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Example 3
4. CARBON DATING Carbon-14 has a decay constant of of thects based on the amount of information
a. a fossil that has lost
95% of its Carbon- 14
b. an animal skeleton that has
95% of its Carbon- 14 remaining

Example $3$ $\newline$ $4$ . CARBON DATING Carbon $-14$ has a decay constant of of thects based on the amount of information $\newline$ a. a fossil that has lost $95 \%$ of its Carbon- $14$ $\newline$ b. an animal skeleton that has $95 \%$ of its Carbon- $14$ remaining

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Interpret confidence intervals for population means

Full solution

Q. Example

3

\newline

4

. CARBON DATING Carbon

-14

has a decay constant of of thects based on the amount of information

\newline

a. a fossil that has lost

95 \%

of its Carbon-

14

\newline

b. an animal skeleton that has

95 \%

of its Carbon-

14

remaining

Calculate Fossil Age: Calculate the age of a fossil that has lost $95$ % of its Carbon $-14$ . Use the formula for exponential decay: $N(t) = N_0 \cdot e^{-kt}$ , where $N(t)$ is the remaining amount of substance, $N_0$ is the initial amount, $k$ is the decay constant, and $t$ is time. Since $95$ % is lost, $N(t) = 0.05 \cdot N_0$ . Let's rearrange the formula to solve for $t$ : $t = \frac{\ln(N(t)/N_0)}{-k}$ . Substitute $k = 0.000121$ and $N(t)/N_0 = 0.05$ into the formula. $N(t)$ $0$ . Calculate the age of the fossil. $N(t)$ $1$ years.
Calculate Skeleton Age: Calculate the age of an animal skeleton that has $95\%$ of its Carbon- $14$ remaining. $\newline$ Use the same formula: $N(t) = N_0 \cdot e^{-kt}$ . $\newline$ Since $95\%$ is remaining, $N(t) = 0.95 \cdot N_0$ . $\newline$ Rearrange the formula to solve for $t$ : $t = \ln(\frac{N(t)}{N_0}) / (-k)$ . $\newline$ Substitute $k = 0.000121$ and $\frac{N(t)}{N_0} = 0.95$ into the formula. $\newline$ $t = \ln(0.95) / (-0.000121)$ . $\newline$ Calculate the age of the animal skeleton. $\newline$ $14$ $0$ years.

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