Several years ago, park rangers discovered that there were no longer any trout in Trout Lake.
So, in 2015 park rangers reintroduced trout to the lake. They released 250 adult trout into its waters. In 2016, there were 265 trout living in Trout Lake.
Assuming the population of trout in Trout Lake has grown exponentially, which expression best models the population growth? Let t represent the number of years since 2015.
265(1.06)*
250(1.06)*
250 + 15t
250(1.15)*
According to the expression you picked, what was the projected trout population in 2020?
Round your answer to the nearest whole number.
BLANK trout
335 was the trout.
Apologies for the mistake. Using the expression 250(1.06)t, the projected trout population in 2020 would be 250(1.06)^5 ≈ 250(1.338225) ≈ 335 trout.
The expression that best models the population growth of trout in Trout Lake is 250(1.06)t.
To find the projected trout population in 2020, we need to substitute t = 5 (2020 - 2015) into the expression:
250(1.06)^5 ≈ 250(1.338225) ≈ 334 trout
Therefore, the projected trout population in 2020 is approximately 334 trout.