Question

Radioactive elements such as phosphorus-32 and vanadium-48 decay into other elements over time. The function P(t)=80(0.953)t represents the amount of a sample of phosphorus-32, in grams, after t

days. The function V(t)=70(0.958)t represents the amount of a sample of vanadium-48, in grams, after t
days. How does the rate of decay for phosphorus-32 compare to the decay rate of vanadium-48?(1 point)
1. The sample of phosphorus-32 decays at a faster rate because it decays by 80% per day, and the sample of vanadium-48 decays by 70% per day.
2. The sample of vanadium-48 decays at a faster rate because its decay rate is 95.8% per day, and the decay rate for the sample of phosphorus-32 is 95.3% per day.
3. The sample of phosphorus-32 decays at a faster rate because it decays by 4.7% per day, and the sample of vanadium-48 decays by 4.2% per day.
4. The sample of phosphorus-32 decays at a faster rate because there are 80
grams in the sample, but there are only 70
grams of the vanadium-48 sample.

Answer 1

The correct answer is option 3.

Explanation:

The percentage of the specimen that breaks down within a specific time period provides information about the rate of decay. The amount still present after a day for the Phosphorus-32 sample is 80(0.953)1 = 76.24 grams. The daily decay rate is (80-76.24)/80 or 4.7%.

The amount still present after a day for the Vanadium-48 sample is 70(0.958)1 = 66.95 grams. The daily decay rate is (70-66.95)/70 or 4.2%.

As a result, the Phosphorus-32 sample degrades more quickly than the Vanadium-48 sample, which degrades at a rate of 4.2% each day. Option 3 recognizes this disparity in the right way.