The number of certain type of bacteria increases according to the model $P ( t ) = 100 e ^ { 0.01896 t }$ where t is time (in hours)
a)Find P(0).
b)Find P(5).
c)Find P(10).
d)Find P(24).
A)a) $P ( 0 ) = 100$ b) $P ( 5 ) \approx 109.94$ c) $P ( 10 ) \approx 120.88$ d) $P ( 24 ) \approx 157.62$
B)a) $P ( 0 ) \approx 100$ b) $P ( 5 ) \& 119.94$ c) $P ( 10 ) \approx 102.88$ d) $\mathrm { P } ( 24 ) \approx 257.62$
C)a) $\mathrm { P } ( 0 ) = 120.56$ b) $P ( 5 ) = 100.94$ c) $P ( 10 ) \approx 620.88$ d) $P ( 24 ) \approx 517.62$
D)a) $\mathrm { P } ( 0 ) = 109.78$ b) $P ( 5 ) \approx 100.94$ c) $P ( 10 ) \approx 150.88$ d) $P ( 24 ) \approx 357.62$
E)a) $P ( 0 ) = 105$ b) $P ( 5 ) \approx 119.94$ c) $P ( 10 ) \approx 120.08$ d) $P ( 24 ) \approx 157.52$

Question

The number of certain type of bacteria increases according to the model $P ( t ) = 100 e ^ { 0.01896 t }$ where t is time (in hours) 
a)Find P(0).
b)Find P(5).
c)Find P(10).
d)Find P(24).
A)a) $P ( 0 ) = 100$ b) $P ( 5 ) \approx 109.94$ c) $P ( 10 ) \approx 120.88$ d) $P ( 24 ) \approx 157.62$
B)a) $P ( 0 ) \approx 100$ b) $P ( 5 ) \& 119.94$ c) $P ( 10 ) \approx 102.88$ d) $\mathrm { P } ( 24 ) \approx 257.62$
C)a) $\mathrm { P } ( 0 ) = 120.56$ b) $P ( 5 ) = 100.94$ c) $P ( 10 ) \approx 620.88$ d) $P ( 24 ) \approx 517.62$
D)a) $\mathrm { P } ( 0 ) = 109.78$ b) $P ( 5 ) \approx 100.94$ c) $P ( 10 ) \approx 150.88$ d) $P ( 24 ) \approx 357.62$
E)a) $P ( 0 ) = 105$ b) $P ( 5 ) \approx 119.94$ c) $P ( 10 ) \approx 120.08$ d) $P ( 24 ) \approx 157.52$

Quizplus · Accepted Answer

a) $P(0) = 100e^{0.01896*0} = 100$. b) $P(5) = 100e^{0.01896*5} \approx 109.94$. c) $P(10) = 100e^{0.01896*10} \approx 120.88$. d) $P(24) = 100e^{0.01896*24} \approx 157.62$.

The Number of Certain Type of Bacteria Increases According to the Model