Math  /  Algebra

QuestionThe bones of a newly discovered dinosaur weigh 170 pounds and measure 9 feet, with a 6 -inch claw on one toe of each hind foot. The age of the dinosaur was estimated using a radioactive substance dating of rocks surrounding the bones. A=A0ektA02=A0e1.25k12=e1.25kln(12)=1.25k0.55452=k\begin{aligned} A & =A_{0} e^{k t} \\ \frac{A_{0}}{2} & =A_{0} e^{1.25 k} \\ \frac{1}{2} & =e^{1.25 k} \\ \ln \left(\frac{1}{2}\right) & =1.25 k \\ -0.55452 & =k \end{aligned}
Substitute.
Divide both sides of the equation by A0A_{0}.
Take the natural logarithm on both sides and simplify. Solve for kk. (Round to five decimal places as needed.) b. Analysis of the rocks surrounding the dinosaur bones indicated that 94.4%94.4 \% of the original amount of radioactive substance was still present. Let A=0.944 A0\mathrm{A}=0.944 \mathrm{~A}_{0} in the model in part ( a ) and estimate the age of the bones of the dinosaur.
The estimated age of these dinosaur bones is about \square billion years. (Round to four decimal places as needed.)

Studdy Solution
Use the exponential decay model to estimate the age of the bones.
Given: A=0.944A0A = 0.944 A_0
Substitute into the decay model: 0.944A0=A0e0.55452t0.944 A_0 = A_0 e^{-0.55452 t}
Divide both sides by A0 A_0 : 0.944=e0.55452t0.944 = e^{-0.55452 t}
Take the natural logarithm on both sides: ln(0.944)=0.55452t\ln(0.944) = -0.55452 t
Solve for t t : t=ln(0.944)0.55452t = \frac{\ln(0.944)}{-0.55452}
Calculate: t=0.057160.554520.1031t = \frac{-0.05716}{-0.55452} \approx 0.1031
The estimated age of these dinosaur bones is about 0.1031 \boxed{0.1031} billion years.

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