P(t) = 100(0.5)^t/5730 :: What percent of carbon is remaining for t=0? Explain the answer in context of Dr. Libby's method in your conclusion.

In 1950, a team of chemists led by Dr. W.F. Libby developed a method for determining the age of any natural specimen, up to approximately 60,000 years of age. Dr. Libby’s method is based on the fact that all living materials contain traced of carbon-14. His method involves measuring the percent of carbon-14 that remains when a specimen is found. The percent of carbon-14 that remains in a specimen after various numbers of years is shown in the table below.

Years Carbon-14 Remaining (%)

5 730 50.0

11 460 25.0

17 190 12.5

22 920 6.25

28 650 3.125

34 380 1.5625

a) what percent of carbon is remaining for *t* = 0? what does this mean in the context of Dr.Libby's method?

b) Draw a graph of the function *P(t) = *100(0.5)^* t/5730*, using the given table of values.

c) What is the reasonable domain for *P(t)?* What is the reasonable range?

d) Determine thge approximate age of spicemen, given that *P(t) =* 70

e) Draw the graph of inverse function.

f) What information does the inverse function provide?

g) What are the domain and the range of the inverse function?

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What percent of carbon is remaining for t=0?

From the reading t is the number of years that have passed since the subject died. So if t=0, this is the year that the subject died and all of the carbon is still present, so P(0)=100%.

Alternatively, you could substitute t=0 into the formula:

`P(t)=100(0.5)^(t/5730)`

so `P(0)=100(.5)^(0/5730)=100(1)=100`

The graph:

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