Question

the bones of a newly discovered dinosaur weigh 170 pounds and measure 9 feet, with a 6 - inch claw on one foot. the age of the dinosaur was estimated using a radioactive substance dating of rocks surrounding the bones. complete parts a and b.
a. the radioactive substance decays exponentially with a half - life of approximately 1.34 billion years. use the fact that after 1.34 billion years a given amount of the radioactive substance will have decayed to half the original amount to show that the decay model for the radioactive substance is given by (a = a_0e^{-0.5177t}), where (t) is in billions of years.
to show that the decay model for the radioactive substance is (a = a_0e^{-0.5177t}), substitute the values of (a) and (t) in the exponential decay model, (a = a_0e^{kt}).
substitute (\boxed{t = 1.34}) and (a=\frac{a_0}{2}) (or other relevant substitution steps as per the problem context).

Explanation:

Step1: Understand the Half - Life Concept

In radioactive decay, after one half - life, the amount of the radioactive substance is half of the original amount. The half - life of the substance is 1.34 billion years. So when \(t = 1.34\) (since \(t\) is in billions of years), the amount \(A\) should be \(\frac{A_0}{2}\) (because half of the original substance \(A_0\) remains after one half - life).
The exponential decay model is given as \(A=A_0e^{- 0.5177t}\). We need to substitute \(t = 1.34\) into this model and check if \(A=\frac{A_0}{2}\).

Step2: Substitute \(t = 1.34\) into the Model

Substitute \(t = 1.34\) into the formula \(A = A_0e^{-0.5177t}\). So we have \(A=A_0e^{-0.5177\times1.34}\).
Calculate the exponent: \(- 0.5177\times1.34=-0.5177\times(1 + 0.34)=-0.5177-0.5177\times0.34=-0.5177 - 0.176018=-0.693718\).
Now, \(e^{-0.693718}\approx\frac{1}{2}\) (since \(e^{\ln(1/2)}=\frac{1}{2}\) and \(\ln(1/2)\approx - 0.6931\), which is close to \(-0.693718\)). So \(A = A_0\times\frac{1}{2}=\frac{A_0}{2}\), which shows that the decay model \(A = A_0e^{-0.5177t}\) is correct for the radioactive substance with a half - life of 1.34 billion years. And for the substitution in the box (where we need to find what \(A\) is when \(t = 1.34\) in the context of half - life), since after one half - life (\(t = 1.34\)) the amount is half of the original, the value of \(A\) when \(t = 1.34\) is \(\frac{A_0}{2}\).

Answer:

\(\frac{A_0}{2}\)