SOLUTION: 1)IN A CLASS OF 80 STUDENTS 53 STUDY ART,60 STUDY BIOLOGY,36 STUDY ART AND BIOLOGY,34 STUDY ART AND CHEMISTRY 6 STUDY BIOLOGY ONLY AND 18 STUDY BIOLOGY BUT NOT CHEMISTRY. ILLUSTRAT

Algebra ->  sets and operations -> SOLUTION: 1)IN A CLASS OF 80 STUDENTS 53 STUDY ART,60 STUDY BIOLOGY,36 STUDY ART AND BIOLOGY,34 STUDY ART AND CHEMISTRY 6 STUDY BIOLOGY ONLY AND 18 STUDY BIOLOGY BUT NOT CHEMISTRY. ILLUSTRAT      Log On


   



Question 999463: 1)IN A CLASS OF 80 STUDENTS 53 STUDY ART,60 STUDY BIOLOGY,36 STUDY ART AND BIOLOGY,34 STUDY ART AND CHEMISTRY 6 STUDY BIOLOGY ONLY AND 18 STUDY BIOLOGY BUT NOT CHEMISTRY. ILLUSTRATE THE INFORMATION IN A VENN DIAGRAM. DETERMINE THE NUMBER OF STUDENTS WHO STUDY ART ONLY, b) CHEMISTRY

2)IN A CLASS OF 100 STUDENT 37 STUDY MATH,32 STUDY PHYSIC and 31 study chemistry,9 study math and physic,12 study math and chemistry,11 study physic and chemistry and 28 student study none of the three subject.Illustrate the information in a venn diagram and find the number of student who study all the three subject (b) the number of student who study only math,only physic and only chemistry (c) how many student study exactly two subject.
3)In a class 30 offer biology,21 offer chemistry and 22 offer physic.15 offer physic and biology,10 offer physic and chemistry,13 offer biology and chemistry,2 offer physic only,3 offer chemistry only and 7 offer biology only.Illustrate this in a vent diagram,find the number of student who offer all three subject (b)in the class (c)if a student is selected at random,what is the probability that he studies either physic or chemistry.
4)In a survey of the 100 out-patients who reported at a hospital one day , it was found out that 70 complained of fever,50 complain of stomach ache and 30 were injured. All 100 patients had at least one of the complaints and 44 had exactly two of the complaints. How many patients had all the three complaint?

Answer by solver91311(24713) About Me  (Show Source):
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Set F is fever, set S is stomach ache, set I is injury. n(F) + n(S) + n(I) = 150, but that counts n(F & S) + n(S & I) + n(F & I) twice. Hence 44 have been counted twice, makes 106. But that also counts n(F & S & I) three times, hence 2 times n(F & S & I) must equal 6. Then the number who reported all three complaints must be 6 divided by 2, or 3.

Next time you post, read the instructions: 1 problem per post. Also, don't type in all caps -- it is both annoying and rude.

John

My calculator said it, I believe it, that settles it