answer:The expectation of a function f(x) when x is from a Laplace distribution with mean mu and scale parameter b is given by the integral: E[f(x)] = int_{-infty}^{infty} f(x) cdot frac{1}{2b} e^{-frac{|x-mu|}{b}} dx For x sim text{LaplaceDistribution}[0, 0.9], where mu = 0 and b = 0.9, we calculate the expectation of the given function: E[-3.9x^2 + 3.8x - 1.3] = int_{-infty}^{infty} (-3.9x^2 + 3.8x - 1.3) cdot frac{1}{1.8} e^{-frac{|x|}{0.9}} dx After evaluating this integral, we find the expectation to be -7.62.

answer:Let the length of the rope be L meters. Using the Pythagorean theorem, we have: ``` L^2 = (2 + x)^2 + h^2 ``` where x is the horizontal distance from the pole to the point where the rope touches the building, and h is the height of the building above the top of the pole. Using similar triangles, we have: ``` h / x = 6 / 2 ``` Solving for h, we get: ``` h = 3x ``` Substituting this into the Pythagorean theorem equation, we get: ``` L^2 = (2 + x)^2 + (3x)^2 ``` Simplifying, we get: ``` L^2 = 10x^2 + 4x + 4 ``` To find the minimum value of L, we differentiate the equation and set it equal to zero: ``` 20x + 4 = 0 ``` Solving for x, we get: ``` x = -0.2 ``` Since x cannot be negative, we reject this solution. Therefore, there is no minimum value for L. This means that the shortest rope that can be stretched from the top of the pole to the roof of the building is infinitely long.

answer:The angular velocity of the arm can be calculated as: ``` ω = (1.5 * 2π rad) / 0.44 s = 21.4 rad/s ``` The tangential velocity of the hand is then: ``` v = ω * L = 21.4 rad/s * 0.8 m = 17.1 m/s ``` Converting to mph: ``` v = 17.1 m/s * (3600 s / 1 h) * (1 mi / 1609.34 m) = 38.4 mph ```

answer:After orthogonalizing the vectors using the Gram-Schmidt process, we obtain the following orthogonal set: begin{align*} u_1 &= frac{1}{sqrt{141}}left{4,10,-5right} u_2 &= frac{1}{sqrt{183206}}left{-145,225,1336right} u_3 &= frac{1}{sqrt{11694}}left{95,-26,100right} end{align*} The normalized vectors are written in the form of u_i = frac{v_i - sum_{j=1}^{i-1} langle v_i, u_j rangle u_j}{|v_i - sum_{j=1}^{i-1} langle v_i, u_j rangle u_j|}, where langle cdot, cdot rangle denotes the inner product and | cdot | represents the Euclidean norm.