Let's carefully analyze each proposed multiplication and determine if it's possible, then we'll calculate the result if it's feasible.
### First proposed multiplication: [tex]\((3.1 \, \text{dL}) \cdot (0.30 \, \text{L})\)[/tex]
For this calculation, remember that 1 liter (L) is equal to 10 deciliters (dL). To perform the multiplication, we need to consider the unit conversion:
[tex]\[
3.1 \, \text{dL} \times 0.30 \, \text{L}
\][/tex]
[tex]\[
= 3.1 \, \text{dL} \times 0.30 \times 10 \, \text{dL} \quad \text{(since 1 L = 10 dL)}
\][/tex]
[tex]\[
= 3.1 \times 3.0 \, \text{dL}^2
\][/tex]
[tex]\[
= 9.3 \, \text{dL}^2
\][/tex]
So, the multiplication is possible, and the result is [tex]\(0.93 \, \text{dL}^2\)[/tex].
### Second proposed multiplication: [tex]\((3.0 \, \text{g}^3) \cdot (0.013 \, \text{kg})\)[/tex]
In this case, we are trying to multiply units of different dimensions: cubic grams ([tex]\(\text{g}^3\)[/tex]) and kilograms (kg). These units are not compatible for multiplication in a meaningful physical context. Therefore, this multiplication is not possible, and no result needs to be calculated.
### Third proposed multiplication: [tex]\((9.0 \, \text{mL}) \cdot (6.0 \, \text{mL})\)[/tex]
Here we are trying to multiply two volumes, both in milliliters (mL). Multiplying two volumes together would result in a new unit of [tex]\(\text{mL}^2\)[/tex], which is not a standard or meaningful unit of measure for volume. Therefore, this multiplication is not considered possible in practical terms.
### Conclusion
Based on this analysis, we can fill in the table as follows:
[tex]\[
\begin{array}{|c|c|c|}
\hline
\text{proposed multiplication or division} & \text{Is this possible?} & \text{result} \\
\hline
(3.1 \, \text{dL}) \cdot (0.30 \, \text{L}) & \text{yes} & 0.93 \, \text{dL}^2 \\
\hline
\left(3.0 \, \text{g}^3\right) \cdot (0.013 \, \text{kg}) & \text{no} & \\
\hline
(9.0 \, \text{mL}) \cdot (6.0 \, \text{mL}) & \text{no} & \\
\hline
\end{array}
\][/tex]
