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… University, each faculty/school has its own International Relations Office (IRO). For FPN exchange students the FPN IRO is their main point of contact before, during and after their exchange period at FPN. The FPN IRO assists exchange students with all the formalities for registration and supplies all the information needed to make a stay successful. You can contact us via email: fpn-iro-incoming@maastrichtuniversity.nl Postal address Maastricht University Faculty of Psychology and … instructions. Courses In the course overview you can find all the information you need about FPN bachelor’s courses, such as the course coordinator(s), learning outcomes and competences, number of ECTS, literature and prerequisites. Course selection When selecting courses, please note that some courses will assume knowledge of topics covered in previous courses, called prerequisites. If you do not meet these requirements, you are not allowed to select these courses. Students are responsible for making sure that they have the required knowledge to take courses that state certain prerequisites. Master’s courses (Fall only) Students who completed their Bachelor’s programme (180 ECTS) can submit a request to the Board of Examiners to take up to two Master’s …

… University, each faculty/school has its own International Relations Office (IRO). For FPN exchange students the FPN IRO is their main point of contact before, during and after their exchange period at FPN. The FPN IRO assists exchange students with all the formalities for registration and supplies all the information needed to make a stay successful. You can contact us via email: fpn-iro-incoming@maastrichtuniversity.nl Postal address Maastricht University Faculty of Psychology and … instructions. Courses In the course overview you can find all the information you need about FPN bachelor’s courses, such as the course coordinator(s), learning outcomes and competences, number of ECTS, literature and prerequisites. Course selection When selecting courses, please note that some courses will assume knowledge of topics covered in previous courses, called prerequisites. If you do not meet these requirements, you are not allowed to select these courses. Students are responsible for making sure that they have the required knowledge to take courses that state certain prerequisites. Master’s courses (Fall only) Students who completed their Bachelor’s programme (180 ECTS) can submit a request to the Board of Examiners to take up to two Master’s …

… is consistent throughout all steps of the application process, as well as across the different study programmes and faculties; · Translating the majority of communication written for applicants from English to Dutch; · Consulting and deliberating with faculties (admission officers, M&C) and project members (communications, IT, processes) on how to best set up application processes and accompanying communication. Profile candidate: We are looking for a candidate who: · Is looking to contribute … as well as admission officers; · Can work from a general UM perspective and in the UM-wide interest, as opposed to serving the needs of their own organisational unit; · And above all: is keen to participate in the Applicant Journey project and feels they will enjoy being involved Knowledge of and/or experience with the admissions process is a plus, but is not required. What can you expect in return? If you join the Applicant Journey project, you: · Will be working in a diverse and very … .. hrs on Wednesday .. hrs on Thursday .. hrs on Friday .. In consultation Start and end date of the project? Start date: as soon as possible End date: October 2021 End date publication project platform*: Not applicable, contact person will indicate when publication can be removed. * after this date the project will be removed from the website. Expected results: Financing of project / temporary work (mark with X): .. Financing from department of project provider X Processed internally by the …

… as follows. d = 0: 0, 0, 0, 0, 0 d = 1: 1, 0, 0, 0, 0 d = 2: 2, 0, 0, 0, 0 d = 3: 3, 0, 0, 0, 0 d = 4: 4, 0, 0, 0, 0 d = 5: 1, 1, 1, 1, 1 d = 6: 2, 1, 1, 1, 1 d = 7: 3, 1, 1, 1, 1 d = 8: 4, 1, 1, 1, 1 d = 9: 5, 1, 1, 1, 1 and in general for d = 5x+ y, with x = 0, 1, . . . and y = 0, 1, 2, 3, 4: d = 5x+ 0: x+ 0, x, x, x, x d = 5x+ 1: x+ 1, x, x, x, x d = 5x+ 2: x+ 2, x, x, x, x d = 5x+ 3: x+ 3, x, x, x, x d = 5x+ 4: x+ 4, x, x, x, x This way we use 21 balls of weight 0, 22 balls of weight 1, 23 balls … Han does not have it). Then, for each card he has played, Han knows the card Lando needs to have to win at the next turn. Moreover, Han knows the cards of Lando (all the numbers between 1 and 24 except the cards he had at the beginning and the cards they played). As Lando started Han has one more card than Lando when he plays, so he can choose a card such that Lando cannot win the next turn. Hence, Lando cannot win. Then, either Han wins before he plays his last card, or he plays his last card, and the sum 1 + 2 + ...+ 24 = 24 ⇤ 25/2 is divisible by 25. So Han …

