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COMM_V 205 101 102 103 2025W1 Exam B2 (R) 2025W1 Final - Requires Respondus LockDown Browser

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Suppose you have a data frame df about everyone enrolled in a class with 30 students. As it just so happens, no two students in the class share the same last name. In other words, there is no more than one student with the same last name in the class. There are three columns in the data frame, which are all character data type: UBC_ID, lastname, and firstname. Assume that UBC_ID can uniquely identify each person in this dataset. That is, each student has a unique UBC_ID. You open the dataset, and then run the following code: dfa <- df %>% arrange(lastname, firstname) Right after that, you run the following line of code: dfb <- df %>% arrange(lastname, UBC_ID) [ Select ] True False Not enough information  The order of the observations in dfa is the same as the order of the observations in dfb. If not enough information is provided, choose “not enough information”.  

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We start by restating the scenario: there are 30 students with unique last names, and we have three character columns UBC_ID, lastname, and firstname. UBC_ID uniquely identifies each person. The code sorts as follows: dfa <- df %>% arrange(lastname, firstname) dfb <- df %>% arrange(lastname, UBC_ID) Option analysis: Opt......Login to view full explanation

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