Answer to Question #326076 in Algebra for Baylea

Question

An element with mass 820 grams decays by 26.8% per minute. How much of the element is remaining after 18 minutes, to the nearest 10th of a gram?

Answer 1

Let $m_0 = 820 \space g,$

$\alpha = 26.8\space\% = 0.268,$

$T = 18\space min,$

$m_t$ - mass we have after time of decay $t$

We have:

$m_{t + 1} = m_t(1 - \alpha)$

This is a geometric progression with a common ratio equal to $(1 - \alpha)$

Its general term is represented by the formula $m_t = m_0(1 - \alpha)^t$

$m_T = m_0(1 - \alpha)^T =\\ = 820\cdot(1 - 0.268)^{18}\space g = 3.0\space g$