Help with part b please # Exponential Growth Model of S. pneumoniae BacteriaThis problem involves modeling the growth of a strain of *S. pneumoniae* bacteria. The growth is described by the exponential function:\[ A(t) = A_0 e^{kt} \]- \( A_0 \) is the initial number of bacteria.- \( t \) is the elapsed time in minutes.- \( k \) is the growth constant.## Given Data- Initial bacteria count (\( A_0 \)): 1200- Doubling time: 22 minutes### (a) Finding the Growth Constant \( k \)To find \( k \), we use the fact that the population doubles in 22 minutes.Equation: \( A(22) = 2A_0 \)Substituting in \( A(t) = A_0 e^{kt} \):\[ 2 \times 1200 = 1200 \times e^{22k} \]This simplifies to:\[ 2 = e^{22k} \]Solving for \( k \) gives:\[ \ln(2) = 22k \implies k = \frac{\ln(2)}{22} \]Thus, \( k \) is calculated and rounded to four decimal places.### (b) Deriving the Function for \( A(t) \)Substitute \( k \) back into the original equation \( A(t) = 1200 e^{kt} \) to find the function for the number of bacteria at any time \( t \).### (c) Time Until Population Reaches 8500To find the time it takes for the bacteria count to reach 8500, set up the equation:\[ 8500 = 1200 e^{kt} \]Solve for \( t \):\[ \frac{8500}{1200} = e^{kt} \]\[ \ln\left(\frac{8500}{1200}\right) = kt \]\[ t = \frac{\ln\left(\frac{8500}{1200}\right)}{k} \]Calculate the time \( t \) and round to the nearest minute.### Diagram DetailsThere are handwritten calculations showing the setup for solving parts (a) through (c). The calculations follow the logical sequence of solving exponential growth problems, ensuring understanding of each step.This educational explanation aims to help

We have to provide answer according the question.We have given pneumoniae modeled by the equation At=A0ekt where, A0 is the intial number of bacteria and t is the elapsed time measured in minute. We have given A0=1200 then equation become At=1200ekt and doubling time of this bacterium is 22 minute so, at 22 minute number of bacteria will be 2400 This implies that 2400=1200e22k Take log on both sides ln2400=ln1200+lne22kln2400-ln1200=lne22kln24001200=22kln2=22k0.69314=22kk=0.0315 So, growth constant k=0.0315 (b) After t minutes number of bacteria given by the function At=1200e0.0315t (c) We have to find t when the number of bacteria are 8500 As we know after t minutes number of bacteria given by At=1200e0.0315t So, At=8500 Therefore, 8500=1200e0.0315t85001200=e0.0315t8512=e0.0315t Take log on both sides ln8512=lne0.0315t1.9577=0.0315tt=1.95770.0315t=62.15≈62 minute