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Exercise
May 01'23
Answer
Solution: D
Note that
[[math]]
\operatorname{E}[X] = \int_{-2}^0 \frac{-x^2}{10} dx + \int_0^4 \frac{x^2}{10} dx = \frac{-x^3}{30} \Big |_{-2}^0 + \frac{x^3}{30} \Big |_0^4 = \frac{28}{15}.
[[/math]]
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