Ask Question
7 January, 17:42

A city us growing at the rate of 0.9% annually. if there were 3619000 residents in yhe city in 1994, find how many to the nearest ten thousand are living in that city in 2000. use y=3619000 (2.7) ^0.009t

+4
Answers (1)
  1. 7 January, 21:24
    0
    For this case we have the following equation:

    y = 3619000 (2.7) ^ 0.009t

    We must evaluate the equation for the year 2000.

    Therefore, we must replace the following value of t:

    t = 2000 - 1994

    t = 6

    Substituting we have:

    y = 3619000 (2.7) ^ (0.009 * 6)

    y = 3818407.078

    Round to the nearest ten thousand:

    y = 3820000

    Answer:

    3820000 residents are living in that city in 2000
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A city us growing at the rate of 0.9% annually. if there were 3619000 residents in yhe city in 1994, find how many to the nearest ten ...” 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