Question

someone please help!!
The half-life of carbon-14 is 5,730 years. Assuming you start with 100% of carbon-14, what is the expression for the percent, P(t), of carbon-14 that remains in an organism that is t years old and what is the percent of carbon-14 remaining (rounded to the nearest whole percent) in an organism estimated to be 12,000 years old?
idk if the answer is P(t)=100(0.5)^5,730t, 23% remaining or P(t)=100(0.5)^t/5730, 23 % remaining

Answer 1

The exponential equation for half life is P(t)=A0(0.5)^t/H, where p(t) is the percent of carbon-14 remaining, A0 is the initial amount (100%), t is the age of organism in years, and H is half time

Answer 2

there was two other answers p(t)= 5,730(0.5)^100/t, 5,690 remaining
and p(t)=100(0.5)^5,730/t, 77% remaining but i don't think those are the answers

Answer 3

@mathmale

Answer 4

I also got about 23 percent when I plugged in 12000 for t

Answer 5

so did I but i am confused with those two answers

Answer 6

12000/5730=2.0942 half-lives of carbon 14 expired
2^-2.0942 =23.42% of original amount of carbon 14 remaining.

Answer 7

is it p(t)=100(0.5)^t/5,730, 23 % remaining?

Answer 8

@Michele_Laino

Answer 9

I think that the remaining percentage, is:
\[\Large p\left( t \right) = 100{e^{ - t/\tau }}\]
where \(\tau =5730\)

Answer 10

so, we can write:
\[\Large p\left( {12000} \right) = 100 \cdot {e^{ - 12000/5730}}\]

Answer 11

I got this result: \(p=12 \%\) (rounded value)

Answer 12

my answer leads to p(t)=100(0.5)^t/730, 23% remaining