AD growth medium is inoculated with 1,000 bacteria, which grow at a rate of \( 15 \% \) each day, What is the population of the oulture 6 days after inoculation? \( y=1,000(1,15)^{6} ; 2,313 \) bacteria \( y=1,000(1.15)^{\%} ; 2,600 \) bacteria \( y=1,000(1.5)^{5} ; 7,594 \) bacteria \( y=1,000(1.5)^{6} ; 11,391 \) bacteria

To determine the population of bacteria 6 days after inoculation, we'll use the formula for exponential growth: \[ y = y_0 \times (1 + r)^t \] Where: - \( y_0 \) is the initial population (1,000 bacteria) - \( r \) is the growth rate per period (15% or 0.15) - \( t \) is the number of periods (6 days) Plugging in the values: \[ y = 1,000 \times (1 + 0.15)^6 \] \[ y = 1,000 \times (1.15)^6 \] Calculating \( (1.15)^6 \): \[ \begin{align*} 1.15^1 & = 1.15 \\ 1.15^2 & \approx 1.3225 \\ 1.15^3 & \approx 1.5209 \\ 1.15^4 & \approx 1.7490 \\ 1.15^5 & \approx 2.0114 \\ 1.15^6 & \approx 2.3131 \\ \end{align*} \] So, \[ y \approx 1,000 \times 2.3131 = 2,313 \text{ bacteria} \] **Correct Answer:** \( y = 1,000(1.15)^{6} ; 2,313 \) bacteria