- Expenential and Logarithmic Functions Writing and evaluating a function modeling continuous exponential growt.. The number of bacteria in a culture decreases according to a continuous exponential decay model. The initial population in a study is 500 bacteria, and there are 290 bacteria left after 9 minutes. (a) Let $t$ be the time (in minutes) since the beginning of the study, and let $y$ be the number of bacteria at time $t$. Write a formula relating $y$ to $t$. Use exact expressions to fill in the missing parts of the formula. Do not use approximations. \[ y=\square e^{\mathbb{Q} t} \] (b) How many bacteria are there 11 minutes after the beginning of the study? Do not round any intermediate computations, and round your answer to the nearest whole number. $\square$ bacteria Explanation Check

Step 1 ：Given the initial population \(P = 500\) bacteria and there are \(290\) bacteria left after \(9\) minutes, we can substitute these values into the formula \(y = P e^{kt}\) to find the decay constant \(k\).

Step 2 ：Substitute \(P = 500\), \(y = 290\), and \(t = 9\) into the formula, we get \(290 = 500 e^{9k}\).

Step 3 ：Solving for \(k\), we get \(e^{9k} = \frac{290}{500} = 0.58\).

Step 4 ：Taking the natural logarithm of both sides, we get \(9k = \ln(0.58)\).

Step 5 ：So, \(k = \frac{\ln(0.58)}{9}\).

Step 6 ：Substituting \(P = 500\) and \(k = \frac{\ln(0.58)}{9}\) back into the formula, we get the formula relating \(y\) to \(t\): \(y = 500 e^{\frac{\ln(0.58)}{9} t}\).

Step 7 ：To find the number of bacteria 11 minutes after the beginning of the study, we substitute \(t = 11\) into the formula: \(y = 500 e^{\frac{\ln(0.58)}{9} \cdot 11}\).

Step 8 ：Evaluating this expression, we get \(y \approx 500 e^{-0.511} \approx 224.5\).

Step 9 ：Rounding to the nearest whole number, we get \(y \approx 225\).

Step 10 ：\(\boxed{225}\) is the approximate number of bacteria 11 minutes after the beginning of the study.