If 14g of a radioactive substance are present initially and 7 yr later only 7 g remain how much of the substance will be present after 9 yr?

Question

Round final answer to three decimal places as needed

Answer 1

7 = 14e^7r

1 = 2e^7r

1/2 = e^7r

ln.5 = 7r

7r = ln.5

r = ln.5/7 = about -0.1

A = 14e^-.1t

A = 14e^-.1(9) = 14e^-.9

= 14(1/2.46) = 14(.4065) = about 5.7

A = 5.7 g left after 9 years