Question

The amount y ( in grams) of a radioactive isotope phosphorus-32 remaining after t days is y=a(0.5)^t/14, where a is the initial amount (in grams) what percent of the phosphorus-32 decays each day

Answer 1

Answer:

Step-by-step explanation:

The way I figured this out is to just pick some starting values for the grams of this element and plug them into the formula using t = 1 day and seeing how much is left.  I chose 2 different starting amounts and came up with the same percentage each time, so it must be correct!  Here's what I did:

First I chose a starting amount, a, of 10 grams.  Plugging into the formula:

$y=10(.5)^{\frac{1}{14} }$

and got that the amount LEFT was 9.5 grams

Then I chose a starting amount, a, of 20 grams.  Plugging into the formula:

$y=20(.5)^{\frac{1}{14}}$

and got that the amount LEFT was 19 grams.

I then asked the algebraic question,"What percent of 10 is 9.5?" which translates to

x% * 10 = 9.5 and

x = 95%  (that's the amount left as a percentage).

and

x% * 20 = 19 and

x = 95%

Since both of those came out the same, that tells me that after 1 day there is still 95% of the element remaining, so 5% decays each day.