Respuesta :
Solution:
Formula for radioactive Decay is given by
[tex]R_{0}= R(1-\frac{S}{100})^t[/tex]
[tex]R_{0}[/tex]= Initial Population
R = Remaining population after time in hours
Rate of Decay = S % per hour
Initial Population = 72 grams
Final population = 15 grams
Rate of Decay = 35 % per hour
Substituting the values to get value of t in hours
[tex]72=15(1-\frac{35}{100})^t\\\\ 4.8= (0.65)^t\\\\ t= -3.64[/tex]→→1 St expression
But taking positive value of t , that is after 3.64 hours the sample of 72 grams decays to 15 grams at the rate of 35 % per hour.
Now , it is also given that, Once the sample reaches a mass of 15 grams, Billy will continually add more of the compound to keep the sample size at a minimum of 15 grams.
Substituting these in Decay Formula
Final Sample = 15 gm
Starting Sample = 15 +k, where k is amount of sample added each time to keep the final sample to 15 grams.
Time is over 3.64 hours i.e new time = 3.64 + t
Rate will remain same i.e 35 % per hour.
[tex]15=(15+k)(1-\frac{35}{100})^{3.64+t}[/tex]→→→ Final expression (Second) , that is inequalities can be used to determine the possible mass of the radioactive sample over time.
Answer:
The required system of inequalities is, [tex]R\geq 72e^{-0.35t}[/tex] and [tex]R\geq 15[/tex]
Step-by-step explanation:
We are given that,
The actual amount of sample remaining (in grams) = R.
Time (in hours) = t
The formula for the radioactive decay given by, [tex]N=N_{0}e^{-kt}[/tex], where k = decay rate
As there are initially at-least 72 grams of the sample, which is decreasing at the rate of 35% = 0.35.
So, [tex]N_0=72[/tex] and k = 0.35
Thus, we get by substituting the values in the formula above,
[tex]R\geq 72e^{-0.35t}[/tex]
Moreover, the minimum size of the sample is 15 grams. So, we have,
[tex]R\geq 15[/tex]
Hence, the required system of inequalities is,
[tex]R\geq 72e^{-0.35t}[/tex]
[tex]R\geq 15[/tex]