… as follows. d = 0: 0, 0, 0, 0, 0 d = 1: 1, 0, 0, 0, 0 d = 2: 2, 0, 0, 0, 0 d = 3: 3, 0, 0, 0, 0 d = 4: 4, 0, 0, 0, 0 d = 5: 1, 1, 1, 1, 1 d = 6: 2, 1, 1, 1, 1 d = 7: 3, 1, 1, 1, 1 d = 8: 4, 1, 1, 1, 1 d = 9: 5, 1, 1, 1, 1 and in general for d = 5x+ y, with x = 0, 1, . . . and y = 0, 1, 2, 3, 4: d = 5x+ 0: x+ 0, x, x, x, x d = 5x+ 1: x+ 1, x, x, x, x d = 5x+ 2: x+ 2, x, x, x, x d = 5x+ 3: x+ 3, x, x, x, x d = 5x+ 4: x+ 4, x, x, x, x This way we use 21 balls of weight 0, 22 balls of weight 1, 23 balls … Han does not have it). Then, for each card he has played, Han knows the card Lando needs to have to win at the next turn. Moreover, Han knows the cards of Lando (all the numbers between 1 and 24 except the cards he had at the beginning and the cards they played). As Lando started Han has one more card than Lando when he plays, so he can choose a card such that Lando cannot win the next turn. Hence, Lando cannot win. Then, either Han wins before he plays his last card, or he plays his last card, and the sum 1 + 2 + ...+ 24 = 24 ⇤ 25/2 is divisible by 25. So Han …

… as follows. d = 0: 0, 0, 0, 0, 0 d = 1: 1, 0, 0, 0, 0 d = 2: 2, 0, 0, 0, 0 d = 3: 3, 0, 0, 0, 0 d = 4: 4, 0, 0, 0, 0 d = 5: 1, 1, 1, 1, 1 d = 6: 2, 1, 1, 1, 1 d = 7: 3, 1, 1, 1, 1 d = 8: 4, 1, 1, 1, 1 d = 9: 5, 1, 1, 1, 1 and in general for d = 5x+ y, with x = 0, 1, . . . and y = 0, 1, 2, 3, 4: d = 5x+ 0: x+ 0, x, x, x, x d = 5x+ 1: x+ 1, x, x, x, x d = 5x+ 2: x+ 2, x, x, x, x d = 5x+ 3: x+ 3, x, x, x, x d = 5x+ 4: x+ 4, x, x, x, x This way we use 21 balls of weight 0, 22 balls of weight 1, 23 balls … Han does not have it). Then, for each card he has played, Han knows the card Lando needs to have to win at the next turn. Moreover, Han knows the cards of Lando (all the numbers between 1 and 24 except the cards he had at the beginning and the cards they played). As Lando started Han has one more card than Lando when he plays, so he can choose a card such that Lando cannot win the next turn. Hence, Lando cannot win. Then, either Han wins before he plays his last card, or he plays his last card, and the sum 1 + 2 + ...+ 24 = 24 ⇤ 25/2 is divisible by 25. So Han …

… as follows. d = 0: 0, 0, 0, 0, 0 d = 1: 1, 0, 0, 0, 0 d = 2: 2, 0, 0, 0, 0 d = 3: 3, 0, 0, 0, 0 d = 4: 4, 0, 0, 0, 0 d = 5: 1, 1, 1, 1, 1 d = 6: 2, 1, 1, 1, 1 d = 7: 3, 1, 1, 1, 1 d = 8: 4, 1, 1, 1, 1 d = 9: 5, 1, 1, 1, 1 and in general for d = 5x+ y, with x = 0, 1, . . . and y = 0, 1, 2, 3, 4: d = 5x+ 0: x+ 0, x, x, x, x d = 5x+ 1: x+ 1, x, x, x, x d = 5x+ 2: x+ 2, x, x, x, x d = 5x+ 3: x+ 3, x, x, x, x d = 5x+ 4: x+ 4, x, x, x, x This way we use 21 balls of weight 0, 22 balls of weight 1, 23 balls … Han does not have it). Then, for each card he has played, Han knows the card Lando needs to have to win at the next turn. Moreover, Han knows the cards of Lando (all the numbers between 1 and 24 except the cards he had at the beginning and the cards they played). As Lando started Han has one more card than Lando when he plays, so he can choose a card such that Lando cannot win the next turn. Hence, Lando cannot win. Then, either Han wins before he plays his last card, or he plays his last card, and the sum 1 + 2 + ...+ 24 = 24 ⇤ 25/2 is divisible by 25. So Han …