Question

Use the data in the table to find each of the following.A. city with greater population density San Diego or Los AngelesB. population of Sacramento, given that the population density is 4,792.1 persons/MIC. area of San Francisco, given that the population density is 17,855 persons/mi

Answer 1

The population density of a place is defined as follows:

`(Population)/(Area)\text{ = Density}`

A. Let us calculate the density of San Diego . The population is 1356000 and area 372.4. So the density is

`(1356000)/(327.4)\text{ = }4141.72persons/mi^2`

In the case of Los Angeles we get

`(3884000)/(503)=7721.67persons/mi^2`

So, it is clear that Los Angeles has a greater density than San Diego.

B. We have the following equality for Sacramento. Let x be the population of Sacramento. Then

`(x)/(100.1)=4792.1`

So we need to solve this equation for x. To do so, simply multiply by 100.1 on both sides, so we get

`x\text{ = 4792.1}\cdot100.1\text{ = 479689.21}`

So population of sacramento is about 479690 people.

C. In here, we are missing the area. So let x be the area of San Francisco. We will have the following equation

`17855\text{ = }(837442)/(x)`

So, we multiply both sides by x. We get

`17855x\text{ = 837442}`

Finally, we divide both sides by 17855. So we get

`x\text{ = }(837442)/(17855)\text{ = }46.90mi^2`

So the area of San francisco is about 46.90 squared miles.