Answer:
<em>1.5%</em><em> will remain at 6.00pm.</em>
Step-by-step explanation:
The function for exponential decay is,

where,
y(t) = the future amount,
A = initial amount,
r = rate of growth,
t = time.
A researcher measures 200 counts per minute coming from a radioactive source at noon.
At 3:00pm, i.e after 3 hours,she finds that this has dropped to 25 counts per minute.
Putting the values,






-ve sign means the amount is decreasing.
As we have to find the amount at 6.00pm i.e after 6 hours, so

As number of bacteria can't be in decimal.
So the percentage will be,
