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Is a(t) = 1200(1 + .03)t a decay function?

Is a(t) = 1200(1 + .03)t a decay function?
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Is a(t) = 1200(1 + .03)t a decay function?

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User Mohsin Mehmood
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7.8k points

1 Answer

5 votes

Answer:

Explanation:

No, it is an exponential growth function because 1.03>1. An exponential decay function would have a rate less than 1.

answered
User Shawntell
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