Question

A homecoming parade will run 3 miles long (1 mile= 5280ft) and 8 feet deep on both sides of the street. How many people could be at the parade? ( rule of thumb is, 6 people can fit in a 25ft^2 area)

Answer 1

Answer:

The number of people that could be at the parade is 30,413 people.

$N\approx30,413\text{ people}$

Explanation:

Given that;

The homecoming parade will run 3 miles long;

$\begin{gathered} l=3\text{ miles}=3\times5280\text{ ft/mile} \\ l=15,840ft \end{gathered}$

And 8 feet deep on both sides of the street;

$b=8ft$

The area of the homecoming parade is;

$\begin{gathered} A=l\times b \\ A=15,840ft\times8ft \\ A=126,720ft^2 \end{gathered}$

The number of people that could be at the parade is;

$N=A\times population\text{ density}$

Population density is given as 6 people per 25ft^2

$Population\text{ density}=\frac{6}{25}\text{people per ft}$

The number of people that could be at the parade will be;

$\begin{gathered} N=A\times population\text{ density} \\ N=126,720ft^2\times\frac{6}{25}\text{ people per ft} \\ N=30,412.8 \\ N\approx30,413\text{ people} \end{gathered}$

Therefore, The number of people that could be at the parade is 30,413 people.

$N\approx30,413\text{ people}$