There are 189 bacteria in 5 hours
There are 13382588 bacteria in 1 day
There are [tex]10(1.8)^{168}[/tex] bacteria in 1 week
Solution:
Given that,
A type of bacteria has a very high exponential growth rate of 80% every hour
There are 10 bacteria
The increasing function is given as:
[tex]y = a(1+r)^t[/tex]
Where,
y is future value
a is initial value
r is growth rate
t is time period
From given,
a = 10
[tex]r = 80 \5 = \frac{80}{100} = 0.8[/tex]
Determine how many will be in 5 hours
Substitute t = 5
[tex]y = 10(1 + 0.8)^5\\\\y = 10(1.8)^5\\\\y = 10 \times 18.89568\\\\y \approx 188.96[/tex]
y = 189
Thus, there are 189 bacteria in 5 hours
Determine how many will be in 1 day ?
1 day = 24 hours
Substitute t = 24
[tex]y = 10(1 + 0.8)^{24}\\\\y = 10(1.8)^{24}\\\\y = 10 \times 1338258.84\\\\y = 13382588.45\\\\y \approx 13382588[/tex]
Thus, there are 13382588 bacteria in 1 day
Determine how many will be in 1 week
1 week = 168
Substitute t = 168
[tex]y = 10(1 + 0.8)^{168}\\\\y = 10(1.8)^{168}[/tex]
Thus there are [tex]10(1.8)^{168}[/tex] bacteria in 1 week
