a colony of ants doubles every 20 days. if the colony has 350 ants today, how many are/were present in 50 days?
a. in 50 days?
b. 55 days ago?
c. in 30 days?
x days from today, the number of ants n, is
n = 350 * 2^(x/20)
Now just plug in your values for n.
To answer these questions, we need to use the growth formula for exponential growth:
N = N₀ * (2^(t/d))
Where:
N = Final population
N₀ = Initial population
t = Time period
d = Doubling period
Let's calculate the answers to the given questions:
a. In 50 days:
N = 350 * (2^(50/20))
N = 350 * (2^(2.5))
N ≈ 350 * 5.6568
N ≈ 1770.18
Therefore, approximately 1770 ants are/were present in 50 days.
b. 55 days ago:
We can use the time period as negative because it represents "ago" in the past.
N = 350 * (2^(-55/20))
N = 350 * (2^(-2.75))
N ≈ 350 * 0.2679
N ≈ 93.77
Therefore, approximately 94 ants are/were present 55 days ago.
c. In 30 days:
N = 350 * (2^(30/20))
N = 350 * (2^(1.5))
N ≈ 350 * 2.8284
N ≈ 992.94
Therefore, approximately 993 ants are/were present in 30 days.