An old bone contains 80% of its original carbon-14. use the half-life model to find the age of the bone

This is an exponential decay problem. These problems are of the form:

y = c * ekt

y = amount of substance left

c = original amount of substance (Here, set this to 100%, or 1.)

e = exponential constant (~2.718)

k = rate of decay constant (We need to figure this out.)

t = time, in years (When t = 0, y = 1 (all of the carbon-14). When t = 5730, y = 0.5 (one half-life has passed). When t = the answer we're trying to find, y = 0.98 (98% of the carbon-14).)

First, we must find the value of k, the rate of decay constant. We know that after 5730 years (t = 5730), one-half of the carbon-14 will remain (y = 0.5).

y = c * ekt

0.5 = 1 * ek * 5730

0.5 = e5730k

To get rid of the e, take the natural logarithm (ln) of both sides:

ln(0.5) = ln(e5730k)

ln(0.5) = 5730k

ln(0.5)/5730 = k

ln(0.5) is a negative number, so our rate of decay constant will be negative. This is a little "sanity check", because radioactive decay means the amount of substance goes down over time. If you get a positive value of k, then you made a mistake somewhere.

Now that we have the rate of decay constant, we can find the value of t (in years) that will yield 98% of the carbon-14 remaining.

y = c * ekt

0.98 = 1 * e[ln(0.5)/5730]t

0.98 = e[ln(0.5)/5730]t

To get rid of the e, take the natural logarithm (ln) of both sides:

ln(0.98) = ln(e[ln(0.5)/5730]t)

ln(0.98) = [ln(0.5)/5730]t

ln(0.98) = ln(0.5)t / 5730

Solve for t:

5730 * ln(0.98) = ln(0.5)t

5730 * ln(0.98) / ln(0.5) = t

Put that into your calculator, to get t ~ 167 years. Both ln(0.98) and ln(0.5) are negative, so the negatives will cancel out to yield a positive number (another "sanity check").

Step-by-step explanation:

see below

step-by-step explanation:

7.   there are 10 data points in the set.   the middle point is between 5 and 6.   the middle point in the top half is 8   and the middle point in the bottom half is 3. to find the iqr , take point 8 and subtract point 3

mid term   point 8 = 100   point 3 = 88

iqr = 100-88 = 12

final   point 8 = 93   point 3 = 78

iqr = 93 - 78 = 15

the final has the greatest iqr

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5.   the   iqr is a better measure of spread for the movies and basketball games.   since the basketball game has an outlier, we do not want to use standard deviation for measurement.  

3.   the exam median is much higher than the class median.   the median line is farther to the right on the exams

2.   there is no way to tell what the means are from the box and whisker plots.