Question

Four different towns in the same state grew in size due to a revitalization effort. The population of town A grew from 22,000 to 24,500 in 4 years. The population of town B grew from 31,200 to 34,770 in 6 years. The population of town C grew from 18,000 to 22,000 in 5 years. The population of town D grew from 32,000 to 34,100 in 3 years. Which town had the greatest average change in population over the corresponding time period?(1 point)

Answer 1

Answer:

Town C.

Step-by-step explanation:

Average change of population with time for town A = $\frac{24,500 - 22,000}{4} = \frac{2,500}{4} = 625$

Average change of population with time for town B = $\frac{34,770 - 31,200}{6} = \frac{3,570}{6} = 595$

Average change of population with time for town C = $\frac{22,000 - 18,000}{5} = \frac{4,000}{5} = 800$

Average change of population with time for town D = $\frac{34,100 - 32,000}{3} = \frac{2,100}{3} = 700$

Town C had the greatest average change in population over time (800/yr).