Data Representation and Interpretation Practice Questions — ACT

The solubility of oxygen ( $O_2$ ) in water at various temperatures is recorded in the table below:

Water Temperature (°C)	Dissolved $O_2$ Solubility (mg/L)
0	14.6
10	11.3
20	9.1
30	7.5
40	6.4

Based on the data table, determine whether the following statement is true or false: A line graph representing these data would feature a line or curve with a negative slope, showing that dissolved oxygen solubility decreases as water temperature increases.

True

False

Show answer & explanation

Answer: True

Answer

True

The data table displays an inverse relationship between water temperature and dissolved oxygen solubility. When plotted on a coordinate plane with water temperature on the x-axis and dissolved oxygen solubility on the y-axis, the resulting curve trends downward from left to right. This downward trend corresponds to a negative slope, making the statement true.

Step-by-Step Solution

1

Analyze the relationship between the two variables in the table.

As water temperature increases (

0 \rightarrow 10 \rightarrow 20 \rightarrow 30 \rightarrow 40

°C), the solubility of dissolved oxygen decreases (

14.6 \rightarrow 11.3 \rightarrow 9.1 \rightarrow 7.5 \rightarrow 6.4

mg/L).

Understanding the trend in the numerical data is the first step in translating the table to a graphical format.

2

Determine how this relationship translates to a line graph.

An inverse relationship where the dependent variable (

y

, solubility) decreases as the independent variable (

x

, temperature) increases translates to a line or curve that falls from left to right.

Identifying that a falling line represents a negative slope allows us to evaluate the accuracy of the statement.

Key Concept

Translating tabular data trends into graphical representations, specifically identifying how inverse relationships correspond to negative slopes.

Question 2Question →

Geothermal geologists analyzed fluid samples collected from 4 distinct wells (Well W, Well X, Well Y, and Well Z). Table 1 lists the measured reservoir temperature in degrees Celsius ( $^\circ\text{C}$ ) and the dissolved silica ( $SiO_2$ ) concentration in milligrams per liter ( $\text{mg/L}$ ) for each well.

Table 1:

Well	Reservoir Temperature ( $^\circ\text{C}$ )	Dissolved $SiO_2$ ( $\text{mg/L}$ )
Well W	140	180
Well X	200	400
Well Y	170	250
Well Z	230	520

Figure 1 displays the fluid viscosity ( $\eta$ , in centipoise, $\text{cP}$ ) as a function of dissolved $SiO_2$ concentration at four different temperature curves:
- At $140^\circ\text{C}$ : $\eta = 0.005 \times (\text{dissolved } SiO_2) + 0.10$
- At $170^\circ\text{C}$ : $\eta = 0.003 \times (\text{dissolved } SiO_2) + 0.20$
- At $200^\circ\text{C}$ : $\eta = 0.002 \times (\text{dissolved } SiO_2) + 0.40$
- At $230^\circ\text{C}$ : $\eta = 0.001 \times (\text{dissolved } SiO_2) + 0.50$

Based on Table 1 and Figure 1, rank the geothermal wells in order from lowest fluid viscosity to highest fluid viscosity.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest fluid viscosity is Well Y, Well W, Well Z, and Well X.

By looking up each well's specific reservoir temperature and dissolved

SiO_2

concentration in Table 1 and substituting those values into the corresponding temperature equation from Figure 1, the calculated viscosities are

0.95\text{ cP}

for Well Y,

1.00\text{ cP}

for Well W,

1.02\text{ cP}

for Well Z, and

1.20\text{ cP}

for Well X. Sorting these from lowest to highest yields the sequence: Well Y, Well W, Well Z, Well X.

Step-by-Step Solution

1

Extract the temperature and dissolved

SiO_2

concentration for each well from Table 1.

Well W:

140^\circ\text{C}

,

180\text{ mg/L}

; Well X:

200^\circ\text{C}

,

400\text{ mg/L}

; Well Y:

170^\circ\text{C}

,

250\text{ mg/L}

; Well Z:

230^\circ\text{C}

,

520\text{ mg/L}

.

Both temperature and concentration are necessary to select the corresponding line equation from Figure 1.

2

Apply the viscosity formulas from Figure 1 corresponding to each well's temperature.

Well W (

140^\circ\text{C}

):

\eta = 0.005(180) + 0.10 = 0.90 + 0.10 = 1.00\text{ cP}

; Well X (

200^\circ\text{C}

):

\eta = 0.002(400) + 0.40 = 0.80 + 0.40 = 1.20\text{ cP}

; Well Y (

170^\circ\text{C}

):

\eta = 0.003(250) + 0.20 = 0.75 + 0.20 = 0.95\text{ cP}

; Well Z (

230^\circ\text{C}

):

\eta = 0.001(520) + 0.50 = 0.52 + 0.50 = 1.02\text{ cP}

.

Synthesizing data from both sources determines the numerical fluid viscosity for each well.

3

Arrange the resulting viscosity values from smallest to largest.

0.95\text{ cP (Well Y)} < 1.00\text{ cP (Well W)} < 1.02\text{ cP (Well Z)} < 1.20\text{ cP (Well X)}

.

The question specifies ranking from lowest viscosity to highest viscosity.

Key Concept

Synthesizing values from a data table with temperature-dependent functional relationships in a graph.

Question 3Question →

A researcher conducted three trials to measure the mass of dissolved solute in $100\text{ mL}$ of water at $25^\circ\text{C}$ :

Trial	Mass of Dissolved Solute (g)
Trial 1	12.0
Trial 2	14.0
Trial 3	16.0

True or False: The average mass of dissolved solute across all three trials is $14.0\text{ g}$ .

True

False

Show answer & explanation

Answer: True

Answer

True. The calculated mean of the three experimental trials is exactly 14.0 g.

Adding the three trial measurements (

12.0\text{ g} + 14.0\text{ g} + 16.0\text{ g} = 42.0\text{ g}

) and dividing by the total number of trials (

3

) gives

\frac{42.0\text{ g}}{3} = 14.0\text{ g}

. Thus, the statement is true.

Step-by-Step Solution

1

Retrieve the mass values for Trial 1, Trial 2, and Trial 3 from the data table.

Trial 1 =

12.0\text{ g}

, Trial 2 =

14.0\text{ g}

, Trial 3 =

16.0\text{ g}

.

Identify all data points needed to calculate the arithmetic mean.

2

Sum the retrieved mass values.

12.0\text{ g} + 14.0\text{ g} + 16.0\text{ g} = 42.0\text{ g}

.

