Suppose a bacteria culture begins with 45 bacteria and after 6 hours the population has grown to 68 bacteria. a) Write a continuous growth function \( P(t)=P_{0} e^{k \cdot t} \) that models the population over time. To do this you will need to solve for the rate of growth. \( P(t)= \) Use \( t \) as your variable and keep at least 4 significant figures on the growth rate. b) According to your model, what is the population after 9 hours? \( P(9)=\square \) bacteria Round to the nearest whole number. c) How long will it take for the population to reach 170 bacteria? \( t=\square \) hours Give your answer accurate to at least 2 decimal places.

To model the bacteria population growth, we can start with the function \( P(t) = P_0 e^{kt} \). Here, \( P_0 = 45 \) (the initial population) and \( P(6) = 68 \). First, we need to find the growth rate \( k \). From the data, we can substitute into the equation: \[ 68 = 45 e^{6k} \] Now, divide both sides by 45: \[ \frac{68}{45} = e^{6k} \] Taking the natural logarithm of both sides gives: \[ \ln\left(\frac{68}{45}\right) = 6k \] Calculating the left side: \[ \ln\left(\frac{68}{45}\right) \approx 0.4551 \] Now solve for \( k \): \[ k \approx \frac{0.4551}{6} \approx 0.07585 \] Thus, the growth function is: \[ P(t) = 45 e^{0.07585t} \] Now, to find \( P(9) \): \[ P(9) = 45 e^{0.07585 \cdot 9} \] Calculating \( 0.07585 \cdot 9 \approx 0.68265 \): \[ P(9) = 45 e^{0.68265} \approx 45 \cdot 1.9803 \approx 89.1135 \] Rounding to the nearest whole number gives: \[ P(9) \approx 89 \text{ bacteria} \] For part (c), we need to find \( t \) when \( P(t) = 170 \): \[ 170 = 45 e^{0.07585t} \] Divide both sides by 45: \[ \frac{170}{45} = e^{0.07585t} \] Taking the natural logarithm: \[ \ln\left(\frac{170}{45}\right) = 0.07585t \] Calculating this gives: \[ \ln\left(\frac{170}{45}\right) \approx 1.0912 \] Now solve for \( t \): \[ t = \frac{1.0912}{0.07585} \approx 14.39 \] So, it will take about: \[ t \approx 14.39 \text{ hours} \] **Final Answers:** a) \( P(t) = 45 e^{0.07585t} \) b) \( P(9) = 89 \) bacteria c) \( t = 14.39 \) hours