Answer:
The volume of the tumor experimented a decrease of 54.34 percent.
Step-by-step explanation:
Let suppose that tumor has an spherical geometry, whose volume ([tex]V[/tex]) is calculated by:
[tex]V = \frac{4\pi}{3}\cdot R^{3}[/tex]
Where [tex]R[/tex] is the radius of the tumor.
The percentage decrease in the volume of the tumor ([tex]\%V[/tex]) is expressed by:
[tex]\%V = \frac{\Delta V}{V_{o}} \times 100\,\%[/tex]
Where:
[tex]\Delta V[/tex] - Absolute decrease in the volume of the tumor.
[tex]V_{o}[/tex] - Initial volume of the tumor.
The absolute decrease in the volume of the tumor is:
[tex]\Delta V = V_{o}-V_{f}[/tex]
[tex]\Delta V = \frac{4\pi}{3}\cdot (R_{f}^{3}-R_{o}^{3})[/tex]
The percentage decrease is finally simplified:
[tex]\%V = \left[1-\left(\frac{R_{f}}{R_{o}}\right)^{3} \right]\times 100\,\%[/tex]
Given that [tex]R_{o} = R[/tex] and [tex]R_{f} = 0.77\cdot R[/tex], the percentage decrease in the volume of tumor is:
[tex]\%V = \left[1-\left(\frac{0.77\cdot R}{R}\right)^{3} \right]\times 100\,\%[/tex]
[tex]\%V = 54.34\,\%[/tex]
The volume of the tumor experimented a decrease of 54.34 percent.