Calculate the total mass recorded across all three trials.

3

Divide the total mass by the number of trials (

3

).

\frac{42.0\text{ g}}{3} = 14.0\text{ g}

.

Obtain the average value per trial.

Key Concept

Calculating the average of data points collected from experimental trials

Question 4Question →

A microbiologist measured the bacterial growth rate (in $\text{cells/mL}\cdot\text{hr}$ ) of a newly isolated strain across three independent trials under four different incubator temperatures, as presented in the table below:

Temperature ( $^\circ\text{C}$ )	Trial 1 ( $\text{cells/mL}\cdot\text{hr}$ )	Trial 2 ( $\text{cells/mL}\cdot\text{hr}$ )	Trial 3 ( $\text{cells/mL}\cdot\text{hr}$ )
25	140	155	125
30	280	310	250
35	420	460	380
40	210	240	180

True or False: The mean growth rate at $35^\circ\text{C}$ exceeds the mean growth rate at $30^\circ\text{C}$ by more than $130\text{ cells/mL}\cdot\text{hr}$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is True because the mean growth rate at

35^\circ\text{C}

(

420\text{ cells/mL}\cdot\text{hr}

) exceeds the mean growth rate at

30^\circ\text{C}

(

280\text{ cells/mL}\cdot\text{hr}

) by

140\text{ cells/mL}\cdot\text{hr}

, which is greater than

130\text{ cells/mL}\cdot\text{hr}

.

The mean growth rate at

35^\circ\text{C}

is

420\text{ cells/mL}\cdot\text{hr}

and at

30^\circ\text{C}

is

280\text{ cells/mL}\cdot\text{hr}

. The difference is

140\text{ cells/mL}\cdot\text{hr}

, which is strictly greater than

130\text{ cells/mL}\cdot\text{hr}

.

Step-by-Step Solution

1

Calculate the average growth rate for

35^\circ\text{C}

across all three trials.

\text{Mean}_{35} = \frac{420 + 460 + 380}{3} = \frac{1260}{3} = 420\text{ cells/mL}\cdot\text{hr}

Determining the representative data point for the

35^\circ\text{C}

temperature condition requires finding the mean.

2

Calculate the average growth rate for

30^\circ\text{C}

across all three trials.

\text{Mean}_{30} = \frac{280 + 310 + 250}{3} = \frac{840}{3} = 280\text{ cells/mL}\cdot\text{hr}

Determining the representative data point for the

30^\circ\text{C}

temperature condition requires finding the mean.

3

Subtract the mean rate at

30^\circ\text{C}

from the mean rate at

35^\circ\text{C}

.

420 - 280 = 140\text{ cells/mL}\cdot\text{hr}

Finding the increase in average growth rate between the two temperature conditions.

4

Compare the calculated difference (

140\text{ cells/mL}\cdot\text{hr}

) to the value given in the statement (

130\text{ cells/mL}\cdot\text{hr}

).

140 > 130

, so the statement is True.

Verifying whether the statement's inequality holds.

Key Concept

Calculating multi-trial averages and evaluating differences between data sets.

Question 5Question →

Environmental scientists investigated the bioremediation of crude oil-contaminated soil using a specialized bacterial consortium. Table 1 shows the measured soil moisture content and temperature at four distinct monitoring sites (Site 1, Site 2, Site 3, and Site 4).

Table 1: Soil Characteristics at Monitoring Sites

Monitoring Site	Soil Moisture Content (%)	Soil Temperature (°C)
Site 1	15	25
Site 2	30	35
Site 3	45	15
Site 4	60	25

Table 2 displays the hydrocarbon degradation rate by the bacterial consortium measured across various soil moisture contents and incubation temperatures in laboratory testing.

Table 2: Hydrocarbon Degradation Rate (mg/kg/day)

Incubation Temperature (°C)	15% Moisture	30% Moisture	45% Moisture	60% Moisture
15	10	25	40	20
25	30	65	85	50
35	15	35	55	30

Based on Table 1 and Table 2, rank the four monitoring sites in order from the site with the HIGHEST estimated hydrocarbon degradation rate to the site with the LOWEST estimated hydrocarbon degradation rate.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

Site 4, Site 3, Site 2, Site 1

To rank the sites from highest to lowest degradation rate, combine the site parameters from Table 1 with the degradation rate values in Table 2. Site 4 (60% moisture at 25°C) yields 50 mg/kg/day. Site 3 (45% moisture at 15°C) yields 40 mg/kg/day. Site 2 (30% moisture at 35°C) yields 35 mg/kg/day. Site 1 (15% moisture at 25°C) yields 30 mg/kg/day. Ordering these values from highest to lowest gives Site 4, Site 3, Site 2, and Site 1.

Step-by-Step Solution

1

Extract moisture and temperature values for each site from Table 1

Site 1: 15% moisture, 25°C; Site 2: 30% moisture, 35°C; Site 3: 45% moisture, 15°C; Site 4: 60% moisture, 25°C.

Identify the independent variable conditions for each site.

2

Cross-reference each site's parameters with Table 2 to determine hydrocarbon degradation rates

Site 1 degradation rate = 30 mg/kg/day; Site 2 degradation rate = 35 mg/kg/day; Site 3 degradation rate = 40 mg/kg/day; Site 4 degradation rate = 50 mg/kg/day.

Find the specific target values corresponding to each site's combined conditions.

3

Arrange the sites from highest degradation rate to lowest degradation rate

Site 4 (50 mg/kg/day) > Site 3 (40 mg/kg/day) > Site 2 (35 mg/kg/day) > Site 1 (30 mg/kg/day).

Fulfill the ordering requirement from highest rate to lowest rate.

Key Concept

Comparing Multiple Data Sources

Estimated Time:1m 30s

Question 6Question →

A group of microbiologists measured the enzymatic reaction rate ( $R$ , in $\text{nmol/min}\cdot\text{mg}$ ) of a newly discovered halophilic bacterial strain across various salinity levels ( $S$ , in parts per thousand, ppt) at three fixed temperatures ( $20^\circ\text{C}$ , $30^\circ\text{C}$ , and $40^\circ\text{C}$ ). The results are recorded in the table below.

