Question

Score on last try: 0.5 of 1 pts. See Details for more.> Next questionGet a similar question You can retry this question belowIn April 1986, a flawed reactor design played a part in the Chernobyl nuclear meltdown. Approximately14252 becqurels (Bas), units of radioactivity, were initially released into the environment. Only areaswith less than 800 Bqs are considered safe for human habitation. The function f(x) = 14252(0.5) ádescribes the amount, f(x), in becqurels, of a radioactive element remaining in the area z yearsafter 1986.

Answer 1

The given function is:

`f(x)=14252(0.5)^{(x)/(32)}`

Where f(x) is the amount of a radioactive element remaining (in becqurels) in the area x years after 1986.

We need to find f(120) in order to determine the amount of becqurels in 2106. So:

`\begin{gathered} f(120)=14252(0.5)^{(120)/(32)} \\ f(120)=14252(0.5)^(3.75) \\ f(120)=14252*0.074 \\ f(120)=1059.3 \end{gathered}`

The amount of becqurels in 2106 would be 1059.3

This value is greater than 800, and the problem says only areas with less than 800 Bqs are considered safe for human habitation. Thus, this area is not safe, because, by 2106, the radioactive element remaining in the area is greater than 800 Bqs.