Question

sun exposure (%) stem mass (g) stem volume (ml)
30 275 1100
45 415 1215
60 563 1425
75 815 1610
90 954 1742

a. convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters (m³).

b. calculate the density of the samples using the equation d = m/v.
\td = density
\tm = mass (kg)
\tv = volume (m³)

c. convert the density values to scientific notation.

d. select the graph that best represents the data.

a b
effects of sunlight on plant stems effects of sunlight on plant stems
c d
effects of sunlight on plant stems effects of sunlight on plant stems

Answer:

Part a: Unit Conversion

Step 1: Convert mass (g to kg)

We know that \(1\space kg = 1000\space g\), so to convert grams to kilograms, we divide by 1000.

• For \(275\space g\): \(\frac{275}{1000}= 0.275\space kg\)
• For \(415\space g\): \(\frac{415}{1000}= 0.415\space kg\)
• For \(563\space g\): \(\frac{563}{1000}= 0.563\space kg\)
• For \(815\space g\): \(\frac{815}{1000}= 0.815\space kg\)
• For \(954\space g\): \(\frac{954}{1000}= 0.954\space kg\)

Step 2: Convert volume (mL to \(m^3\))

We know that \(1\space mL = 1\space cm^3\) and \(1\space m = 100\space cm\), so \(1\space m^3=(100)^3\space cm^3 = 10^6\space cm^3=10^6\space mL\). To convert mL to \(m^3\), we divide by \(10^6\).

• For \(1100\space mL\): \(\frac{1100}{10^6}= 1.1\times 10^{- 3}\space m^3\)
• For \(1215\space mL\): \(\frac{1215}{10^6}= 1.215\times 10^{-3}\space m^3\)
• For \(1425\space mL\): \(\frac{1425}{10^6}= 1.425\times 10^{-3}\space m^3\)
• For \(1610\space mL\): \(\frac{1610}{10^6}= 1.61\times 10^{-3}\space m^3\)
• For \(1742\space mL\): \(\frac{1742}{10^6}= 1.742\times 10^{-3}\space m^3\)

Part b: Calculate Density

The formula for density is \(d=\frac{m}{v}\)

Step 1: For sun exposure \(30\%\)

\(m = 0.275\space kg\), \(v=1.1\times 10^{-3}\space m^3\)
\(d=\frac{0.275}{1.1\times 10^{-3}}=\frac{275}{1.1}= 250\space kg/m^3\)

Step 2: For sun exposure \(45\%\)

\(m = 0.415\space kg\), \(v = 1.215\times 10^{-3}\space m^3\)
\(d=\frac{0.415}{1.215\times 10^{-3}}\approx\frac{415}{1.215}\approx 341.56\space kg/m^3\)

Step 3: For sun exposure \(60\%\)

\(m = 0.563\space kg\), \(v=1.425\times 10^{-3}\space m^3\)
\(d=\frac{0.563}{1.425\times 10^{-3}}\approx\frac{563}{1.425}\approx 395.09\space kg/m^3\)

Step 4: For sun exposure \(75\%\)

\(m = 0.815\space kg\), \(v = 1.61\times 10^{-3}\space m^3\)
\(d=\frac{0.815}{1.61\times 10^{-3}}=\frac{815}{1.61} = 506.21\space kg/m^3\)

Step 5: For sun exposure \(90\%\)

\(m = 0.954\space kg\), \(v=1.742\times 10^{-3}\space m^3\)
\(d=\frac{0.954}{1.742\times 10^{-3}}\approx\frac{954}{1.742}\approx 547.64\space kg/m^3\)

Part c: Convert to Scientific Notation

• For \(250\space kg/m^3\): \(250 = 2.5\times 10^{2}\space kg/m^3\)
• For \(341.56\space kg/m^3\): \(341.56\approx 3.42\times 10^{2}\space kg/m^3\)
• For \(395.09\space kg/m^3\): \(395.09\approx 3.95\times 10^{2}\space kg/m^3\)
• For \(506.21\space kg/m^3\): \(506.21\approx 5.06\times 10^{2}\space kg/m^3\)
• For \(547.64\space kg/m^3\): \(547.64\approx 5.48\times 10^{2}\space kg/m^3\)

Part d: Select the Graph

We analyze the variables:

• The independent variable can be sun exposure (%), and the dependent variables can be mass, volume, or density.
• Looking at the trend, as sun exposure increases, mass, volume, and density (calculated) also increase.
• Graph A: If the x - axis is sun exposure (%) and y - axis is stem mass (g) (or kg after conversion), the trend of increasing mass with increasing sun exposure matches.
• Graph B: x - axis is stem volume (mL) and y - axis is sun exposure (%). The trend of sun exposure increasing with volume is not the typical way to represent the data (usually independent variable on x - axis).
• Graph C: The x - axis and y - axis labels are not clear from the description, but from the context, it's less likely.
• Graph D: The x - axis is sun exposure (%) and y - axis is density. The trend of increasing density with sun exposure also matches, but we need to check the axes. However, if we consider the original data, when we plot sun exposure vs mass (or density), the linear trend (approximate) is best represented by a graph where the x - axis is sun exposure and y - axis is the dependent variable (mass, density etc.). Among the options, if we assume that Graph A has x - axis as sun exposure and y - axis as mass (which increases with sun exposure), Graph A is a better representation. If we consider density, since density also increases with sun exposure, but from the initial data, the mass - sun exposure and volume - sun exposure have a more direct linear trend (visually). So the best graph is likely Graph A (assuming the axes are sun exposure on x and mass/volume/density on y with increasing trend).

Final Answers

Part a:

Sun exposure (%)	Stem mass (kg)	Stem volume (\(m^3\))
45	0.415	\(1.215\times 10^{-3}\)
60	0.563	\(1.425\times 10^{-3}\)
75	0.815	\(1.61\times 10^{-3}\)
90	0.954	\(1.742\times 10^{-3}\)

Part b:

Sun exposure (%)	Density (\(kg/m^3\))
45	\(\approx341.56\)
60	\(\approx395.09\)
75	506.21
90	\(\approx547.64\)

Part c:

Sun exposure (%)	Density (Scientific Notation)
45	\(\approx3.42\times 10^{2}\)
60	\(\approx3.95\times 10^{2}\)
75	\(5.06\times 10^{2}\)
90	\(\approx5.48\times 10^{2}\)

Part d:

The best graph is Graph A (assuming the axes are set such that the independent variable (sun exposure) is on the x - axis and the dependent variable (mass/volume/density) is on the y - axis with an increasing trend).