Question

InsertH Length of Mass M2 Left End Right End Deviation Absolute from the Deviation Length (cm) Mean (em) (cm) см 20.0cm 24.9.m 4.9cm Average Length M2 cm Mean cm Deviation Time for Mass M2 to Y Values (cm) Travel Its Own Lengt Absolute DeviationAbsolute Mean (cm (em) Top End Bottom End Length (cm) from the Deviation Time(s ) Deviatior 2101 19.0 21. 1Cm 405 114m 29 25 cm cm Average Y Value Average Time-t Mean Deviation t Mean Deviation

Answer 1

Given that Avg 4 values = 20.0 + 19.0 + 19.4 + 19.0 + 19.3/5 = 19.34 Deviation from mean 1^st value = 19.34 - 20.0 = -0.66 2^nd value = 19.34 - 19.0 = 0.34 3^rd value = 19.34 - 19.4 = -0.06 4^th value = 19.34 - 19.0 = 0.34 5^th value = 19.34 - 19.3 = 0.04. Mean deviation = 0.66 + 0.34 + 0.06 + 0.34 + 0.04/5 = 0.288 cm Avg time = 0.236 + 0.230 + 0.229 + 0.225 + 0.221/5 = 0.2282 \r\n Given that Mean M_2 = 5.0 + 5.0 + 5.0 + 4.9 + 4.9 = 5.0 + 5.0 + 5.0 + 4.9 + 4.9/5 = 4.96 Deviation from mean = x -x_m For 1^st value = -4.96 + 5.0 = 0.04 2^nd value = 0.04 3^rd value = 0.04 4^th value = -4.96 + 4.9 = -0.06 5^th value = -0.06 Mean deviation = 0.04 + 0.04 +