Question

A species of extremely rare, deep water fish are slowly becoming extinct. If there are a total of 821 of this type of fish and there are 15% fewer fish each month, how many will there be in half a year? Exponential growth or decay?: Write a function that represents the situation: - Initial amount: -Growth/Decay rate: -Answer:

A. 249 fish
B. 455 fish
C. 614 fish
D. 700 fish

Answer 1

Using an exponential decay function to model the decline in a rare deep water fish population, the estimated number of fish after six months is approximately 310, which is closest to answer choice A. 249 fish.

Step-by-step explanation:

The question deals with an exponential decay of a rare deep water fish population. To find the number of fish after half a year (six months), we use the function that represents exponential decay:

Initial amount, A(0) = 821 fish

Decay rate per month, r = 15%

The function to model this decay is:

A(t) = A(0) × (1 - r)^t

For half a year (t = 6), the function becomes:

A(6) = 821 × (1 - 0.15)^6

Calculating the value,

A(6) = 821 × (0.85)^6
A(6) = 821 × 0.377149
A(6) ≈ 310 (rounded to the nearest whole number)

Since none of the multiple choice answers exactly matches our calculation, there might be a misunderstanding or a typo in the question. However, if we're to choose the closest value, the answer would be: Answer: A. 249 fish (though the actual calculation gives us approximately 310 fish)