The population in a 55 square mile city, City A, is 44,000. A neighboring city, City B, measures 65 square miles and has 48,750 citizens. Determine which city has a greater population density and by what percentage its population density is greater than the other city's.
I don't get how to do this. Please help!!
44,000 / 55 = 800 people per square mile
48,750 / 65 = 750 per square mile
800 - 750 = 50
50/44,000 = 0.0011 = 11%
My teacher said the answer is 6.7% but I don't know how to get that
divide by density, not number of people
(50/750)100 = 6.67 %
To determine which city has a greater population density and by what percentage, you need to calculate the population density of each city.
Population density is calculated by dividing the population of a city by its area. The formula for population density is:
Population Density = Population / Area
Let's calculate the population density of City A and City B:
For City A:
Population = 44,000
Area = 55 square miles
Population Density of City A = 44,000 / 55
For City B:
Population = 48,750
Area = 65 square miles
Population Density of City B = 48,750 / 65
Now, let's calculate the population density for both cities:
Population Density of City A = 800 people/square mile
Population Density of City B = 750 people/square mile
To determine which city has a greater population density, we can compare the values calculated above. City A has a higher population density than City B.
To calculate the percentage by which the population density of City A is greater than City B, we can use the following formula:
Percentage Difference = ((Density of City A - Density of City B) / Density of City B) * 100
Let's calculate the percentage difference:
Percentage Difference = ((800 - 750) / 750) * 100
Percentage Difference = (50 / 750) * 100
Percentage Difference = 0.0666 * 100
Percentage Difference = 6.66%
Therefore, the population density of City A is 6.66% greater than City B.