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Zorluk: OrtaCalculating with Data

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 μmol/m2/s150\text{ }\mu\text{mol/m}^2\text{/s}) and High Intensity (800 μmol/m2/s800\text{ }\mu\text{mol/m}^2\text{/s})—at a controlled temperature of 25C25^\circ\text{C}. The measured rates are shown in the table below:

Plant SpeciesTranspiration Rate at Low Intensity (mg H2O/cm2/hr\text{mg H}_2\text{O/cm}^2\text{/hr})Transpiration Rate at High Intensity (mg H2O/cm2/hr\text{mg H}_2\text{O/cm}^2\text{/hr})
Species W1.21.24.84.8
Species X2.52.56.56.5
Species Y0.80.83.23.2
Species Z1.51.55.55.5

Based on the table, what was the average increase in transpiration rate (in mg H2O/cm2/hr\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. A
    1.5 mg H2O/cm2/hr1.5\text{ mg H}_2\text{O/cm}^2\text{/hr}
  2. 3.5 mg H2O/cm2/hr3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}Cevap
  3. C
    4.0 mg H2O/cm2/hr4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}
  4. D
    5.0 mg H2O/cm2/hr5.0\text{ mg H}_2\text{O/cm}^2\text{/hr}

Cevap

The average increase in transpiration rate across the four species is 3.5 mg H2O/cm2/hr3.5\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 (4.81.2=3.64.8 - 1.2 = 3.6), Species X (6.52.5=4.06.5 - 2.5 = 4.0), Species Y (3.20.8=2.43.2 - 0.8 = 2.4), and Species Z (5.51.5=4.05.5 - 1.5 = 4.0). Adding these four increases yields a total of 14.0 mg H2O/cm2/hr14.0\text{ mg H}_2\text{O/cm}^2\text{/hr}. Dividing this sum by 44 species gives an average increase of 3.5 mg H2O/cm2/hr3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}.

Adım Adım Çözüm

1
Calculate the difference between High Intensity and Low Intensity transpiration rates for each species.
Species W: 4.81.2=3.6 mg H2O/cm2/hr4.8 - 1.2 = 3.6\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species X: 6.52.5=4.0 mg H2O/cm2/hr6.5 - 2.5 = 4.0\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species Y: 3.20.8=2.4 mg H2O/cm2/hr3.2 - 0.8 = 2.4\text{ mg H}_2\text{O/cm}^2\text{/hr}; Species Z: 5.51.5=4.0 mg H2O/cm2/hr5.5 - 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 mg H2O/cm2/hr3.6 + 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 (44) to determine the mean increase.
14.0/4=3.5 mg H2O/cm2/hr14.0 / 4 = 3.5\text{ mg H}_2\text{O/cm}^2\text{/hr}.
Yields the average (mean) change in transpiration rate per species.

Anahtar Kavram

Calculating mean changes and rate differences from tabular data
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