Rational number line locations (ACT). Game on.

The interval is open: $-\dfrac{19}{4} = -4.75$ and $\dfrac{11}{2} = 5.5$, so values must satisfy $-4.75 < x < 5.5$. Check each choice: $-5 < -4.75$ (outside), $-4$ is between the bounds, $\dfrac{11}{2}$ equals the upper endpoint (not strict), and $-\dfrac{19}{4}$ equals the lower endpoint (not strict).

Compare positions: $\dfrac{9}{5} = 1.8$, $-\dfrac{2}{3} \approx -0.67$, $\dfrac{5}{8} = 0.625$, and $-\dfrac{7}{12} \approx -0.58$. The greatest is $\dfrac{9}{5}$.

Distance from zero is absolute value: $\left|-\dfrac{23}{8}\right| = \dfrac{23}{8}$.

Convert the bounds: $-\dfrac{83}{7} \approx -11.86$ and $\dfrac{67}{4} = 16.75$. Strictly between means integers $n$ with $-\dfrac{83}{7} < n < \dfrac{67}{4}$, so $n$ runs from $-11$ through $16$, including $0$. Count with $16 - (-11) + 1$: subtracting endpoints gives the gap between them; the $+1$ counts both $-11$ and $16$ inclusive.

Convert: $-\dfrac{91}{5} = -18.2$ and $-\dfrac{12}{4} = -3$. Integers $n$ with $-18.2 < n < -3$ are $-18, -17, \ldots, -4$. Count with $-4 - (-18) + 1$: subtracting endpoints gives the gap between them; the $+1$ counts both $-18$ and $-4$ inclusive.

Convert: $\dfrac{7}{3} \approx 2.33$ and $\dfrac{47}{5} = 9.4$. Integers $n$ with $2.33 < n < 9.4$ are $3, 4, 5, 6, 7, 8, 9$. Count with $9 - 3 + 1$: subtracting endpoints gives the gap between them; the $+1$ counts both $3$ and $9$ inclusive.

$-\dfrac{29}{6} \approx -4.83$. Compare distances: $|{-5} - (-4.83)| \approx 0.17$ and $|{-4} - (-4.83)| \approx 0.83$. The closest integer is $-5$.

$-\dfrac{11}{2} = -5.5$, so integers greater than $-5.5$ start at $-5$; the closed upper bound includes $4$. Integers: $-5, -4, \ldots, 0, \ldots, 4$; $0$ is an integer and must be counted. Total: $4 - (-5) + 1 = 10$; the $+1$ counts both $-5$ and $4$ inclusive.