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A population of bacteria grows so that its rate of change (in cells per hour) at time…
MATHEMATICS · Single correct · CBSE · Mathematics · Integrals · Integrals
Problem
A population of bacteria grows so that its rate of change (in cells per hour) at time
t
t
t
hours is given by
d
P
d
t
=
40
e
−
0.2
t
\dfrac{dP}{dt}=40e^{-0.2t}
d
t
d
P
=
40
e
−
0.2
t
. If the population is
100
100
100
cells at
t
=
0
t=0
t
=
0
, what is
P
(
5
)
P(5)
P
(
5
)
? (a)
100
+
200
e
100+\dfrac{200}{e}
100
+
e
200
(b)
300
−
200
e
300-\dfrac{200}{e}
300
−
e
200
(c)
100
+
40
e
100+\dfrac{40}{e}
100
+
e
40
(d)
300
+
200
e
300+\dfrac{200}{e}
300
+
e
200
Related theory
Applications
Applications of Integration
Fundamental theorem of calculus
Definite integrals
Integrals
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