Ask Question
7 April, 13:20

In 2018 the state of Arizona has a population of 7,123,898 this is an increase from 2010 when a population of 6,392,017 if the area of Arizona covers 113,998MI squared then what is the difference in population density between 2010 in 2018

+3
Answers (1)
  1. 7 April, 16:49
    0
    The difference in population density between 2010 and 2018 is of 6.42 hab/MI²

    Step-by-step explanation:

    The population density is the number of habitants divided by the area.

    Arizona:

    Has an area of 113,998MI².

    2010:

    6,392,017 habitants, so the density was 6,392,017/113,998 = 56.07 hab/MI².

    2018:

    7,123,898 habitants, so the density was 7,123,898/113,998 = 62.49 hab/MI²

    Difference:

    62.49 - 56.07 = 6.42 hab/MI²

    The difference in population density between 2010 and 2018 is of 6.42 hab/MI²
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “In 2018 the state of Arizona has a population of 7,123,898 this is an increase from 2010 when a population of 6,392,017 if the area of ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers