Question

pls ans question 5,6,7 & 8

w Help andout 2016 update.pdf (pase 8 of 11) Highilight Rotate Markup Exercises For questions (1)- (8) use the following information. You have measured the length of a table to be 205.0 cm, 205.8 cm, 205.4 cm, 204.6 cm, and 204.9 cm five independent times. You measured the width of the same table to be 60.1 cm, 60.4 cm, 60.2 cm, 60.0 cm, and 60.5 cm five independent times. 1) Calculate the mean length L of the table. 2) Calculate the standard deviation of the mean length ơL of the table. 3) Calculate the mean width W of the table. 4) Calculate the standard deviation of the mean width ơw of the table. 5) Calculate the area A Lx W of the table. 6) Using the correct equation for propagation of error, calculate the uncertainty of the area ơAof the table. 7) Calculate the perimeter P- 2L+2W of the table 8) Using the correct equation for propagation of error, calculate the uncertainty of the perimeter 하 of the table. 9) Renort the mean length. mean width. arca, and nerimeter.includine their

Answer 1

Mean length of table = 205.0 cm + 205.8 cm + 205 .4 cm + 204 .8 cm + 204 .9 cm/5 = 1025.7 cm/5 = 205.14 cm (iii) mean width table = 60.1cm + 60.14 cm + 60.2 cm + 60.0 cm + 60.5 cm/5 301.2 cm/5 = 60.24 (ii) standard deviation of mean length for sigma_L of the table sigma = squareroot 1/v sigma^N_i = 1 (x_i - mu)^2 mu = 205.14cm X_1 = 205.8 cm - (x_i - mu)^2 - (205.0 - 205 - 14)^2 - 0.28 X_2 = 205. 8 cm - (x_i - mu)^2 - (205.8 - 205.14)^2 - 0.4356 X_3 = 205. 4 cm - (x_i - mu)^2 - (205.4 - 205.14)^2 - 0.0676 X_4 = 204. 61 cm - (x_i - mu)^2 - (205.6 - 205.14)^2 - 0.2116 X_5 = 204. 9 cm - (x_i - mu)^2 - (204.9 - 205.14)^2 - 0.0576 1.0524 sigma_L = squareroot 1/5 times 1.0524 = 0.4588 (iv) Header deviation of mean median sigma_w of the table sigma_w = squareroot 1/5 times 0.172 = squareroot 0.0344 = 0.1855 cm