Answer:
[tex]\textsf{D)} \quad y=\dfrac{60}{x}[/tex]
Step-by-step explanation:
The provided graph is a curve that passes through points (1.5, 40), (2, 30), (2.5, 24), (3, 20) and (4, 15).
The equation y = -10x + 55 does not accurately model the data because it is a linear equation, whereas the graph exhibits non-linear behavior.
The equation y = (x - 4)² + 15 does not accurately model the data because it is a quadratic equation with a minimum point at (4, 15). Although the graphed function intersects point (4, 15), it extends beyond it, indicating that (4, 15) is not the absolute minimum.
The equation y = 120(1/2)ˣ does not accurately model the data because when x = 4, y = 7.5, implying it should pass through the point (4, 7.5), whereas the graphed function passes through point (4, 15).
Therefore, the function that best models the data is:
[tex]\Large\boxed{\boxed{y=\dfrac{60}{x}}}[/tex]
To check this, substitute each x-coordinate into the function and solve for y:
[tex]x=1.5\implies y=\dfrac{60}{1.5}=40\\\\\\x=2\implies y=\dfrac{60}{2}=30\\\\\\x=2.5\implies y=\dfrac{60}{2.5}=24\\\\\\x=3\implies y=\dfrac{60}{3}=20\\\\\\x=4\implies y=\dfrac{60}{4}=15[/tex]
