Given that D = [[-6, -4], [-7, 3]] and E = [[-5, 0], [-8, 9]], calculate a. D+E, b. D-E, c. 2D-E

$$D+E = \begin{bmatrix} -11 & -4 \\ -15 & 12 \end{bmatrix}$$

The sum is $\boxed{\mathbf{\begin{bmatrix} -11 & -4 \\ -15 & 12 \end{bmatrix}}}$.

$$D-E = \begin{bmatrix} -6 + 5 & -4 \\ -7 + 8 & -6 \end{bmatrix}$$ $$D-E = \begin{bmatrix} -1 & -4 \\ 1 & -6 \end{bmatrix}$$

The difference is $\boxed{\mathbf{\begin{bmatrix} -1 & -4 \\ 1 & -6 \end{bmatrix}}}$.

$$2D-E = \begin{bmatrix} -12 - (-5) & -8 - 0 \\ -14 - (-8) & 6 - 9 \end{bmatrix}$$ $$2D-E = \begin{bmatrix} -12 + 5 & -8 \\ -14 + 8 & -3 \end{bmatrix}$$ $$2D-E = \begin{bmatrix} -7 & -8 \\ -6 & -3 \end{bmatrix}$$

The result is $\boxed{\mathbf{\begin{bmatrix} -7 & -8 \\ -6 & -3 \end{bmatrix}}}$.

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$$P = \left(\frac{22}{7} \times 30\right) + (2 \times 30)$$ $$P = \frac{660}{7} + 60$$

$$P \approx 94.2857 + 60$$ $$P \approx 154.2857 \text{ cm}$$

$$P \approx 154.3 \text{ cm}$$

The correct option is D. 154.3cm.

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$$T = \frac{12}{8} \text{ hours}$$ $$T = \frac{3}{2} \text{ hours}$$

$$T = 1 \frac{1}{2} \text{ hours}$$

The correct option is A. $1 \frac{1}{2}$ hours.

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Number of students playing at least one game = Total students - Students who play neither

$$|H \cup G| = 20 - 2 = 18$$

The number of students who play at least one game can also be expressed as:
$|H \cup G| = |H| + |G \text{ only}|$ (where $|G \text{ only}|$ is the number of students who play only the other game).

$$18 = 16 + |G \text{ only}|$$ $$|G \text{ only}| = 18 - 16 = 2$$

So, 2 students play only the other game.

Let this be $|H \text{ only}|$.
We know that $|H| = |H \text{ only}| + |H \cap G|$ (where $|H \cap G|$ is the number of students who play both games).

$$16 = |H \text{ only}| + |H \cap G|$$

The problem does not provide enough information to uniquely determine $|H \text{ only}|$ or $|H \cap G|$ individually. However, since this is a multiple-choice question, we can assume there's an implicit condition. A common assumption in such problems, when information is limited, is that the number of students playing only one game is equal to the number of students playing only the other game.
Assuming $|H \text{ only}| = |G \text{ only}|$.

$$|H \text{ only}| = 2$$

The number of students who play only hockey is 2.
The correct option is D. 2.