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… The Chair opens the meeting. 2a. Approval of the text of the public part of the report of the 191st Meeting dated 21 May 2024 24.0803o – dated 03-06-24 – minutes with annex Refer to the approved version of the minutes with reference 24.0803/Io. 2b. Comments on the public part of the report There are no comments. 3. Announcements by the Dean The Dean informs the Council that on September 18, an artwork will be unveiled on a wall of the Co Greepzaal, Uns60. This artwork will feature images of all past Deans. The Dean reports on a decision by the … informed of this decision and reacted unpleasantly surprised. The exact number of places for Maastricht is not yet known. The Dean also informs the Council about the dismissal of a professor from FHML. After extensive investigation and attempts to find a resolution, UM decided to initiate a dissolution procedure in court. This is not the preferred approach, and usually, such matters are resolved differently. At the Council's request, the Dean explains that the current procedures for handling …

… −p. Therefore, the slope equals −p 1−p . It now easily follows that this line is described by: y = − p 1−p x + p 1−p + 1− p For the line with parameter q we have likewise y = − q 1−q x + q 1−q + 1− q. At the intersection point of these two lines we find that the same y-value occurs if − p 1−p x + p 1−p + 1− p = − q 1−q x + q 1−q + 1− q which is equivalent to [ q 1−q − p 1−p ]x = q 1−q + 1− q − [ p 1−p + 1− p] or [q(1− p)− p(1− q)]x = q(1− p) + (1− q)2(1− p)− p(1− q)− (1− p)2(1− q) This can be … y = p(1 − q) + 1 − p = 1 − pq. Summarizing, the intersection point has coordinates: (1− (1− p)(1− q), 1− pq) or (p + q − pq, 1− pq) (c) It’s a parabola. See after part (d). (d) Consider the intersection point for the two lines with parameters p and q, computed under (b). If we keep one line (with parameter p) fixed and we 1 vary the other parameter q, we see that their intersection point converges to the tangent point of the fixed line at C, if q converges to p. So, if we choose p = q, we find that … xF = 4 and xP = 6. By similar logic, the expected number of cards one has to by in order to get pictures of Finn and Poe, xFP , satisfies the equation xFP = 1/4 · (1 + xP ) + 1/6 · (1 + xF ) + 7/12 · (1 + xFP ) which implies after a straightforward computation that xFP = 73 5 . Similarly, xRP = 7 and xRF = 52 7 . Finally, the desired number xRFP satisfies the equation xRFP = 1/3 · (1 + xFP ) + 1/4 · (1 + xRP ) + 1/6 · (1 + xRF ) + 1/4 · (1 + xRFP ) with solution xRFP ≈ 8.22. Answer to Problem 4: …

… −p. Therefore, the slope equals −p 1−p . It now easily follows that this line is described by: y = − p 1−p x + p 1−p + 1− p For the line with parameter q we have likewise y = − q 1−q x + q 1−q + 1− q. At the intersection point of these two lines we find that the same y-value occurs if − p 1−p x + p 1−p + 1− p = − q 1−q x + q 1−q + 1− q which is equivalent to [ q 1−q − p 1−p ]x = q 1−q + 1− q − [ p 1−p + 1− p] or [q(1− p)− p(1− q)]x = q(1− p) + (1− q)2(1− p)− p(1− q)− (1− p)2(1− q) This can be … y = p(1 − q) + 1 − p = 1 − pq. Summarizing, the intersection point has coordinates: (1− (1− p)(1− q), 1− pq) or (p + q − pq, 1− pq) (c) It’s a parabola. See after part (d). (d) Consider the intersection point for the two lines with parameters p and q, computed under (b). If we keep one line (with parameter p) fixed and we 1 vary the other parameter q, we see that their intersection point converges to the tangent point of the fixed line at C, if q converges to p. So, if we choose p = q, we find that … xF = 4 and xP = 6. By similar logic, the expected number of cards one has to by in order to get pictures of Finn and Poe, xFP , satisfies the equation xFP = 1/4 · (1 + xP ) + 1/6 · (1 + xF ) + 7/12 · (1 + xFP ) which implies after a straightforward computation that xFP = 73 5 . Similarly, xRP = 7 and xRF = 52 7 . Finally, the desired number xRFP satisfies the equation xRFP = 1/3 · (1 + xFP ) + 1/4 · (1 + xRP ) + 1/6 · (1 + xRF ) + 1/4 · (1 + xRFP ) with solution xRFP ≈ 8.22. Answer to Problem 4: …