summary>A thermometer was placed at different distances from a heat lamp.
| Distance (cm) | 10 | 20 | 40 | 80 |
|---|---|---|---|---|
| Temperature (°C) | 48 | 40 | 32 | 26 |
Question 1: Estimate the temperature at 30 cm.
The neighbors of 30 cm are 20 cm and 40 cm (not 10 and 20). 30 is halfway between them, so the temperature is halfway between 40 and 32: 36°C.
Question 2: At about what distance would the temperature be 29°C?
29°C lies between 32°C (40 cm) and 26°C (80 cm), exactly halfway. Halfway from 40 cm to 80 cm is 60 cm.
Notice that equal temperature steps take larger and larger distance steps. The spacing of the table is a clue that the relationship is not linear over the whole range, which is why you should always interpolate between the nearest rows.
</details> <details> <summary><b>Example 2: Extending a trend, and its limit</b></summary>A water tank was drained through a valve.
| Time (min) | 0 | 5 | 10 | 15 |
|---|---|---|---|---|
| Water remaining (L) | 120 | 105 | 90 | 75 |
Question: If the trend continues, when will the tank be empty?
- Pattern: the tank loses 15 L every 5 min, a rate of 3 L/min.
- Gap: 75 L remain at 15 min.
- Time needed: 75 ÷ 3 = 25 more minutes.
- Clock time: 15 + 25 = 40 min.
Answering 25 min reports only the extra time. And the pattern cannot continue past 40 min: a tank cannot hold negative water, so any prediction of "−15 L at 45 min" is impossible.
</details>Worked Examples
<details> <summary><b>Example 1: Which trials isolate which variable?</b></summary>Students grew bean seedlings under lamps for 14 days.
| Trial | Light color | Light (h/day) | Water (mL/day) | Height (cm) |
|---|---|---|---|---|
| 1 | white | 12 | 20 | 9.1 |
| 2 | red | 12 | 20 | 10.4 |
| 3 | red | 16 | 20 | 12.8 |
| 4 | blue | 16 | 30 | 12.2 |
Dependent variable: height, the only value measured. The other three columns were set by the students.
Effect of light color alone: Trials 1 and 2 differ only in color (white vs. red), so red light added 1.3 cm under these conditions.
Effect of hours of light alone: Trials 2 and 3 differ only in hours (12 vs. 16), so 4 extra hours added 2.4 cm.
Effect of water: Trial 4 differs from Trial 3 in both color and water, so it is confounded. Nothing can be concluded about water. To fix it, add a trial with red, 16 h, 30 mL: compared with Trial 3, it changes only the water.
</details> <details> <summary><b>Example 2: Using the right control to judge a claim</b></summary>Volunteers touched a lab surface, then either did not wash, washed with plain soap, or washed with antibacterial soap. Their fingertips were pressed onto nutrient plates.
| Group | Mean bacterial colonies |
|---|---|
| No washing (control) | 180 |
| Plain soap | 60 |
| Antibacterial soap | 45 |
Claim 1: "Antibacterial soap removes 135 more colonies than plain soap." Contradicted. The 135 comes from comparing with no washing (180 − 45). Compared with plain soap, antibacterial soap left only 60 − 45 = 15 fewer colonies.
Claim 2: "Antibacterial soap is the only soap that reduces bacteria." Contradicted. Plain soap cut colonies from 180 to 60.
Claim 3: "Antibacterial soap works better on every type of bacterium." Can't tell. The study counted colonies but never identified types of bacteria.
</details>Worked Example: One Passage, Start to Finish
<details> <summary><b>Rainwater pH downwind of a power plant</b></summary>Passage: Researchers collected rainwater at five distances downwind of a coal-burning power plant and measured its pH. Lower pH means more acidic water.
| Distance downwind (km) | 0 | 5 | 10 | 20 | 40 |
|---|---|---|---|---|---|
| Rainwater pH | 4.2 | 4.5 | 4.9 | 5.3 | 5.6 |
Routine (about 20 seconds): Distance was chosen (independent); pH was measured (dependent). The distances are unevenly spaced, so be careful with neighbors.
Question 1: As distance increases, the rainwater pH:
Every value is higher than the one before, so pH increases. Per km, the gains are 0.06, 0.08, 0.04, and 0.015, so it rises more slowly at large distances. A choice of "decreases" reverses the trend; "increases, then decreases" has no support. (About 30 seconds.)
Question 2: Assuming linear change between measurements, the pH at 15 km would be closest to:
The neighbors of 15 km are 10 km and 20 km, not 5 and 10. Halfway between 4.9 and 5.3 is 5.1. (About 30 seconds.)
Question 3: Which conclusion is best supported?
The researchers only observed pH at different distances; they did not change the plant's output. The safe conclusion is that rain was more acidic closer to the plant. A choice saying the data prove the plant causes the acidity overreaches, and one saying rain "becomes neutral (pH 7) beyond 40 km" extrapolates past the data. (About 45 seconds.)
Total: under two minutes for three questions, with time banked for harder ones.
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