Salinity ( $S$ , ppt)	Rate at $20^\circ\text{C}$	Rate at $30^\circ\text{C}$	Rate at $40^\circ\text{C}$
10	14.0	22.0	10.0
20	20.0	32.0	16.0
30	26.0	42.0	22.0
50	38.0	62.0	34.0

Based on linear interpolation and extrapolation of the data trends, arrange the four estimated reaction rates described below in order from lowest to highest.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest to highest estimated reaction rate is: the estimated rate at

30^\circ\text{C}

and

5\text{ ppt}

(

17.0\text{ nmol/min}\cdot\text{mg}

), followed by the estimated rate at

40^\circ\text{C}

and

25\text{ ppt}

(

19.0\text{ nmol/min}\cdot\text{mg}

), followed by the estimated rate at

20^\circ\text{C}

and

60\text{ ppt}

(

44.0\text{ nmol/min}\cdot\text{mg}

), and finally the estimated rate at

30^\circ\text{C}

and

40\text{ ppt}

(

52.0\text{ nmol/min}\cdot\text{mg}

).

To place the estimated rates in correct order from lowest to highest, calculate each value using the constant linear rates of change shown in the table. At

30^\circ\text{C}

and

5\text{ ppt}

, extrapolating below

10\text{ ppt}

yields

17.0\text{ nmol/min}\cdot\text{mg}

. At

40^\circ\text{C}

and

25\text{ ppt}

, interpolating halfway between

20

and

30\text{ ppt}

yields

19.0\text{ nmol/min}\cdot\text{mg}

. At

20^\circ\text{C}

and

60\text{ ppt}

, extrapolating above

50\text{ ppt}

yields

44.0\text{ nmol/min}\cdot\text{mg}

. At

30^\circ\text{C}

and

40\text{ ppt}

, interpolating halfway between

30

and

50\text{ ppt}

yields

52.0\text{ nmol/min}\cdot\text{mg}

. Comparing these numerical values (

17.0 < 19.0 < 44.0 < 52.0

) produces the correct sequence.

Step-by-Step Solution

1

Determine the linear relationship (slope) for each temperature column.

For

20^\circ\text{C}

, the rate increases by

0.6\text{ nmol/min}\cdot\text{mg}

per ppt salinity. For

30^\circ\text{C}

, the rate increases by

1.0\text{ nmol/min}\cdot\text{mg}

per ppt salinity. For

40^\circ\text{C}

, the rate increases by

0.6\text{ nmol/min}\cdot\text{mg}

per ppt salinity.

Establishing the rate of change allows calculation of values between existing data points (interpolation) and beyond existing data points (extrapolation).

2

Calculate the numerical value for each item.

Item 1 (

30^\circ\text{C}, 40\text{ ppt}

):

42.0 + 10(1.0) = 52.0\text{ nmol/min}\cdot\text{mg}

.
Item 2 (

20^\circ\text{C}, 60\text{ ppt}

):

38.0 + 10(0.6) = 44.0\text{ nmol/min}\cdot\text{mg}

.
Item 3 (

40^\circ\text{C}, 25\text{ ppt}

):

16.0 + 5(0.6) = 19.0\text{ nmol/min}\cdot\text{mg}

.
Item 4 (

30^\circ\text{C}, 5\text{ ppt}

):

22.0 - 5(1.0) = 17.0\text{ nmol/min}\cdot\text{mg}

.

Obtaining exact numerical estimates enables precise relative comparison.

3

Sort the items by calculated value from lowest to highest.

17.0

(Item 4) <

19.0

(Item 3) <

44.0

(Item 2) <

52.0

(Item 1).

Direct numerical comparison determines the correct sequence.

Key Concept

Interpolation and Extrapolation of Data Trends

Estimated Time:1m 30s

Question 7Question →

Materials engineers measured the electrical conductivity ( $\sigma$ , in $10^6\text{ S/m}$ ) of a copper-nickel alloy at various temperatures ( $T$ , in K). The measured values are recorded in the table below:

Temperature ( $T$ , K)	Electrical Conductivity ( $\sigma$ , $10^6\text{ S/m}$ )
300	14.2
400	11.8
500	9.4
600	7.0
700	4.6

Based on the data, if the linear relationship between temperature and electrical conductivity continues beyond $700\text{ K}$ , what is the predicted electrical conductivity of the alloy at $800\text{ K}$ ?

2.2 \times 10^6\text{ S/m}

3.4 \times 10^6\text{ S/m}

4.6 \times 10^6\text{ S/m}

7.0 \times 10^6\text{ S/m}

Show answer & explanation

Answer: $2.2 \times 10^6\text{ S/m}$

Answer

The predicted electrical conductivity at

800\text{ K}

is

2.2 \times 10^6\text{ S/m}

.

The electrical conductivity decreases by a constant rate of

2.4 \times 10^6\text{ S/m}

for every

100\text{ K}

increase in temperature. At

700\text{ K}

, the conductivity is

4.6 \times 10^6\text{ S/m}

. Extrapolating to

800\text{ K}

requires subtracting

2.4 \times 10^6\text{ S/m}

from

4.6 \times 10^6\text{ S/m}

, giving

2.2 \times 10^6\text{ S/m}

.

Step-by-Step Solution

1

Determine the change in electrical conductivity per

100\text{ K}

temperature increase.

Between

300\text{ K}

and

400\text{ K}

, conductivity decreases by

14.2 - 11.8 = 2.4 \times 10^6\text{ S/m}

. Checking other intervals confirms a constant rate of

-2.4 \times 10^6\text{ S/m}

per

100\text{ K}

.

Establishing the linear rate of change is necessary for linear extrapolation.

2

Apply the linear trend to extrapolate from

700\text{ K}

to

800\text{ K}

.

4.6 - 2.4 = 2.2 \times 10^6\text{ S/m}

.

Extrapolation involves extending the observed trend by one temperature interval (

100\text{ K}

) beyond the measured range.

Key Concept

Linear Extrapolation

Estimated Time:1m 0s

Question 8Question →

Soil ecologists evaluated nitrogen mineralization by measuring nitrate production rates (in $\text{mg/kg/day}$ ) across 4 depth zones in two distinct forest management plots (Plot X and Plot Y). The results are summarized in Table 1 below.

Depth Zone	Plot X Nitrate Rate ( $\text{mg/kg/day}$ )	Plot Y Nitrate Rate ( $\text{mg/kg/day}$ )
0–10 cm	5.0	9.0
10–20 cm	3.5	6.5
20–30 cm	2.0	4.0
30–40 cm	1.5	2.5

Based on Table 1, if a composite sample is created using equal masses of soil from all 4 depth zones (0–40 cm), by how much does the average nitrate production rate of Plot Y exceed the average nitrate production rate of Plot X?

