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Searching for :BiologyDate:2018-11-21 06:47:57Done by:Anonymous user(Visitor)Searching for :a chain line ABC crosses a river B and C being on the near and distant banks respectively. points B and A are on the same bank and distance between them is 20m. two perpendiculare AD= 25m and BC=16m are constructed at A and B such that DEC are in one line.determine BC the width of riverDate:2018-11-21 06:33:10Done by:Anonymous user(Visitor)

Searching for :a chain line ABC crosses a river B and C being on the near and distant banks respectively. points B and A are on the same bank and distance between them is 20m. two perpendiculare AD= 25m and BC=16m are constructed at A and B such that DEC are in one line.determine BC the width of riverDate:2018-11-21 06:32:51Done by:Anonymous user(Visitor)

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### Notes

**BMS college of engineering**

**University Visvesvaraya College of Engineering, Bangalore**

**The Oxford College of Engineering, Bangalore**

**BMS Institute of Technology and Management, Bangalore**

**CE 2453 PROJECT WORK**

**Fr. Conceicao Rodrigues College of Engineering, Mumbai**

**Sri Sairam College Of Engineering, Bangalore**

**CE2259 SURVEY PRACTICAL II**

**Dayanand Sagar College of Engineering, Bangalore**

**MVJ College of Engineering, Bangalore**

## In a survey of 260 college students, the following data were obtained:

->In a survey of 260 college students, the following data were obtained: 64had taken a mathematics course, 94 had taken a computer science course,58 had taken a business course, 28 had taken both a mathematics and a business course, 26 had taken both a mathematics and a computer science course, 22 had taken both a computer science and a business course, and 14 had taken all three typesi.How many of these students had taken none of the three courses?ii. How many had taken only a computer science courses ? -By:Nikhil-bharadwaj

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## Answers

Solution:Given: U=260 |A|=64 |B|=94 |C|=58 |A ∩ C| = 28 |A ∩ B| = 26 |B ∩ C| = 22 |A ∩ B ∩ C| = 14

|A U B U C| =? |B1| =?

|A U B U C| = A + B + C – |A ∩ B| – |B ∩ C| – |A ∩ C| + |A ∩ B ∩ C|

= 64 + 94 + 58 – 26 – 22 – 28 + 14

= 154

i. |A U B U C| = U – |A U B U C|

= 260 – 154

= 106

ii.|B1| = |B| - |B ∩ C| - |B ∩ A| + |A ∩ B ∩ C|

= 94 – 22 – 26 + 14 = 60Nikhil-bharadwaj

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