Question

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Answer 1

Given function:

`\large \text{$ n(t)=73e^(0.02t) $}`

where:

• t is the number of years since 1993
• n(t) is the rat population measured in millions

To calculate the rat population in 1993, substitute
`t = 0` into the function:

`\large \begin{aligned}\implies n(0) & =73e^(0.02(0))\\& = 73 \cdot 1\\& = 73 \sf \: million\end{aligned}`

To calculate a prediction of the rat population in 2009, first determine the value of t by subtracting the initial year of 1993 from the given year of 2009:

`\large \text{$ \implies t=2009-1993=16 $}`

Substituting the found value of t into the function to find the predicted number of rats in 2009:

`\large \begin{aligned}\implies n(16) & =73e^(0.02(16))\\& = 73e^(0.32)\\& = 100.5303268...\\ & = 100.5 \sf \: million\:(1\:dp)\end{aligned}`

Answer 2

See below ~

Explanation:

What is the rat population in 1993?

⇒ Number of years since = 0 ⇒ t = 0

⇒ Apply in the formula

⇒ n(0) = 73e^(0.02 × 0)

⇒ n(0) = 73e⁰

⇒ n(0) = 73,000,000 rats

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What does the model predict the rat population was in the year 2009?

⇒ Number of years since 1993 : 2009 - 1993 = 16 ⇒ t = 16

⇒ Applying in the formula

⇒ n(16) = 73e^(0.02 × 16)

⇒ n(16) = 73e^(0.32)

⇒ n(16) = 73 × 1.37712776 × 10⁶

⇒ n(16) = 100.530326 x 10⁶

⇒ n(16) = 100,530,326 rats