Question

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c) (-1\frac{1}{2} \times \frac{3}{4} div \frac{2}{3})

Explanation:

Step1: Convert mixed number to improper fraction

$-1\frac{1}{2}=-\frac{3}{2}$

Step2: Replace in the expression and change division to multiplication (multiply by reciprocal)

$-\frac{3}{2} \times \frac{3}{4} \times \frac{3}{2}$ (Wait, no, reciprocal of $\frac{2}{3}$ is $\frac{3}{2}$, so the expression becomes $-\frac{3}{2} \times \frac{3}{4} \times \frac{3}{2}$? Wait, no, original is $-1\frac{1}{2} \times \frac{3}{4} \div \frac{2}{3}$, so division by $\frac{2}{3}$ is multiplication by $\frac{3}{2}$. So correct step: $-\frac{3}{2} \times \frac{3}{4} \times \frac{3}{2}$? Wait, no, $-1\frac{1}{2}$ is $-\frac{3}{2}$, then multiply by $\frac{3}{4}$, then divide by $\frac{2}{3}$ (multiply by $\frac{3}{2}$). So:
$-\frac{3}{2} \times \frac{3}{4} \times \frac{3}{2}$? Wait, no, $\frac{3}{4} \times \frac{3}{2}$? Wait, no, let's do it step by step. First, $-1\frac{1}{2} = -\frac{3}{2}$. Then the expression is $-\frac{3}{2} \times \frac{3}{4} \div \frac{2}{3}$. Dividing by a fraction is multiplying by its reciprocal, so $\div \frac{2}{3} = \times \frac{3}{2}$. So now the expression is $-\frac{3}{2} \times \frac{3}{4} \times \frac{3}{2}$? Wait, no, $\frac{3}{4} \times \frac{3}{2}$? Wait, no, $-\frac{3}{2} \times \frac{3}{4} = -\frac{9}{8}$, then multiply by $\frac{3}{2}$: $-\frac{9}{8} \times \frac{3}{2} = -\frac{27}{16}$? Wait, no, wait, maybe I made a mistake. Wait, original expression: $-1\frac{1}{2} \times \frac{3}{4} \div \frac{2}{3}$. Let's re-express:

First, convert mixed number: $-1\frac{1}{2} = -\frac{3}{2}$.

Then, division by $\frac{2}{3}$ is multiplication by $\frac{3}{2}$. So the expression becomes:

$-\frac{3}{2} \times \frac{3}{4} \times \frac{3}{2}$? Wait, no, $\frac{3}{4} \times \frac{3}{2}$? Wait, no, $\frac{3}{4} \times \frac{3}{2}$ is $\frac{9}{8}$, then multiply by $-\frac{3}{2}$: $-\frac{3}{2} \times \frac{9}{8} = -\frac{27}{16}$? Wait, but let's check again. Wait, maybe I messed up the reciprocal. Wait, $\div \frac{2}{3}$ is $\times \frac{3}{2}$, so the order is: $(-\frac{3}{2}) \times (\frac{3}{4}) \times (\frac{3}{2})$. Let's compute numerator: $3 \times 3 \times 3 = 27$? Wait, no, $-\frac{3}{2} \times \frac{3}{4} = -\frac{9}{8}$, then $-\frac{9}{8} \times \frac{3}{2} = -\frac{27}{16}$. Wait, but let's do it correctly. Wait, maybe I made a mistake in the reciprocal. Wait, $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$. So here, $\frac{3}{4} \div \frac{2}{3} = \frac{3}{4} \times \frac{3}{2} = \frac{9}{8}$. Then multiply by $-1\frac{1}{2} = -\frac{3}{2}$. So $-\frac{3}{2} \times \frac{9}{8} = -\frac{27}{16}$. Yes, that's correct.

Wait, let's redo the steps properly:

Step1: Convert mixed number to improper fraction

$-1\frac{1}{2} = -\frac{2\times1 + 1}{2}=-\frac{3}{2}$

Step2: Change division to multiplication (reciprocal)

$\div \frac{2}{3} = \times \frac{3}{2}$, so the expression becomes $-\frac{3}{2} \times \frac{3}{4} \times \frac{3}{2}$? Wait, no, the order is $(-1\frac{1}{2}) \times \frac{3}{4} \div \frac{2}{3}$, so it's $(-\frac{3}{2}) \times (\frac{3}{4} \div \frac{2}{3})$? No, order of operations: multiplication and division are left to right. So first, multiply $-1\frac{1}{2}$ and $\frac{3}{4}$, then divide by $\frac{2}{3}$.

So Step1: Multiply $-1\frac{1}{2}$ and $\frac{3}{4}$:

$-1\frac{1}{2} \times \frac{3}{4} = -\frac{3}{2} \times \frac{3}{4} = -\frac{9}{8}$

Step2: Divide the result by $\frac{2}{3}$:

$-\frac{9}{8} \div \frac{2}{3} = -\frac{9}{8} \times \frac{3}{2} = -\frac{27}{16}$

Yes, that's correct. So:

Step1: Convert mixed number

$-1\frac{1}{2} = -\frac{3}{2}$

Step2: Multiply by $\frac{3}{4}$

$-\frac{3}{2} \times \frac{3}{4} = -\frac{9}{8}$

Step3: Divide by $\frac{2}{3}$ (multiply by reciprocal)

$-\frac{9}{8} \times \frac{3}{2} = -\frac{27}{16}$

Answer:

$-\frac{27}{16}$