The number of bacteria in a culture is given by the function where is measured in hours. (a) What is the relative rate of growth of this bacterium population? Your answer is (b) What is the initial population of the culture (at \( \mathrm{t}=0 \) )? Your answer is 900 (c) How many bacteria will the culture contain at time \( \mathrm{t}=5 \) ? Your answer is 3,141

It appears you're working with a bacterial growth model, and you've provided answers for parts (b) and (c). To assist you with part (a), let's deduce the relative rate of growth based on the information provided. **Given:** - **Initial population (P₀) at \( t = 0 \):** 900 bacteria - **Population at \( t = 5 \) hours (P(5)):** 3,141 bacteria Assuming the bacterial growth follows an **exponential growth model**, the population can be described by the equation: \[ P(t) = P_0 \times e^{kt} \] Where: - \( P(t) \) = population at time \( t \) - \( P_0 \) = initial population - \( k \) = relative rate of growth - \( e \) = base of the natural logarithm **Step 1: Plug in the known values to solve for \( k \).** \[ 3,141 = 900 \times e^{5k} \] Divide both sides by 900: \[ \frac{3,141}{900} = e^{5k} \] \[ 3.49 \approx e^{5k} \] **Step 2: Take the natural logarithm of both sides to solve for \( k \).** \[ \ln(3.49) = 5k \] \[ 1.25276 \approx 5k \] \[ k \approx \frac{1.25276}{5} \] \[ k \approx 0.2505 \, \text{per hour} \] **Step 3: Express \( k \) as a percentage.** \[ k \approx 0.2505 \times 100 = 25.05\% \, \text{per hour} \] **Answer to Part (a):** The **relative rate of growth** of the bacterium population is approximately **0.2505 per hour**, or **25.05% per hour**. --- **Summary of All Parts:** - **(a)** Relative rate of growth: **0.2505 per hour** (25.05% per hour) - **(b)** Initial population at \( t = 0 \): **900 bacteria** - **(c)** Population at \( t = 5 \) hours: **3,141 bacteria** If you have the specific growth function or additional details, feel free to share them for a more tailored explanation!