Answer:
a) P(t) = P₀ e⁰•⁰⁵⁴ᵗ
b) 2786 fishes
c) The estimated population will be 24000 in the year 2030.
Step-by-step explanation:
The function representing the population of fish in a lake as it grows exponentially
P(t) = P₀eᵏᵗ
If the base year is taken to be 2000
P₀ = 4780 fishes
k = constant
In 2010, P(t=10) = 8200 fishes, we can now solve for the constant
kt = k × 10 = 10k
8200 = 4780 e¹⁰ᵏ
e¹⁰ᵏ = (8200/4780) = 1.7155
In e¹⁰ᵏ = In 1.7155 = 0.5397
10k = 0.5397
k = 0.054 to 3 d.p
P(t) = P₀ e⁰•⁰⁵⁴ᵗ
b) What was the estimated population in 1990?
1990 is 10 years before 2000, hence, t = -10
P(t) = P₀ e⁰•⁰⁵⁴ᵗ
0.054t = 0.054 × -10 = -0.54
P₀ = 4780
P(t) = 4780 e⁻⁰•⁵⁴ = 4780 × 0.5829 = 2,786.4 = 2786 fishes to the nearest thousand
c) When will the population 24,000?
P(t) = 24000
P₀ = 4780
t = ?
24000 = 4780 e⁰•⁰⁵⁴ᵗ
e⁰•⁰⁵⁴ᵗ = (24000/4780) = 5.021
In e⁰•⁰⁵⁴ᵗ = In 5.021 = 1.6136
0.054t = 1.6136
t = (1.6136/0.054) = 29.88 years = 30 years to the nearest whole number.
Since the base year is 2000, 30 years after that is 2000 + 30 = 2030.
Hope this Helps!!!
