Solution:
Let H represent the height.
Given that the initial height from which the rubber ball is dropped is 800 centimeters, this implies that
![H_1=800](https://img.qammunity.org/2023/formulas/mathematics/college/1i2s0l73808fb3k808zmni0lj12zouflux.png)
It is noticed that after each bounce, it reached 85% of its previous height. This implies that
![H_2=(85)/(100)* H_1=0.85H_1](https://img.qammunity.org/2023/formulas/mathematics/college/k5sse5voncdln5nqy2852f1odxnczu84z7.png)
Similarly,
![H_3=(85)/(100)* H_2=0.85H_2](https://img.qammunity.org/2023/formulas/mathematics/college/yvvrgnkbu79upiurore7043i0vpks7cd0h.png)
Thus, since each height is attained by a common ratio, using the geometric sequence formula expressed as
![\begin{gathered} T_n=ar^(n-1) \\ where \\ T_n\implies H_n \\ a\implies H_1 \\ r\text{ is the common ratio between consecutive terms eevaluated as} \\ (H_2)/(H_1)=(0.85H_1)/(H_1)=0.85 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/biz6k4zkjyosjbzluuo87dyi8lz5xrgwbd.png)
Thus, substituting these parameters into the geometric sequence formula, we have
![\begin{gathered} H_n=H_1(0.85)^(n-1) \\ where \\ H_1=800 \\ \Rightarrow H_n=800(0.85)^(n-1) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qktnwm2znr5xhfmvyakb8chozr58zxgrbm.png)
Hence, the equation model for the height H, for n bounces is
![\begin{gathered} \begin{equation*} H_n=800(0.85)^(n-1) \end{equation*} \\ where \\ n=1,2,3,... \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/86u9oggctg3ny93yndfmujqoe6r13kib0y.png)
The third option is the correct answer.