EXERCISE 2 ANALYSIS - Bio 205L EXERCISE 2 ANALYSIS 1(a(14 19 11 14.5 19.5/5 = 15.6 reticle units average length(15.6 r.u(10 m/r.u = 156 m average

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Bio 205L EXERCISE 2 ANALYSIS 1. (a) (14 + 19 + 11 + 14.5 + 19.5)/5 = 15.6 reticle units average length (15.6 r.u.)(10 μm/r.u.) =  156 μm average length (5.5 + 6.5 + 5 + 5.5 + 6)/5 = 5.7 reticle units average width = (5.7 r.u.)(10  μm/r.u.) =  57 μm average width (b) Standard deviation for average length = 3.595136 Standard deviation for average width = 0.570088 (c) 95% confidence level = (standard deviation)(2) length: 7.190272 width: 0.1140176 2. assuming width = height, volume = (4/3)(π)(length)(width)(height) = (4/3)(π)(156 μm )(57 μm )(57 μm ) = 2123063.183 μm 3 3. surface area = (2)(π)(width)(width) + (2)(π)(width)(length) = (2)(π)(57 μm )(57 μm ) + (2)(π)(57 μm )(156 μm ) = 76284.153  μm 2 4. ratio of surface area to volume: (Surface area  μm 2 / volume μm 3 )       = (76284.153 μm 2  / 2123063 μm 3 )       =  0.0359312 μm -1 5. (a) The microscope I used had a built-in illuminator. (b) The microscope I used was a binocular microscope.
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This note was uploaded on 06/15/2008 for the course BIO 205L taught by Professor Hanson during the Fall '07 term at University of Texas at Austin.

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EXERCISE 2 ANALYSIS - Bio 205L EXERCISE 2 ANALYSIS 1(a(14 19 11 14.5 19.5/5 = 15.6 reticle units average length(15.6 r.u(10 m/r.u = 156 m average

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