The amount y (in grams) of the radioactive isotope fermium-253 remaining after t hours is y=a(0.5)t/72, where a is the initial amount (in grams). What percent of the fermium-253 decays each hour? Round your answer to the nearest hundredth of a percent.

Per hour decay of the isotope is 0.96%.

Step-by-step explanation:

Amount of radioactive element remaining after t hours is represented by

where a = initial amount

t = duration of decay (in hours)

Amount remaining after 1 hour will be,

y = 0.9904a

So amount of decay in one hour = a - 0.9904a

                                                      = 0.0096a gms

Percentage decay every hour =

                                                  =

                                                  = 0.958 %

                                                  ≈ 0.96 %

Therefore, per hour decay of the radioactive isotope is 0.96%.

step-by-step explanation:

the true ones are a b and d