Solve the following: 1) 3(1)/(2) + (3(1)/(2) of 3(1)/(2)) of (1)/(8) 2) (0.325 x 2.76 x 100)/(13 x 0.25 x 92)

Here are the solutions to the problems:

1) Using your knowledge of BODMAS, solve the following:

$$3\frac{1}{2} + \left(3\frac{1}{2} \text{ of } 3\frac{1}{2}\right) \text{ of } \frac{1}{8}$$

Step 1: Convert mixed numbers to improper fractions.

$$3\frac{1}{2} = \frac{3 \times 2 + 1}{2} = \frac{7}{2}$$

The expression becomes:

$$\frac{7}{2} + \left(\frac{7}{2} \times \frac{7}{2}\right) \times \frac{1}{8}$$

Step 2: Solve the operation inside the parentheses (multiplication).

$$\left(\frac{7}{2} \times \frac{7}{2}\right) = \frac{49}{4}$$

The expression becomes:

$$\frac{7}{2} + \frac{49}{4} \times \frac{1}{8}$$

Step 3: Perform the multiplication ("of") operation.

$$\frac{49}{4} \times \frac{1}{8} = \frac{49 \times 1}{4 \times 8} = \frac{49}{32}$$

The expression becomes:

$$\frac{7}{2} + \frac{49}{32}$$

Step 4: Perform the addition. Find a common denominator, which is 32.

$$\frac{7}{2} = \frac{7 \times 16}{2 \times 16} = \frac{112}{32}$$ $$\frac{112}{32} + \frac{49}{32} = \frac{112 + 49}{32} = \frac{161}{32}$$

Step 5: Convert the improper fraction to a mixed number.

$$\frac{161}{32} = 5 \frac{1}{32}$$

The solution is $\boxed{5\frac{1}{32}}$.

2) Solve:

$$\frac{0.325 \times 2.76 \times 100}{13 \times 0.25 \times 92}$$

Step 1: Simplify the numerator by multiplying by 100.

$$0.325 \times 2.76 \times 100 = 0.325 \times 276$$

Step 2: Rewrite the expression and simplify by grouping terms.

$$\frac{0.325 \times 276}{13 \times 0.25 \times 92} = \frac{0.325}{0.25} \times \frac{276}{92} \times \frac{1}{13}$$

Step 3: Calculate each fraction.

$$\frac{0.325}{0.25} = \frac{325}{250} = \frac{13 \times 25}{10 \times 25} = \frac{13}{10} = 1.3$$ $$\frac{276}{92} = 3$$

Step 4: Substitute the simplified values back into the expression and multiply.

$$1.3 \times 3 \times \frac{1}{13} = \frac{13}{10} \times 3 \times \frac{1}{13}$$ $$= \frac{3}{10}$$

Step 5: Convert the fraction to a decimal.

$$\frac{3}{10} = 0.3$$

The solution is $\boxed{0.3}$.

3) Given the universal set $\mathcal{U} = \{1, 2, 3, 4, \dots, 20\}$, If Set A = {Multiples of 3} and B = {Factors of 18} Find: