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MIP2602 Assignment 4 (COMPLETE SOLUTIONS) 2025 (750826 ) - DUE 15 August 2025

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Mathematics for Intermediate Phase teachers IV - MIP2602 Assessment 4 2025 (750826 ) - DUE 15 August 2025; 100 % TRUSTED workings, Expert Solved, Explanations and Solutions. For assistance call or W.h.a.t.s.a.p.p us on ...(.+.2.5.4.7.7.9.5.4.0.1.3.2)........... MIP2602 Assessment 4 2025 answers UNISA MIP2602 probability solutions Mathematics for Intermediate Phase Teachers UNISA Mathematics Education assignment help Probability and Data Handling for teachers UNISA Ethical Practices in data analysis MIP2602 probability tree diagram UNISA College of Education assessments Professional practices in data handling UNISA MIP2602 complete answers PDF UNISA assignments with cover page and declaration Statistics education ethical use of technology Data privacy in educational research Probability experimental vs theoretical calculations Mathematics ethics and professionalism in UNISA Complete and correct MIP2602 Assessment 4 2025 answers for UNISA students. Includes probability, data ethics, tree diagrams, and professionalism. Get complete and accurate solutions for UNISA MIP2602 Assessment 4 2025, covering probability, experimental data handling, ethical practices, professionalism, tree diagrams, and more. Includes a sample cover page, declaration, and comprehensive probability calculations. #UNISAAssignments #MIP2602Answers #ProbabilitySolutions #DataHandlingEthics #UNISAMathematicsEducation #StatisticsForTeachers #ProfessionalDataAnalysis #EducationalResearchEthics #CompleteAssignmentAnswers

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Kakoma Luneta Teaching Mathematics

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Course: Mathematics for Intermediate Phase teachers IV (MIP2602)

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MIP2602
ASSIGNMENT 4 2025

UNIQUE NO. 750826
DUE DATE: 15 AUGUST 2025

, Mathematics for Intermediate Phase teachers IV

SECTION A: PROBABILITY

QUESTION 1

1.1

1.1.1 Theoretical probability
Total marbles = 3 red + 4 blue + 2 green + 1 yellow = 10 marbles

 1.1.1.1 Red:
P(Red) = 3/10 = 0.3
 1.1.1.2 Blue:
P(Blue) = 4/10 = 0.4
 1.1.1.3 Green:
P(Green) = 2/10 = 0.2
 1.1.1.4 Yellow:
P(Yellow) = 1/10 = 0.1

1.1.2 Experimental probability

Trial Colour

1 Blue

2 Red

3 Green

4 Red

5 Blue

6 Blue

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