Question

classwork

1. a species of extremely rare, deep water fish are slowly becoming extinct. if there are a total 821 of this type of fish and there are 15% fewer fish each month, how many will there be in half a year?

initial amount =
growth/decay rate:
percent =\tdecimal =
function that represents this situation:

Explanation:

Step1: Identify initial amount, rate, time

Initial amount \( a = 821 \), decay rate \( r = 15\% = 0.15 \), time \( t = 6 \) months (half a year). Decay formula: \( A(t)=a(1 - r)^t \).

Step2: Substitute values into formula

\( A(6)=821(1 - 0.15)^6 = 821(0.85)^6 \).

Step3: Calculate \( 0.85^6 \)

\( 0.85^6\approx0.3771495156 \).

Step4: Multiply by initial amount

\( A(6)\approx821\times0.3771495156\approx309.64 \). Round to nearest whole number.

Answer:

Initial Amount: \( 821 \)
Growth/Decay Rate: Percent = \( 15\% \) (decay), Decimal = \( 0.15 \)
Function: \( A(t)=821(0.85)^t \)
Number of fish in half a year: \( \approx 310 \)