2.5\text{ mg/kg/day}

4.0\text{ mg/kg/day}

5.5\text{ mg/kg/day}

10.0\text{ mg/kg/day}

Show answer & explanation

Answer: $2.5\text{ mg/kg/day}$

Answer

The average nitrate production rate of Plot Y exceeds that of Plot X by

2.5\text{ mg/kg/day}

.

The composite average rate for Plot Y is calculated by summing its four depth zone values (

9.0 + 6.5 + 4.0 + 2.5 = 22.0

) and dividing by

4

, yielding

5.5\text{ mg/kg/day}

. The composite average rate for Plot X is calculated similarly (

5.0 + 3.5 + 2.0 + 1.5 = 12.0

) and divided by

4

, yielding

3.0\text{ mg/kg/day}

. The difference between these two averages is

5.5 - 3.0 = 2.5\text{ mg/kg/day}

.

Step-by-Step Solution

1

Calculate the average nitrate production rate for Plot Y across the 4 depth zones.

Average for Plot Y =

\frac{9.0 + 6.5 + 4.0 + 2.5}{4} = \frac{22.0}{4} = 5.5\text{ mg/kg/day}

.

An equal-mass composite sample over 0–40 cm requires finding the arithmetic mean of all 4 depth intervals.

2

Calculate the average nitrate production rate for Plot X across the 4 depth zones.

Average for Plot X =

\frac{5.0 + 3.5 + 2.0 + 1.5}{4} = \frac{12.0}{4} = 3.0\text{ mg/kg/day}

.

The same averaging procedure must be applied to Plot X to determine its overall composite rate.

3

Subtract the average rate of Plot X from the average rate of Plot Y.

5.5\text{ mg/kg/day} - 3.0\text{ mg/kg/day} = 2.5\text{ mg/kg/day}

.

The question asks by how much Plot Y's composite average exceeds Plot X's composite average.

Key Concept

Calculating mean values from multi-source tabular data and determining net differences between treatments.

Question 9Question →

Atmospheric scientists measured the mixing ratio of methane ( $\text{CH}_4$ , in $\text{ppm}$ ) and carbon monoxide ( $\text{CO}$ , in $\text{ppb}$ ) at various altitudes above sea level ( $\text{km}$ ) during an atmospheric survey, as shown in the table below:

Altitude ( $\text{km}$ )	Methane ( $\text{CH}_4$ , $\text{ppm}$ )	Carbon Monoxide ( $\text{CO}$ , $\text{ppb}$ )
0	1.85	120
3	1.78	95
6	1.65	70
9	1.48	48
12	1.20	30

Statement: If these tabular data are translated into a line graph plotting atmospheric concentration versus altitude ( $0$ to $12\text{ km}$ ), both gases will be represented by curves with negative slopes, and carbon monoxide will show a larger overall percentage decrease than methane across the altitude range.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true because both trace gases decrease in concentration as altitude increases, resulting in negative slopes for both plotted curves, and carbon monoxide undergoes a

75.0\%

decrease compared to a

35.1\%

decrease for methane.

The statement accurately translates the tabular data into graphical features. Both gas concentrations decrease steadily with increasing altitude, establishing negative slopes. Additionally, carbon monoxide experiences a

75.0\%

drop versus methane's

35.1\%

drop, confirming that carbon monoxide has the larger overall percentage decrease.

Step-by-Step Solution

1

Analyze the concentration trend for each gas as altitude increases from

0\text{ km}

to

12\text{ km}

.

Methane decreases from

1.85\text{ ppm}

to

1.20\text{ ppm}

, and carbon monoxide decreases from

120\text{ ppb}

to

30\text{ ppb}

.

Determining whether concentrations increase or decrease establishes the direction of the slope on a concentration versus altitude graph.

2

Determine the slope sign for both curves on a line graph.

Since both concentrations decrease continuously as the horizontal axis (altitude) increases, both curves display negative slopes.

Translating tabular data where the dependent variable decreases while the independent variable increases yields a negative slope.

3

Calculate and compare the percentage decrease for both gases from

0\text{ km}

to

12\text{ km}

.

Methane percentage decrease =

\frac{1.85 - 1.20}{1.85} \times 100\% \approx 35.14\%

. Carbon monoxide percentage decrease =

\frac{120 - 30}{120} \times 100\% = 75.00\%

.

Evaluating relative percentage changes verifies whether carbon monoxide experienced a greater proportional reduction than methane.

Key Concept

Translating numerical tables into graphical slope directions and percentage change comparisons

Question 10Question →

A chemist measured the volume of carbon dioxide ( $CO_2$ ) gas collected during a reaction at three different temperatures over a 10-minute period, recorded in the table below.

Time (min)	Volume at 20°C (mL)	Volume at 30°C (mL)	Volume at 40°C (mL)
0	0	0	0
2	4	9	18
4	8	17	31
6	12	23	38
8	15	27	40
10	17	29	40

Match each reaction temperature condition with the curve characteristic that best describes its dataset when translated into a line graph of Volume vs. Time.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The 20°C reaction condition matches the line with a steady, low positive slope reaching 17 mL without plateauing; the 30°C reaction condition matches the curve with a moderate initial slope flattening near 29 mL; and the 40°C reaction condition matches the steep curve reaching a plateau at 40 mL by minute 8.

Each temperature condition in the table exhibits a unique rate of

CO_2

generation over time. The 20°C data shows steady non-zero growth up to 17 mL, the 30°C data demonstrates decelerating growth reaching 29 mL, and the 40°C data displays rapid initial growth that reaches a constant maximum of 40 mL (plateau) at minute 8.

Step-by-Step Solution

1

Examine the volume trend over time for the 20°C trial in the table.

The volume rises from 0 mL to 17 mL with steady increments of 3–4 mL per 2-minute interval, showing no leveling off.

A constant rate of increase translates directly to a linear trend with a steady positive slope.

2

Examine the volume trend over time for the 30°C trial in the table.

The volume increases from 0 mL to 29 mL, with interval gains decreasing from 9 mL to 2 mL near the end.

Decreasing gains over equal time intervals translate to a curve whose slope flattens gradually over time.

3

Examine the volume trend over time for the 40°C trial in the table.

The volume rises rapidly to 38 mL by minute 6 and remains unchanged at 40 mL at minutes 8 and 10.

An unchanged measurement across consecutive time points translates to a horizontal plateau on a line graph.

