Answer:
The answer is ⇒ No, because Deon starts with less bacteria,
but it grows at a faster rat than Cameron's bacteria
Explanation:
* Lets study the graph and the equation
- At t = 0
# Cameron's population = 200
# Deon population = 100
- At t = 5
# From the equation b(5) = 200(1 + 0.08)^5 ≅ 294
# From the graph b(5) ≅ 200
∴ Cameron's population > Deon's population
- The increase of the Cameron's population ≅ 94
(294 - 200 = 94)
- The increase of the Deon's population ≅ 100
(200 - 100 = 100)
∴ The rat of increase of Deon > The rat of increase of Cameron
- At t = 8
# From the equation b(8) = 200(1 + 0.08)^8 ≅ 370
# From the graph b(8) ≅ 300
∴ Cameron's population > Deon's population
- The increase of the Cameron's population ≅ 76
(370 - 294 = 76)
- The increase of the Deon's population ≅ 100
(300 - 200 = 100)
∴ The rat of increase of Deon > The rat of increase of Cameron
- At t = 11
# From the equation b(11) = 200(1 + 0.08)^11 ≅ 466
# From the graph b(11) ≅ 500
∴ Cameron's population < Deon's population
- The increase of the Cameron's population ≅ 96
(466 - 370 = 96)
- The increase of the Deon's population ≅ 200
(500 - 300 = 200)
∴ The rat of increase of Deon > The rat of increase of Cameron
- At t = 14
# From the equation b(14) = 200(1 + 0.08)^14 ≅ 587
# From the graph b(14) ≅ 700
∴ Cameron's population < Deon's population
- The increase of the Cameron's population ≅ 121
(587 - 466 = 121)
- The increase of the Deon's population ≅ 200
(700 - 500 = 200)
∴ The rat of increase of Deon > The rat of increase of Cameron
* From all these calculations the rate of increase of
Cameron's population is less than the rate of increase
of Deon's population
∴ Cameron is not right because Deon starts with less bacteria,
but it grows at a faster rat than Cameron's bacteria