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KStrontium-90 decays according to the exponential function y = yowhere t is time in years. Match the given question with the correct procedure to answer the question.If the initial amount of Strontium-90 is 300 g, how much will remain after 18 yr?Choose the correct answer below.1OA. Solve Yo-Yo-0.02411-0.0241(30)-0.02411OB. Evaluate y=300 eO.C. Solve 0.75%-YoOD. Evaluate y = 300 -0.0241(18)-0.02411

KStrontium-90 decays according to the exponential function y = yowhere t is time in-example-1

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Given:


y=y_0e^(-0.0241t)

Where t is the time in years.

Initial amount = 300 g

Time, t = 18 years

Let's find the amount that will remain after 18 years.

Given that the function is an exponential decay function, we have:

Initial amount = y0

Final amount = y

time = t

To find the amount that will remain after 18 years, plug in 300 for y0 and 18 for t.

We have:


y=300e^(-0.0241(18))

Therefore, to find the amount remaining after 18 years, we are to evaluate the function below;


\begin{gathered} y=300e^(-0.0241(18)) \\ \\ y=194.4\text{ } \end{gathered}

ANSWER: D


y=300e^(-0.0241(18))

User Alexey Blinov
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