Answer:
Option D.
Step-by-step explanation:
It is given that a radioactive substance has an initial radioactivity level of 40 counts per second. After each week, the level of radioactivity is half of the value of the level from the previous week.
Initial value = 40 counts per second.
Decay factor = [tex]\dfrac{1}{2}[/tex]
The general equation of exponential decay function is
[tex]y=ab^x[/tex]
where, a is initial value and b is decay factor.
Substitute a=40 and [tex]b=\frac{1}{2}[/tex] in the above equation.
[tex]y=40\left(\dfrac{1}{2}\right)^x[/tex]
So, the required function for radioactivity levels of the substance per week is
[tex]R(x)=40\left(\dfrac{1}{2}\right)^x[/tex]
At x=0,
[tex]R(0)=40\left(\dfrac{1}{2}\right)^0=40[/tex]
At x=1,
[tex]R(1)=40\left(\dfrac{1}{2}\right)^1=20[/tex]
At x=2,
[tex]R(2)=40\left(\dfrac{1}{2}\right)^2=40\left(\dfrac{1}{4}\right)=10[/tex]
At x=3,
[tex]R(3)=40\left(\dfrac{1}{2}\right)^3=40\left(\dfrac{1}{8}\right)=5[/tex]
It means, the exponential function passing through the points (0,40), (1,20), (2,10) and (3,5).
Therefore, the correct option is D.
