If we allow $$ 2 \frac{7}{8} $$ in. for the thickness of a course of brick, including mortar joints, how many courses of brick will there be in a wall $$ 51 \frac{3}{4} $$ in. high?

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To determine how many courses of brick are needed for a wall that's $$ 51 \frac{3}{4} $$ inches high, given each course (including mortar) is $$ 2 \frac{7}{8} $$ inches thick, follow these steps:

1. **Convert Mixed Numbers to Improper Fractions:**
   - Thickness per course: $$ 2 \frac{7}{8} $$ inches = $$ \frac{23}{8} $$ inches
   - Total wall height: $$ 51 \frac{3}{4} $$ inches = $$ \frac{207}{4} $$ inches

2. **Calculate the Number of Courses:**
   $$
   \text{Number of courses} = \frac{\text{Total height}}{\text{Thickness per course}} = \frac{\frac{207}{4}}{\frac{23}{8}} = \frac{207}{4} \times \frac{8}{23} = \frac{207 \times 8}{4 \times 23}
   $$

3. **Simplify the Calculation:**
   $$
   \frac{207 \times 8}{4 \times 23} = \frac{1656}{92} = \frac{207}{23} \times \frac{8}{4} = 9 \times 2 = 18
   $$

**Answer:**  
18 courses of brick are needed for the wall.
18 courses of brick are needed for the wall.

If we allow in. for the thickness of a course of brick, including mortar joints, how many courses of brick will there be in a wall in. high?

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