If a single pane of glass obliterates $$ 4 \% $$ of the light passing through it, the percent $$ p $$ of light that passes through $$ n $$ successive panes is given approximately by the function below. $$ p(n)=100(0.96)^{n} $$ (a) What percent of light will pass through 10 panes? (b) What percent of light will pass through 15 panes? (c) Explain the meaning of the base 0.96 in this problem. (a) The percent of light that will pass through 10 panes is approximately $$ \square \% $$. (Round to the nearest whole number as needed.)

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**(a)**  
We use the formula  
$$
p(n)=100(0.96)^n.
$$  
For $$ n=10 $$, we have  
$$
p(10)=100(0.96)^{10}.
$$  
Using a calculator,  
$$
(0.96)^{10} \approx 0.6648,
$$  
so  
$$
p(10)\approx 100 \times 0.6648 \approx 66.48\%.
$$  
Rounded to the nearest whole number, approximately $$ 66\% $$ of the light passes through 10 panes.

**(b)**  
For $$ n=15 $$, we have  
$$
p(15)=100(0.96)^{15}.
$$  
Calculating,  
$$
(0.96)^{15} \approx 0.5421,
$$  
thus  
$$
p(15)\approx 100 \times 0.5421 \approx 54.21\%.
$$  
Rounding to the nearest whole number, approximately $$ 54\% $$ of the light passes through 15 panes.

**(c)**  
The base $$ 0.96 $$ in the function represents the fraction of light that passes through each individual pane. Since a single pane of glass obliterates $$ 4\% $$ of the light, it transmits $$ 96\% $$ of the light. Hence, with each pane, the light is multiplied by $$ 0.96 $$, and after $$ n $$ panes, the transmitted light is given by $$ 100(0.96)^n $$ percent.
**(a)** Approximately $$ 66\% $$ of the light passes through 10 panes. **(b)** Approximately $$ 54\% $$ of the light passes through 15 panes. **(c)** The base $$ 0.96 $$ represents the $$ 96\% $$ of light that passes through each pane.

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