Answer:
d. is the closest one to fit.
Step-by-step explanation:
At x=0 we see that y=1000 from the graph.
We are looking for an initial mass of 1000 from the equations.
[tex]y=A\cdot b^{x}[/tex] tells us at [tex]x=0[/tex], [tex]y=A[/tex].
If you are unsure that plug in 0 for [tex]x[/tex]:
[tex]y=A \cdot b^0[/tex]
[tex]y=A \cdot 1[/tex]
[tex]y=A[/tex]
So we have here for this problem that [tex]A=1000[/tex] since that is what happens at [tex]x=0[/tex].
So far we know the equation is:
[tex]y=1000 \cdot b^{x}[/tex]
To find [tex]b[/tex] we need to look at another point on our graph.
At almost x=4, we see that y is about 500.
Let's plug this into our equation above:
[tex]500=1000 \cdot b^{4}[/tex]
Divide both sides by 1000:
[tex]\frac{1}{2}=b^4[/tex]
Take the fourth root of both sides:
[tex]\sqrt[4]{\frac{1}{2}}=b[/tex]
So our equation is approximately:
[tex]y=1000 \cdot (\sqrt[4]{\frac{1}{2}})^x[/tex]
The closet answer to this is d.
(a) isn't close because that would contain a negative factor which would include some kind of reflection through the x-axis.
(b) isn't close because again it includes a negative factor leading to some kind of reflection through the x-axis.
(c) isn't close because it's initial population is 1/2 and is increasing by factors of 1000.
(d) is the closest because it's initial population is 1000 and is decreasing because of the factors of ~1/2 .
