The distance between $\dfrac{5}{8}$ and $\dfrac{7}{4}$ is $\dfrac{7}{4} - \dfrac{5}{8} = \dfrac{14}{8} - \dfrac{5}{8} = \dfrac{9}{8}$.
Two-thirds of this distance is $\dfrac{2}{3} \cdot \dfrac{9}{8} = \dfrac{6}{8} = \boxed{\dfrac{3}{4}}$.
On a number line, what number is two-thirds of the way from $\dfrac{5}{8}$ to $\dfrac{7}{4}\,?$ Express your answer as a common fraction.
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Let $x$ be the number that is two-thirds of the way from $\frac{5}{8}$ to $\frac{7}{4}$. We can represent this as:
\[\frac{7}{4} - \frac{5}{8} = \frac{2}{3} \cdot (x - \frac{5}{8}).\]
Solving the equation above, we get
\[\frac{7}{4} - \frac{5}{8} = \frac{2}{3} \cdot (x - \frac{5}{8})\]
\[\frac{14}{8} - \frac{5}{8} = \frac{2}{3} \cdot (x - \frac{5}{8})\]
\[\frac{9}{8} = \frac{2}{3} \cdot (x - \frac{5}{8})\]
\[\frac{27}{24} = \frac{16}{24} \cdot (x - \frac{5}{8})\]
\[\frac{27}{24} = \frac{16x - 10}{24}\]
\[27 = 16x - 10\]
\[16x = 37\]
\[x = \frac{37}{16}\]
Therefore, two-thirds of the way from $\frac{5}{8}$ to $\frac{7}{4}$ is $\boxed{\frac{37}{16}}$.
\[\frac{7}{4} - \frac{5}{8} = \frac{2}{3} \cdot (x - \frac{5}{8}).\]
Solving the equation above, we get
\[\frac{7}{4} - \frac{5}{8} = \frac{2}{3} \cdot (x - \frac{5}{8})\]
\[\frac{14}{8} - \frac{5}{8} = \frac{2}{3} \cdot (x - \frac{5}{8})\]
\[\frac{9}{8} = \frac{2}{3} \cdot (x - \frac{5}{8})\]
\[\frac{27}{24} = \frac{16}{24} \cdot (x - \frac{5}{8})\]
\[\frac{27}{24} = \frac{16x - 10}{24}\]
\[27 = 16x - 10\]
\[16x = 37\]
\[x = \frac{37}{16}\]
Therefore, two-thirds of the way from $\frac{5}{8}$ to $\frac{7}{4}$ is $\boxed{\frac{37}{16}}$.