Question

A species of extremely rare deep water fish has an extremely low reproduction rate . If there are 821 of this type of fish and their growth rate is 2% each month , how many will there be in 1/2 year ? In 10 years ? In 100 years ?

Answer 1

This can be model using the exponential growth formula:

`\begin{gathered} y=a\cdot b^x \\ _{\text{ }}where\colon \\ b=1+r \end{gathered}`

a = Initial amount = 821

r = Growth rate per time period = 2% = 0.02

x = time period

So:

`y=821(1.02)^x`

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For x = 1/2 year = 6 months

`\begin{gathered} y=821(1.02)^6 \\ y\approx925 \end{gathered}`

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For x = 10 years = 120 months

`\begin{gathered} y=821(1.02)^(120) \\ y\approx8838 \end{gathered}`

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For x = 100 years = 1200 months

`\begin{gathered} y=821(1.02)^(1200) \\ y\approx1.7*10^(13) \end{gathered}`