The ratio of strontium-90 remaining, p , as a function of years, t , since the nuclear accident is 
Further explanation:
The exponential decay formula can be expressed as follows,

Here, a represents the initial amount, y is the final amount,r is the rate of decay, and x represents the time.
Given:
The rate of decay is 
Explanation:
Consider the initial amount of the radioactive substance be
.
The final amount of the radioactive substance is 
The ratio of decay can be obtained as follows,

The ratio of strontium-90 remaining,
, as a function of years,
, since the nuclear accident is 
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Answer details:
Grade: High School
Subject: Mathematics
Chapter: Exponential decay
Keywords: twenty percent, contaminants, nuclear accident, Chernobyl, strontium-90, exponentially, 2.5% per year, ratio of strontium, p, function of time, radioactive substance, decays, ten years, each year, rate of decay each year, substance.