Question: Table 1: Estimated Values Mean value Standard deviation Number of from GA measurements Length (cm) 170.01 cm 9 30 Width (cm) 711) cm 3 30

in your experiment Table 2: Measured Values Uncertainty Number of measurements measured

Area Areames = 182* 76 Areameasured =+ 3, 8 32 cm percentage

Table 1: Estimated Values Mean value Standard deviation Number of from GA measurements Length (cm) 170.01 cm 9 30 Width (cm) 711) cm 3 30 Record the measured mean values and standard deviations you found in your experiment Table 2: Measured Values Uncertainty Number of measurements measured Area Areames = 182* 76 Areameasured =+ 3, 8 32 cm percentage error in measured Area 1. Errormeasured = ( 182 . 1 ) x($ 6 . 1 ) - 13, 832 * 100% 13832 Ans = 0.186 % . Standard deviation for length = g So the value of length will be 1701 9 the value of bengt width will be lie 71+3 length ( 160 .9 to + 79.1 ) approx 503%% change 74 width ( 68 to 80 ) approx 4.22% change11:56 AM Thu Sep 1 X IMG-20220901-WA0230.jpg 01 100% Table 1 : Estimated Values mean value Standard deviation from GA Number of Meawremew 30 Length (CM) 170. 01cm 30 71.0cm 3 width( to Measured Values Number of Measurements Value Incertanity Length ( am ) 182.Ocm width Icmj 76 . Ocm O. Jem Error in length & width Error in length = 182 - 170.01 * 100% 170 01 Error in length - 6.9# % Error in width = 76- 71 x10% 71 Erron in width = 7 % on. Estimated Area AreaEx = lengthest x whathest ArcaEst = 12 , 070 . 71Cm 2 Helpful Unhelpful Ask Expert Tutors

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