Key Concept

Translating tabular data into qualitative line graph characteristics

Question 11Question →

Hydrogeologists evaluated the hydraulic conductivity ( $K$ , in $\text{m/day}$ ) of four sediment types under three compaction levels ( $1.4\text{ g/cm}^3$ , $1.6\text{ g/cm}^3$ , and $1.8\text{ g/cm}^3$ ). The results are presented in the following table:

Sediment Type	$K$ at $1.4\text{ g/cm}^3$ (m/day)	$K$ at $1.6\text{ g/cm}^3$ (m/day)	$K$ at $1.8\text{ g/cm}^3$ (m/day)
Clay	$0.002$	$0.001$	$0.0005$
Silt	$0.05$	$0.02$	$0.008$
Fine Sand	$4.5$	$2.1$	$0.9$
Coarse Gravel	$120.0$	$85.0$	$50.0$

Statement: If this data were translated into a line graph with compaction density on the horizontal axis and hydraulic conductivity ( $K$ ) on the vertical axis, the lines for all four sediment types would show a positive slope from left to right.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False because higher compaction density leads to lower hydraulic conductivity for all four sediment types, which translates to a negative slope on a line graph.

The statement is incorrect because the tabular data shows an inverse relationship between compaction density and hydraulic conductivity (

K

). As compaction density increases from left to right along the table headers (

1.4 \rightarrow 1.6 \rightarrow 1.8\text{ g/cm}^3

), the corresponding

K

values decrease for all four sediment types. When translated into a line graph with density on the x-axis and

K

on the y-axis, lines that fall from left to right exhibit a negative slope, not a positive slope.

Step-by-Step Solution

1

Identify the independent variable for the horizontal axis and the dependent variable for the vertical axis.

Compaction density (

1.4

,

1.6

,

1.8\text{ g/cm}^3

) is on the horizontal (

x

) axis, and hydraulic conductivity (

K

) is on the vertical (

y

) axis.

Establishing axis assignment is the first step in translating tabular data into graphical form.

2

Examine the trend of hydraulic conductivity (

K

) values as compaction density increases across each row.

For Clay (

0.002 \rightarrow 0.001 \rightarrow 0.0005

), Silt (

0.05 \rightarrow 0.02 \rightarrow 0.008

), Fine Sand (

4.5 \rightarrow 2.1 \rightarrow 0.9

), and Coarse Gravel (

120.0 \rightarrow 85.0 \rightarrow 50.0

),

K

consistently decreases as density increases.

Determining the direction of change in

y

relative to

x

reveals the graphical slope.

3

Translate the observed mathematical trend into a graphical feature (slope direction).

A relationship where

y

decreases as

x

increases forms a downward line from left to right, which defines a negative slope, making the claim of a positive slope false.

Translating inverse data trends to graphs requires correctly associating decreasing values with negative slopes.

Key Concept

Translating Tabular Relationships to Graphical Slopes

Estimated Time:1m 0s

Question 12Question →

An environmental scientist measured the rate of enzymatic breakdown of microplastics ( $\text{mg/L/hr}$ ) by a bacterial strain across five different incubation temperatures ( $^\circ\text{C}$ ). The results are shown in the table below:

Incubation Temperature ( $^\circ\text{C}$ )	Breakdown Rate ( $\text{mg/L/hr}$ )
10	2.0
20	5.5
30	12.0
40	8.5
50	1.0

Which of the following descriptions best characterizes the line graph that accurately translates these experimental results?

A line graph plotting Incubation Temperature (

^\circ\text{C}

) on the x-axis and Breakdown Rate (

\text{mg/L/hr}

) on the y-axis, where the curve rises from a point at

(10, 2.0)

to a peak at

(30, 12.0)

, before declining to

(50, 1.0)

.

A line graph plotting Breakdown Rate (

\text{mg/L/hr}

) on the x-axis and Incubation Temperature (

^\circ\text{C}

) on the y-axis, where the curve moves rightward from

(2.0, 10)

to a peak at

(12.0, 30)

, before returning leftward to

(1.0, 50)

.

A line graph plotting Incubation Temperature (

^\circ\text{C}

) on the x-axis and Breakdown Rate (

\text{mg/L/hr}

) on the y-axis, showing a strictly increasing linear trend from

(10, 2.0)

up to

(50, 12.0)

.

A line graph plotting Incubation Temperature (

^\circ\text{C}

) on the x-axis and Breakdown Rate (

\text{mg/L/hr}

) on the y-axis, where the curve decreases from a maximum at

(10, 12.0)

down to a minimum at

(50, 1.0)

.

Show answer & explanation

Answer: A line graph plotting Incubation Temperature ( $^\circ\text{C}$ ) on the x-axis and Breakdown Rate ( $\text{mg/L/hr}$ ) on the y-axis, where the curve rises from a point at $(10, 2.0)$ to a peak at $(30, 12.0)$ , before declining to $(50, 1.0)$ .

Answer

The correct line graph plots Incubation Temperature on the horizontal x-axis and Breakdown Rate on the vertical y-axis, forming a curve that increases to a peak value of 12.0 mg/L/hr at 30°C and then drops to 1.0 mg/L/hr at 50°C.

The correct graph properly places Incubation Temperature on the x-axis and Breakdown Rate on the y-axis. The data points from the table demonstrate an initial rise in rate up to 30°C (12.0 mg/L/hr), followed by a decline as temperature increases further to 50°C (1.0 mg/L/hr), which corresponds exactly to a peaked curve.

Step-by-Step Solution

1

Identify axis assignments for translating tabular data to a line graph.

The independent variable (Incubation Temperature in °C) belongs on the horizontal x-axis, and the dependent variable (Breakdown Rate in mg/L/hr) belongs on the vertical y-axis.

Standard ACT scientific graphical conventions place controlled/independent variables on the x-axis and measured/dependent variables on the y-axis.

2

Trace the key data coordinates from the table

(x, y)

.

Points are

(10, 2.0)

,

(20, 5.5)

,

(30, 12.0)

,

(40, 8.5)

, and

(50, 1.0)

.

Accurate format translation requires verifying specific numeric values at each data level.

3

Analyze the trend of the data points.

The rate increases from 10°C to 30°C (reaching a peak of 12.0 mg/L/hr), then decreases from 30°C to 50°C.

The curve must visually reflect this unimodal (bell-shaped) trend.

