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[tex]\\ \sf\Rrightarrow 2\dfrac{1}{4}-\dfrac{3}{8}[/tex]

[tex]\\ \sf\Rrightarrow \dfrac{9}{4}-\dfrac{3}{8}[/tex]

[tex]\\ \sf\Rrightarrow \dfrac{18-3}{8}[/tex]

[tex]\\ \sf\Rrightarrow \dfrac{15}{8}[/tex]

[tex]\\ \sf\Rrightarrow 1\dfrac{7}{8}[/tex]

Answer:

[tex]\textsf{D)} \quad 1 \dfrac{7}{8}[/tex]

Step-by-step explanation:

Subtract the used volume of punch from the starting volume of punch.

Convert the mixed number to an improper fraction:

[tex]\begin{aligned} \implies 2\frac{1}{4}-\frac{3}{8} & = \dfrac{2 \times 4+1}{4}-\dfrac{3}{8}\\\\ & = \dfrac{9}{4}-\dfrac{3}{8}\end{aligned}[/tex]

Make the denominators of both fractions the same by multiplying the numerator and the denominator of 9/4 by 2:

[tex]\implies \dfrac{9 \times 2}{4 \times 2}-\dfrac{3}{8} = \dfrac{18}{8}-\dfrac{3}{8}[/tex]

[tex]\textsf{Apply the fraction rule} \quad \dfrac{a}{c}-\dfrac{b}{c}=\dfrac{a-b}{c}:[/tex]

[tex]\begin{aligned} \implies \dfrac{18}{8}-\dfrac{3}{8} & = \dfrac{18-3}{8}\\\\ & = \dfrac{15}{8}\end{aligned}[/tex]

Convert back into a mixed number:

[tex]\begin{aligned}\implies \dfrac{15}{8} & = \dfrac{8+7}{8}\\\\ & = \dfrac{8}{8}+\dfrac{7}{8}\\\\ & = 1\dfrac{7}{8}\end{aligned}[/tex]

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Triangle Midsegment Theorem