\begin{tabular}{l} How does $$ f(t)=1.15^{t} $$ change over the interval from $$ t=4 $$ to $$ t=5 $$ ? \ $$ f(t) $$ decreases by a factor of 1.15 \ $$ f(t) $$ decreases by 1.15 \ $$ f(t) $$ increases by $$ 115 \% $$ \ $$ f(t) $$ increases by a factor of 1.15 \ \hline\end{tabular}

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To determine how the function $$ f(t) = 1.15^{t} $$ changes from $$ t = 4 $$ to $$ t = 5 $$, let's analyze the function's behavior.

1. **Calculate $$ f(4) $$ and $$ f(5) $$:**
   $$
   f(4) = 1.15^{4} \approx 1.7490
   $$
   $$
   f(5) = 1.15^{5} \approx 2.0114
   $$

2. **Determine the Change:**
   $$
   \text{Change Factor} = \frac{f(5)}{f(4)} = \frac{1.15^{5}}{1.15^{4}} = 1.15
   $$
   
   This means that $$ f(t) $$ increases by a factor of **1.15** when $$ t $$ increases from 4 to 5.

3. **Interpretation:**
   - **Increasing by a factor of 1.15** implies that the function's value at $$ t = 5 $$ is **115%** of its value at $$ t = 4 $$, which is a 15% increase.
   - The other options suggesting a decrease are incorrect since the base of the exponential function (1.15) is greater than 1, indicating growth, not decay.

**Correct Statement:**
- $$ f(t) $$ increases by a **factor of 1.15**.

**Answer:**  
$$\boxed{\,f(t)\ \text{increases by a factor of 1.15}\,}$$
$$ f(t) $$ increases by a factor of 1.15.

\begin{tabular}{l} How does \( f(t)=1.15^{t} \) change over the interval from \( t=4 \) to \( t=5 \) ? \\ \( f(t) \) decreases by a factor of 1.15 \\ \( f(t) \) decreases by 1.15 \\ \( f(t) \) increases by \( 115 \% \) \\ \( f(t) \) increases by a factor of 1.15 \\ \hline\end{tabular}

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