Question

Iodine-125 has a daily decay rate of about 1%. How many milligrams of a 500 mg sample will remain after 300 days? Round the answer to two decimal places. 0.60 mg 7.81 mg 24.52 mg 31.25 mg

Answer 1

so.. in this case, the starting amount is the 500mg sample... and the rate of decay, negative rate, is 1%, and at the time, the elapsed days is 0, to t = 0, P = 400

$\bf \qquad \textit{Amount for Exponential change}\\\\ A=P(1\pm r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{starting amount}\to &500\\ r=rate\to 1\%\to \frac{1}{100}\to &0.01\\ t=\textit{elapsed period}\to &300\\ \end{cases} \\\\\\ A=500(1-0.01)^{300}$

Answer 2

Answer:

24.52 mg will remain.

Step-by-step explanation:

Given,

The initial  quantity of Iodine-125  = 500 mg,

Also, it has a daily decay rate of about 1%.

Thus, the final quantity of Iodine-125 after x days,

$A=500(1-\frac{1}{100})^x$

$=500(1-0.001)^x$

$=500(0.999)^x$

For x = 300 days,

The remained quantity would be,

$A=500(0.999)^{300}$

$=24.5204470356$

$\approx 24.52\text{ mg}$

Third option is correct.