The population of a colony of bacteria grew after being counted. The population of the colony is modeled by the function p(t)=2500*e^0.01t, where t is the number of minutes since counted.
a. How many bacteria were present when first counted?
b. How many bacteria were present 1 hour after first counted?
c. When will the colony have a population of 1,000,000?
This exact problem will be on a test tomorrow morning, and I can't recall how to do it. All I am looking for are the answers, and work if possible.
a. How many bacteria were present when first counted?
b. How many bacteria were present 1 hour after first counted?
c. When will the colony have a population of 1,000,000?
This exact problem will be on a test tomorrow morning, and I can't recall how to do it. All I am looking for are the answers, and work if possible.
-
a.-
p(0)=2500e^[0.01(0)]
=2500e^0
=2,500
b.-
p(1)=2500e^[0.01(60)]
=2500e^0.6
=4555
c.-
10^6=2500e^(0.01t)
e^(0.01t)=10^6/2500
e^(0.01t)=400
0.01t=ln(400)
t=ln(400)/0.01
t=599.15 min
=9 h 59 min 8.8 sec
p(0)=2500e^[0.01(0)]
=2500e^0
=2,500
b.-
p(1)=2500e^[0.01(60)]
=2500e^0.6
=4555
c.-
10^6=2500e^(0.01t)
e^(0.01t)=10^6/2500
e^(0.01t)=400
0.01t=ln(400)
t=ln(400)/0.01
t=599.15 min
=9 h 59 min 8.8 sec
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a) set t = 0, the time of first counting and solve for p(0)
b) set t = 60, the number of minutes elapsed one hour after counting, and solve for p(60)
c) set p(t) = 1000000 and solve for t which will be the number of minutes after counting that the population reaches one million.
b) set t = 60, the number of minutes elapsed one hour after counting, and solve for p(60)
c) set p(t) = 1000000 and solve for t which will be the number of minutes after counting that the population reaches one million.