By far the largest and most widespread use of americium-241 is as a component in household and industrial smoke detectors, where a small amount is used in an ionization chamber inside the detector. the half life of americium-241 is 432.2 years. if a smoke detector is buried in a land fill, how many years are required for the original amount of americium-241 to decay to less than 1% of the original amount? round your numerical answer to the nearest tenth of a year. 1,435.7 years 4,322 years 2,871.5 years 43,220 years

2871.5 years

The number of half-lives can be found by solving ...

... (1/2)^x = (1/100)

Taking logs, you have ...

... x·(-log(2)) = -log(100)

... x = log(100)/log(2) ≈ 6.64386

Then the number of years for the desired decay is ...

... (432.2)×6.64386 ≈ 2871.5 . . . . years

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%