Key Concept

Translating Tabular Data into Graphical Representations

Estimated Time:1m 0s

Question 13Question →

Biochemists measured the rate of glucose consumption (in $\text{mM/min}$ ) by a newly isolated bacterial strain across 3 temperature treatments ( $25^\circ\text{C}$ , $35^\circ\text{C}$ , and $45^\circ\text{C}$ ). Each temperature condition was evaluated in 3 separate trials. The results are recorded in the table below:

Temperature ( $^\circ\text{C}$ )	Trial 1 ( $\text{mM/min}$ )	Trial 2 ( $\text{mM/min}$ )	Trial 3 ( $\text{mM/min}$ )
$25$	$1.2$	$1.5$	$1.8$
$35$	$4.0$	$4.6$	$4.6$
$45$	$2.1$	$2.5$	$2.0$

Based on the data provided, what is the average rate of glucose consumption, in $\text{mM/min}$ , for the $35^\circ\text{C}$ treatment across the 3 trials?

Show answer & explanation

Answer: 4.4

Answer

The average rate of glucose consumption at

35^\circ\text{C}

across the 3 trials is

4.4\text{ mM/min}

.

To find the average glucose consumption rate at

35^\circ\text{C}

, locate the row for

35^\circ\text{C}

in the table and sum the values across the 3 trials (

4.0 + 4.6 + 4.6 = 13.2\text{ mM/min}

). Then divide by the number of trials (

3

) to get

\frac{13.2}{3} = 4.4\text{ mM/min}

.

Step-by-Step Solution

1

Extract data points for the

35^\circ\text{C}

row

Trial 1 =

4.0\text{ mM/min}

, Trial 2 =

4.6\text{ mM/min}

, Trial 3 =

4.6\text{ mM/min}

The question specifically asks for the average at

35^\circ\text{C}

.

2

Calculate the sum of the trial values

4.0 + 4.6 + 4.6 = 13.2\text{ mM/min}

Calculating an average requires finding the total sum of all trial measurements first.

3

Divide the sum by the total number of trials

\frac{13.2}{3} = 4.4\text{ mM/min}

Dividing the sum by

3

calculates the arithmetic mean across the 3 trials.

Key Concept

Calculating the average (mean) of data values from a table

Estimated Time:1m 0s

Question 14Question →

A laboratory experiment measured the rate of thermal decomposition (in $\text{mmol/L}\cdot\text{s}$ ) of four organic compounds (Compounds W, X, Y, and Z) across three temperatures ( $T$ , in $\text{K}$ ). The results are shown in the table below:

Compound	Rate at $300\text{ K}$	Rate at $350\text{ K}$	Rate at $400\text{ K}$
Compound W	$1.2$	$2.4$	$4.8$
Compound X	$5.0$	$3.5$	$2.0$
Compound Y	$0.8$	$0.8$	$0.8$
Compound Z	$2.0$	$4.0$	$6.0$

Based on the table, match each compound to its corresponding line or curve characteristic when translated into a rate versus temperature graph.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Compound W matches the concave-up exponential curve; Compound X matches the negative slope of

-0.03\text{ mmol/L}\cdot\text{s}\cdot\text{K}^{-1}

; Compound Y matches the horizontal zero-slope line at

0.8\text{ mmol/L}\cdot\text{s}

; Compound Z matches the positive slope of

0.04\text{ mmol/L}\cdot\text{s}\cdot\text{K}^{-1}

.

Each tabular dataset directly translates to a specific graphical shape or slope: exponential doubling yields a concave-up curve; equal decreases yield a linear negative slope of

-0.03

; constant values yield a zero-slope horizontal line; equal increases yield a linear positive slope of

0.04

.

Step-by-Step Solution

1

Analyze Compound W's data trend

The rates are

1.2

,

2.4

, and

4.8

at

300\text{ K}

,

350\text{ K}

, and

400\text{ K}

.

Dividing consecutive rates yields

\frac{2.4}{1.2} = 2

and

\frac{4.8}{2.4} = 2

, showing exponential growth (concave-up curve with a doubling pattern).

2

Analyze Compound X's data trend and calculate slope

Rate decreases from

5.0

to

3.5

to

2.0

as temperature rises from

300\text{ K}

to

400\text{ K}

.

The rate change per kelvin is

\frac{2.0 - 5.0}{400 - 300} = \frac{-3.0}{100} = -0.03\text{ mmol/L}\cdot\text{s}\cdot\text{K}^{-1}

.

3

Analyze Compound Y's data trend

The rate remains unchanged at

0.8\text{ mmol/L}\cdot\text{s}

across all temperatures.

A constant value across the independent variable axis translates to a horizontal line with zero slope.

4

Analyze Compound Z's data trend and calculate slope

Rate increases linearly from

2.0

to

4.0

to

6.0

.

The slope is

\frac{6.0 - 2.0}{400 - 300} = \frac{4.0}{100} = 0.04\text{ mmol/L}\cdot\text{s}\cdot\text{K}^{-1}

.

Key Concept

Translating Data Formats between Tables and Linear/Non-linear Graphical Features

Question 15Question →

Oceanographers measured the sound velocity in seawater at various depths under two different salinity levels ( $34\text{ ppt}$ and $36\text{ ppt}$ ). The results are recorded in Table 1 below:

Depth (m)	Sound Velocity at 34 ppt (m/s)	Sound Velocity at 36 ppt (m/s)
0	1,520	1,523
200	1,495	1,498
600	1,475	1,478
1,000	1,470	1,473
1,500	1,480	1,483
2,000	1,490	1,493

Statement: If the data in Table 1 were translated into a two-line graph plotting Sound Velocity (m/s) on the y-axis against Depth (m) on the x-axis, both curves would reach a minimum value at a depth of $1,000\text{ m}$ , and the curve representing $36\text{ ppt}$ salinity would be positioned strictly above the curve representing $34\text{ ppt}$ salinity across all measured depths.

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

The statement accurately reflects the conversion of the tabular data into a line graph. In Table 1, both sound velocity columns drop to their lowest values (

1,470\text{ m/s}

and

1,473\text{ m/s}

) at a depth of

1,000\text{ m}

before rising again, creating a V-shaped curve minimum at

x = 1,000\text{ m}

. Additionally, at every listed depth, the value for

36\text{ ppt}

is higher than for

34\text{ ppt}

, placing the

36\text{ ppt}

curve vertically above the

34\text{ ppt}

curve.

Step-by-Step Solution

1

Analyze the trend along the vertical axis variable (Sound Velocity) as a function of the horizontal axis variable (Depth) for both datasets.

