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A quantity with an initial value of 7600 decays exponentially at a rate of 55% every 7 days. What is the value of the quantity after 2 weeks, to the nearest hundredth?

1 Answer

3 votes

Answer:

1539

Explanation:

We solve the above question using the Exponential decay formula

= A(t) = Ao(1 - r) ^t

Ao = Initial Amount invested = 7600

r = Decay rate = 55% = 0.54

t = time in weeks = 2

Hence:

A(t) = 7600(1 - 0.55)²

A(t) = 7600 × (0.45)²

A(t) = 1539

Therefore, the value of the quantity after 2 weeks is 1539

User Jason Baker
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