Answer:
<h2>A.</h2>
Step-by-step explanation:
The given table is
Days Bees
0 10,000
10 7,500
20 5,600
30 4,200
40 3,200
50 2,400
Where
represents days and
represents bees.
The exponential function that models this problem must be like
, which represenst an exponential decary, because in this case, the number of bees decays.
We nned to use one points, to find the rate of decay. We know that
, because it starts with 10,000 bees.
Let's use the points (10, 7500)
![y=a(1-r)^{x}\\7500=10000(1-r)^{10}](https://tex.z-dn.net/?f=y%3Da%281-r%29%5E%7Bx%7D%5C%5C7500%3D10000%281-r%29%5E%7B10%7D)
Solving for
, we have
![\frac{7500}{10000}=(1-r)^{10} \\(1-r)^{10} =0.75](https://tex.z-dn.net/?f=%5Cfrac%7B7500%7D%7B10000%7D%3D%281-r%29%5E%7B10%7D%20%5C%5C%281-r%29%5E%7B10%7D%20%3D0.75)
Using logarithms, we have
![ln((1-r)^{10}) =ln(0.75)\\10 \times ln(1-r)=ln(0.75)\\ln(1-r)=\frac{ln(0.75)}{10} \approx -0.03\\e^{ln(1-r)}=e^{-0.03}\\1-r =e^{-0.03}\\r=-e^{-0.03}+1 \approx 1.97](https://tex.z-dn.net/?f=ln%28%281-r%29%5E%7B10%7D%29%20%3Dln%280.75%29%5C%5C10%20%5Ctimes%20ln%281-r%29%3Dln%280.75%29%5C%5Cln%281-r%29%3D%5Cfrac%7Bln%280.75%29%7D%7B10%7D%20%5Capprox%20-0.03%5C%5Ce%5E%7Bln%281-r%29%7D%3De%5E%7B-0.03%7D%5C%5C1-r%20%3De%5E%7B-0.03%7D%5C%5Cr%3D-e%5E%7B-0.03%7D%2B1%20%5Capprox%201.97)
Replacing all values in the model, we have
![y=10000(1-1.97)^{x}\\y=10000(0.97)^{x}](https://tex.z-dn.net/?f=y%3D10000%281-1.97%29%5E%7Bx%7D%5C%5Cy%3D10000%280.97%29%5E%7Bx%7D)
Therefore, the right answer is the first choice, that's the best approximation to this situation.