Answer:
![\boxed{\text{30 800 people}}](https://img.qammunity.org/2020/formulas/mathematics/high-school/7e34k0dxzbtfiiblpkyxx2fhl0jpy9sw4w.png)
Explanation:
1. Area of the crowd
The crowd fills a rectangle that is 2 mi long and 7 ft wide.
(a) Convert miles to feet.
![\text{l} = \text{2 mi} * \frac{\text{5280 ft}}{\text{1 mi}} = \text{10 560 ft}](https://img.qammunity.org/2020/formulas/mathematics/high-school/rexh3knbi2l2v6olxc1k012zetd3qvh6y5.png)
(b) Calculate the area
A = lw = 10 560 ft × 7 ft = 73 920 ft²
2. Size of crowd
15 people occupy an area of 6 ft × 6 ft = 36 ft²
![\text{Size of crowd} = \text{73 920 ft}^(2)* \frac{\text{15 people}}{\text{36 ft}^(2)} = \textbf{30 800 people}\\\text{The size of the crowd is $\boxed{\textbf{30 800 people}}$}](https://img.qammunity.org/2020/formulas/mathematics/high-school/hldqx5ta99dgej07p3w4xxzu959gayma8m.png)