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3 years ago

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The amount of carbon-14 present in animal bones t years after the animal's death is given by P(t)= -0.00012097t. How old is a

n ivory tusk that has lost 34% of its carbon-14?

1 answer:

Tamiku [17]3 years ago

5 0

I believe the equation you gave is wrong because the standard form of equation for C-14 decay is in the form of:

A = Ao e^-kt

So I think the right form of equation is (correct me if I’m wrong):

P(t) = Po e^(-0.00012097t)

Where,

Po = initial value of C-14 at t = 0

t = time elapsed

Since it is given that:

P = (1 – 0.34) Po

P = 0.66 Po

Therefore, t is:

0.66 Po = Po e^(-0.00012097 t)

0.66 = e^(-0.00012097 t)

taking ln of both side:

ln 0. 66 = -0.00012097 t

t = - ln 0. 66 / 0.00012097

t = 3,434.86 years

<span>Therefore the ivory tusk is about 3,435 years old.</span>

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Answer:

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c) 30.85% of the scores are expected to be less than 450.

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Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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In this problem, we have that:

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This is 1 subtracted by the pvalue of Z when X = 600. So

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$Z = \frac{600 - 500}{100}$

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$Z = 1$ has a pvalue of 0.8413.

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2.28% of the scores are expected to be greater than 700.

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Pvalue of Z when X = 450. So

$Z = \frac{X - \mu}{\sigma}$

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30.85% of the scores are expected to be less than 450.

d. Between 450 and 600

pvalue of Z when X = 600 subtracted by the pvalue of Z when X = 450. So

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$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{600 - 500}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413.

X = 450

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{450 - 500}{100}$

$Z = -0.5$

$Z = -0.5$ has a pvalue of 0.3085.

0.8413 - 0.3085 = 0.5328

53.28% of the scores are expected to be between 450 and 600.

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