The change in a sample of bacteria, P, is represented by the function P(t)=80(3)t6

Question

, where t is the number of hours.

Which of the following statements best describes the initial number of bacteria and the change in the number of bacteria over time?

(1 point)
Responses

The initial number of bacteria is 240. The number of bacteria triples 6 times per hour.
The initial number of bacteria is 240. The number of bacteria triples 6 times per hour.

The initial number of bacteria is 80. The number of bacteria triples every hour.
The initial number of bacteria is 80. The number of bacteria triples every hour.

The initial number of bacteria is 80. The number of bacteria triples 6 times per hour.
The initial number of bacteria is 80. The number of bacteria triples 6 times per hour.

The initial number of bacteria is 80. The number of bacteria triples every 6 hours.
The initial number of bacteria is 80. The number of bacteria triples every 6 hours.

Answer 1

To analyze the function \( P(t) = 80(3)^{t/6} \), we can break it down to determine the initial number of bacteria and the rate at which the bacteria change over time.

1. Initial number of bacteria: To find the initial number of bacteria, we evaluate \( P(0) \):

   \[ P(0) = 80(3)^{0/6} = 80(3)^0 = 80 \times 1 = 80 \]

   So, the initial number of bacteria is 80.

2. Change over time: The term \( 3^{t/6} \) indicates how the bacteria multiply over time. It can be interpreted that for every 6 hours (since \( t \) is measured in hours), the bacteria triple. To clarify:

   • If \( t = 6 \) hours, \( P(6) = 80(3)^{6/6} = 80(3)^1 = 80 \times 3 = 240 \).
   • If \( t = 12 \) hours, \( P(12) = 80(3)^{12/6} = 80(3)^2 = 80 \times 9 = 720 \).
   • Thus, every time \( t \) increases by 6 hours, the number of bacteria triples.

Based on this analysis, the correct statement is:

The initial number of bacteria is 80. The number of bacteria triples every 6 hours.