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anonymous
 one year ago
a culture started with 3,000 bacteria after 5 hours, it grew 3,900. bacteria. predict how many bacteria will be present after 9 hours. eound your answer to the nearet whole number.
P=Ae^kt
anonymous
 one year ago
a culture started with 3,000 bacteria after 5 hours, it grew 3,900. bacteria. predict how many bacteria will be present after 9 hours. eound your answer to the nearet whole number. P=Ae^kt

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hm... I actually thought of this as a exponential funtions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its exponential growth

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hurry pleaseee i need help last question!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0OHHHHH i remember this. okay, so, any ideas

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use P(0) find the value of A Then plug P(5) and find the value of K now u have your function P complete Just plug the P(9) and there you go

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what ? srry confused i really need help like this is the last question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't really know, this is just a guess, but 5,400 bacteria, maybe?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[P(x)=Ae^{kt}\] this is the equation that models the population of bacteria We have as initial data that we started with 3000 bacteria, this means \[P(0)=3000\] If we calculate \[P(0)=A{e}^{k(0)}=A=3000\] we are able to find the value of A. Problem also says that after 5 hours, bacteria population is 3900 which is \[P(5)=3900\] Evaluating t=5 on our main equation we would have \[P(5)=3000{e}^{5k}=3900\] Solving this equation for k \[{e}^{5k}=1.3\] \[\ln {e}^{5k}=\ln 1.3\] \[5k=\ln 1.3\] \[k=\frac{ \ln 1.3 }{ 5 }\] We just found the value of k So our main equation should be \[P(t)=3000{e}^{\frac{ \ln1.3 }{ 5 }t}\] So if we want to know what's the population of bacteria after 9 hours, we just evaluate t=9 in the equation \[P(9)=3000{e}^{\frac{ \ln 1.3 }{ 5 } (9)}=4810.8\] which is roughly 4810 bacteria
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