For

34\text{ ppt}

, velocities are

1,520 \rightarrow 1,495 \rightarrow 1,475 \rightarrow 1,470 \rightarrow 1,480 \rightarrow 1,490

. The lowest point (minimum) occurs at

1,000\text{ m}

(

1,470\text{ m/s}

). For

36\text{ ppt}

, velocities are

1,523 \rightarrow 1,498 \rightarrow 1,478 \rightarrow 1,473 \rightarrow 1,483 \rightarrow 1,493

. The lowest point also occurs at

1,000\text{ m}

(

1,473\text{ m/s}

).

Determining where the graph reaches its lowest point requires identifying the minimum y-value for each x-value series.

2

Compare the relative vertical height (y-values) of the two salinity series across all depth levels (x-values).

At depth

0\text{ m}

:

1,523 > 1,520

; at

200\text{ m}

:

1,498 > 1,495

; at

600\text{ m}

:

1,478 > 1,475

; at

1,000\text{ m}

:

1,473 > 1,470

; at

1,500\text{ m}

:

1,483 > 1,480

; at

2,000\text{ m}

:

1,493 > 1,490

.

A curve with higher y-values at every x-value will be graphed higher up (above) another curve.

3

Synthesize the graphical behavior described in the statement with the tabular evidence.

Both curves reach a minimum at

1,000\text{ m}

and the

36\text{ ppt}

line remains strictly higher on the y-axis than the

34\text{ ppt}

line at all points. Thus, the statement is true.

The textual description of the proposed graph matches the tabular data perfectly in both shape (minimum location) and relative alignment.

Key Concept

Translating tabular data to graphical trends by matching values to axes, extrema, and relative curve placements.

Question 16Question →

A team of plant physiologists investigated the transpiration rate of four plant species (Species W, X, Y, and Z) under two distinct light conditions—Low Intensity ( $150\text{ }\mu\text{mol/m}^2\text{/s}$ ) and High Intensity ( $800\text{ }\mu\text{mol/m}^2\text{/s}$ )—at a controlled temperature of $25^\circ\text{C}$ . The measured rates are shown in the table below:

Plant Species	Transpiration Rate at Low Intensity ( $\text{mg H}_2\text{O/cm}^2\text{/hr}$ )	Transpiration Rate at High Intensity ( $\text{mg H}_2\text{O/cm}^2\text{/hr}$ )
Species W	$1.2$	$4.8$
Species X	$2.5$	$6.5$
Species Y	$0.8$	$3.2$
Species Z	$1.5$	$5.5$

Based on the table, what was the average increase in transpiration rate (in $\text{mg H}_2\text{O/cm}^2\text{/hr}$ ) across all four plant species when light condition was increased from Low Intensity to High Intensity?

1.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

5.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

Show answer & explanation

Answer: $3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}$

Answer

The average increase in transpiration rate across the four species is

3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

.

To find the average increase, first subtract the Low Intensity transpiration rate from the High Intensity transpiration rate for each of the four species: Species W (

4.8 - 1.2 = 3.6

), Species X (

6.5 - 2.5 = 4.0

), Species Y (

3.2 - 0.8 = 2.4

), and Species Z (

5.5 - 1.5 = 4.0

). Adding these four increases yields a total of

14.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

. Dividing this sum by

4

species gives an average increase of

3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

.

Step-by-Step Solution

1

Calculate the difference between High Intensity and Low Intensity transpiration rates for each species.

Species W:

4.8 - 1.2 = 3.6\text{ mg H}_2\text{O/cm}^2\text{/hr}

; Species X:

6.5 - 2.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

; Species Y:

3.2 - 0.8 = 2.4\text{ mg H}_2\text{O/cm}^2\text{/hr}

; Species Z:

5.5 - 1.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

.

Determines the specific rate increase for each trial.

2

Sum the calculated rate increases across all four species.

3.6 + 4.0 + 2.4 + 4.0 = 14.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

.

Aggregates the individual increases to calculate the total change.

3

Divide the total sum of increases by the number of species (

4

) to determine the mean increase.

14.0 / 4 = 3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}

.

Yields the average (mean) change in transpiration rate per species.

Key Concept

Calculating mean changes and rate differences from tabular data

Estimated Time:1m 0s

Question 17Question →

Astrophysicists measured the transit depth (percentage of starlight blocked) for four exoplanets orbiting a host star at two observation wavelengths ( $0.5\ \mu\text{m}$ and $1.5\ \mu\text{m}$ ), as recorded in the table below:

Exoplanet	Transit Depth at $0.5\ \mu\text{m}$ (%)	Transit Depth at $1.5\ \mu\text{m}$ (%)
Exoplanet 1	1.2	1.8
Exoplanet 2	2.5	2.1
Exoplanet 3	0.8	1.6
Exoplanet 4	3.0	3.6

True or False: The mean transit depth across all four exoplanets measured at $1.5\ \mu\text{m}$ is $0.4\%$ greater than the mean transit depth measured at $0.5\ \mu\text{m}$ .

True

False

Show answer & explanation

Answer: True

Answer

The statement is True.

The mean transit depth at 1.5 μm is 2.275%, and the mean transit depth at 0.5 μm is 1.875%. Subtracting 1.875% from 2.275% gives exactly 0.4%, making the true/false statement correct.

Step-by-Step Solution

1

Calculate the mean transit depth at 0.5 μm.

Sum = 1.2 + 2.5 + 0.8 + 3.0 = 7.5%. Mean = 7.5 / 4 = 1.875%.

Determines the baseline average value for the first wavelength column.

2

Calculate the mean transit depth at 1.5 μm.

Sum = 1.8 + 2.1 + 1.6 + 3.6 = 9.1%. Mean = 9.1 / 4 = 2.275%.

Determines the target average value for the second wavelength column.

3

Calculate the difference between the two averages.

2.275% - 1.875% = 0.4%.

Compares the calculated difference against the claimed value of 0.4%.

Key Concept

Calculating and comparing average values from structured tabular data

Question 18Question →

A team of marine biologists studied the acoustic activity of the snapping shrimp (*Alpheus heterochaelis*) under controlled laboratory conditions. They measured the average snap rate (in snaps per minute) across three water temperatures and two salinity levels ( $30\text{ PSU}$ and $36\text{ PSU}$ ).

