1. **State the problem:** We have a random variable $X$ uniformly distributed on the interval $(-3,5)$, and another random variable $Y = e^{-X/3}$. We want to find the expected value $E(Y)$. 2. **Recall the formula for expectation:** For a continuous random variable $X$ with probability density function (pdf) $f_X(x)$, the expectation of a function $g(X)$ is given by $$E(g(X)) = \int_{-\infty}^{\infty} g(x) f_X(x) \, dx.$$ Since $X$ is uniform on $(-3,5)$, its pdf is $$f_X(x) = \frac{1}{5 - (-3)} = \frac{1}{8} \quad \text{for } x \in (-3,5).$$ 3. **Set up the integral for $E(Y)$:** Here, $g(x) = e^{-x/3}$, so $$E(Y) = E\left(e^{-X/3}\right) = \int_{-3}^5 e^{-x/3} \cdot \frac{1}{8} \, dx = \frac{1}{8} \int_{-3}^5 e^{-x/3} \, dx.$$ 4. **Evaluate the integral:** Use substitution or integrate directly: $$\int e^{-x/3} dx = -3 e^{-x/3} + C.$$ So, $$\int_{-3}^5 e^{-x/3} dx = \left[-3 e^{-x/3}\right]_{-3}^5 = -3 e^{-5/3} + 3 e^{1} = 3 \left(e^{1} - e^{-5/3}\right).$$ 5. **Calculate $E(Y)$:** $$E(Y) = \frac{1}{8} \times 3 \left(e^{1} - e^{-5/3}\right) = \frac{3}{8} \left(e - e^{-5/3}\right).$$ 6. **Interpretation:** The expected value $E(Y)$ is the average value of $Y$ when $X$ is uniformly distributed on $(-3,5)$. It depends on the exponential function evaluated at the interval endpoints. **Final answer:** $$\boxed{E(Y) = \frac{3}{8} \left(e - e^{-5/3}\right)}.$$