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

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

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

Step-by-step explanation:

Given that;

The homecoming parade will run 3 miles long;

`\begin{gathered} l=3\text{ miles}=3*5280\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* b \\ A=15,840ft*8ft \\ A=126,720ft^2 \end{gathered}`

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

`N=A* population\text{ density}`

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

`Population\text{ density}=(6)/(25)\text{people per ft}`

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

`\begin{gathered} N=A* population\text{ density} \\ N=126,720ft^2*(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}`