The decay constant for 14C is 0.00012. A 4050-year-old wooden chest is found by archaeologists. What percentage of the original 14C would you expect to find in the wooden chest? (Express your answer as a percentage rounded to one decimal place.)

The radioactive decay constant or  k = ln (.5) / Half-Life

Half-Life =  -.693147 / .00012

Half-Life = -5,776.225  years

We'll call beginning amount as 100%

Ending Amount = Beginning Amount / 2^n (where n = # of half-lives)

n = 4,050 / 5,776.225 = 0.7011499725

Ending Amount = 100% / 2^0.7011499725

Ending Amount = 100% / 1.625800202

Ending Amount = 61.5081729452%

Ending Amount = 61.5 % (rounded)

Step-by-step explanation:

d

step-by-step explanation:

answer:

125 yards

explanation:

75+50=125. that is my answer