answer:We define the stopping time tau = min(tau_{-a}, tau_b), assuming it is finite almost surely. Utilize the exponential martingale M_t = exp(lambda B_t - lambda^2 t/2) for t geq 0. By the Doob's optional stopping theorem, we have E[M_{twedgetau}] = 1. Expanding this expectation, we get: begin{align*} E[M_{twedgetau}] &= E[e^{lambda B_{tau} - lambda^2 tau}1_{{tau leq t}}] + E[e^{lambda B_{t} - lambda^2 t}1_{{tau geq t}}]. end{align*} As t to infty, the first term converges to E[e^{lambda B_{tau} - lambda^2 tau}] by the dominated convergence theorem. The second term converges to 0 due to the boundedness of B_{twedgetau}. Thus, we have E[e^{lambda B_{tau} - lambda^2 tau}] = 1. Since B_{tau} = -a on {tau = tau_{-a}} and B_{tau} = b on {tau = tau_{b}}, we can write: begin{align*} 1 &= e^{-lambda a}E[e^{-lambda^2 tau_{-a}/2}1_{{tau = tau_{-a}}}] + e^{lambda b}E[e^{-lambda^2 tau_{b}/2}1_{{tau = tau_{b}}}]. end{align*} Similarly, with the modified martingale tilde{M}_t = exp(-lambda B_t - lambda^2 t/2), we get another equation: begin{align*} 1 &= e^{lambda a}E[e^{-lambda^2 tau_{-a}/2}1_{{tau = tau_{-a}}}] + e^{-lambda b}E[e^{-lambda^2 tau_{b}/2}1_{{tau = tau_{b}}}]. end{align*} Solving this system for E[e^{-lambda^2 tau_{-a}/2}1_{{tau = tau_{-a}}}], we have: [E[e^{-lambda^2 tau_{-a}/2}1_{{tau = tau_{-a}}}] = frac{sinh(lambda b)}{sinh(lambda(a+b))}.] To find E(tau_{-a}1_{{tau_{-a}<tau_b}}), we take the derivative with respect to mu = lambda^2 and let lambda approach 0: [E(tau_{-a}1_{{tau_{-a}<tau_b}}) = lim_{lambda to 0} frac{partial}{partial mu} left(frac{sinh(lambda b)}{sinh(lambda(a+b))}right).] Note that the almost sure finiteness of tau and the justification for the derivative under the integral sign are important steps in this derivation, which need to be properly established.

answer:Using the formula relating centripetal force and kinetic energy: #ΔF = (2/r) * ΔKE# where: * ΔF is the change in centripetal force * r is the radius of the track * ΔKE is the change in kinetic energy Solving for ΔKE: #ΔKE = (r/2) * ΔF# #ΔKE = (5 m/2) * 10 N# #ΔKE = 25 J# Therefore, the car's kinetic energy will change by #25 J#.

answer:The multiplication of the scalar -frac{2}{25} with the matrix results in: left( begin{array}{ccc} -frac{2}{25} cdot 7 & -frac{2}{25} cdot 6 & -frac{2}{25} cdot (-2) -frac{2}{25} cdot 0 & -frac{2}{25} cdot 5 & -frac{2}{25} cdot 6 -frac{2}{25} cdot 10 & -frac{2}{25} cdot (-10) & -frac{2}{25} cdot (-6) end{array} right) Simplifying each element, we get: left( begin{array}{ccc} -frac{14}{25} & -frac{12}{25} & frac{4}{25} 0 & -frac{2}{5} & -frac{12}{25} -frac{4}{5} & frac{4}{5} & frac{12}{25} end{array} right)

answer:The approach you've used for parts c and d is correct. You effectively utilized the information given and applied it to the Venn diagram to calculate the probabilities. You calculated P(A cup B cup C) initially, which encapsulates all the possibilities, and then used it to find the probabilities for parts c and d. For part c, you calculated P(C cap (overline{A} cup overline{B})) = P(A cup B cup C) - P(A cup B) = 0.85 - 0.63 = 0.22. For part d, your method to find the probability of selecting exactly one option is correct: P = 3P(A cup B cup C) - P(A cup B) - P(A cup C) - P(B cup C) = 3 times 0.85 - 0.63 - 0.77 - 0.8 = 0.35. Since you've correctly identified the relevant data and applied it to the Venn diagram, your solutions for parts c and d are valid. You were efficient in focusing on the necessary information instead of being distracted by the additional percentages provided.