Table 1

Water Temperature ( $^\circ\text{C}$ )	Snap Rate at $30\text{ PSU}$ (snaps/min)	Snap Rate at $36\text{ PSU}$ (snaps/min)
$18$	$45$	$50$
$22$	$65$	$85$
$26$	$109$	$130$

Based on Table 1, what was the average rate of increase in snap rate (in snaps/min per $1^\circ\text{C}$ ) for shrimp kept at a salinity of $36\text{ PSU}$ as the water temperature increased from $18^\circ\text{C}$ to $26^\circ\text{C}$ ?

8.0\text{ snaps/min per }^\circ\text{C}

10.0\text{ snaps/min per }^\circ\text{C}

40.0\text{ snaps/min per }^\circ\text{C}

80.0\text{ snaps/min per }^\circ\text{C}

Show answer & explanation

Answer: $10.0\text{ snaps/min per }^\circ\text{C}$

Answer

The average rate of increase in snap rate for shrimp at

36\text{ PSU}

from

18^\circ\text{C}

to

26^\circ\text{C}

is

10.0\text{ snaps/min per }^\circ\text{C}

.

To find the average rate of increase per

1^\circ\text{C}

, determine the difference in snap rate at

36\text{ PSU}

between

18^\circ\text{C}

and

26^\circ\text{C}

, which is

130 - 50 = 80\text{ snaps/min}

. Then divide by the temperature span of

26 - 18 = 8^\circ\text{C}

, yielding

80 / 8 = 10.0\text{ snaps/min per }^\circ\text{C}

.

Step-by-Step Solution

1

Locate the snap rates for

36\text{ PSU}

at

18^\circ\text{C}

and

26^\circ\text{C}

in Table 1.

At

18^\circ\text{C}

, snap rate =

50\text{ snaps/min}

; at

26^\circ\text{C}

, snap rate =

130\text{ snaps/min}

.

The question asks specifically about the

36\text{ PSU}

salinity condition.

2

Calculate the total change in snap rate and total change in temperature.

Change in snap rate =

130 - 50 = 80\text{ snaps/min}

; Change in temperature =

26 - 18 = 8^\circ\text{C}

.

Rate of change requires dividing the change in the dependent variable by the change in the independent variable.

3

Divide the change in snap rate by the change in temperature.

80\text{ snaps/min} / 8^\circ\text{C} = 10.0\text{ snaps/min per }^\circ\text{C}

.

This yields the average change per

1^\circ\text{C}

increase.

Key Concept

Calculating rate of change from tabular data

Estimated Time:1m 15s

Question 19Question →

A group of biochemical researchers measured the enzymatic breakdown rate of cellulose (in $\text{mg/L}\cdot\text{min}$ ) by three bacterial strains (Strain X, Strain Y, and Strain Z) across three different pH environments ( $5.0$ , $6.0$ , and $7.0$ ) at a constant temperature of $37^\circ\text{C}$ . The results are recorded in the table below:

Strain	Breakdown Rate at pH $5.0$	Breakdown Rate at pH $6.0$	Breakdown Rate at pH $7.0$
Strain X	$12$	$18$	$24$
Strain Y	$15$	$25$	$35$
Strain Z	$8$	$14$	$20$

True or False: The average enzymatic breakdown rate across all three pH environments for Strain Y is $5\text{ mg/L}\cdot\text{min}$ greater than the average breakdown rate for Strain X.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False because the average breakdown rate for Strain Y (

25\text{ mg/L}\cdot\text{min}

) exceeds that of Strain X (

18\text{ mg/L}\cdot\text{min}

) by

7\text{ mg/L}\cdot\text{min}

, not

5\text{ mg/L}\cdot\text{min}

.

The correct response is False because computing the mean breakdown rates yields

18\text{ mg/L}\cdot\text{min}

for Strain X and

25\text{ mg/L}\cdot\text{min}

for Strain Y. Subtracting

18

from

25

gives a difference of

7\text{ mg/L}\cdot\text{min}

, rendering the statement claiming a difference of

5\text{ mg/L}\cdot\text{min}

mathematically incorrect.

Step-by-Step Solution

1

Calculate the average breakdown rate for Strain X across all three pH levels.

Average for Strain X =

\frac{12 + 18 + 24}{3} = \frac{54}{3} = 18\text{ mg/L}\cdot\text{min}

.

To determine the overall mean performance of Strain X across the tested environments.

2

Calculate the average breakdown rate for Strain Y across all three pH levels.

Average for Strain Y =

\frac{15 + 25 + 35}{3} = \frac{75}{3} = 25\text{ mg/L}\cdot\text{min}

.

To determine the overall mean performance of Strain Y across the tested environments.

3

Subtract the average rate of Strain X from the average rate of Strain Y.

Difference =

25 - 18 = 7\text{ mg/L}\cdot\text{min}

.

To test the claim that the average for Strain Y is

5\text{ mg/L}\cdot\text{min}

greater than Strain X.

Key Concept

Calculating mean values from tabular experimental data and finding differences between summary statistics.

Question 20Question →

Bioacousticians measured the echolocation click rate (in clicks per minute, $\text{clicks/min}$ ) of a harbor porpoise at four different water depths across three 15-minute observation trials. The results are summarized in the table below:

Water Depth (m)	Trial 1 (clicks/min)	Trial 2 (clicks/min)	Trial 3 (clicks/min)
10	120	115	125
20	140	148	138
30	185	170	185
40	210	205	215

Based on the table, what is the average echolocation click rate, in $\text{clicks/min}$ , across all three trials at a depth of 30 m?

Show answer & explanation

Answer: 180

Answer

The average echolocation click rate at a depth of 30 m across all three trials is 180 clicks/min.

To calculate the average echolocation click rate at 30 m depth, locate the row for 30 m, add the click rates from the three trials (

185 + 170 + 185 = 540

), and divide by the number of trials (

540 \div 3 = 180\text{ clicks/min}

).

Step-by-Step Solution

1

Locate the depth row for 30 m in the provided data table.

Retrieved values for Trial 1 (185), Trial 2 (170), and Trial 3 (185).

The question specifically requests calculations for the 30 m depth.

2

Sum the recorded click rates across the three trials.

185 + 170 + 185 = 540\text{ clicks/min}

.

Obtaining the sum is required before dividing by the total count of trials.

3

Divide the calculated sum by 3.

\frac{540}{3} = 180\text{ clicks/min}

.

Calculating the arithmetic mean yields the average click rate.

Key Concept

Calculating the arithmetic mean from tabular data

Estimated Time:1m 0s

Data Representation and Interpretation

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Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Answer

Step-by-Step Solution

Key Concept