Question

1. Ten steel specimens were submerged in seawater at various temperatures, and the corrosion rates were measured. The results are presented in the following table. Temperature ("C) Corrosion (mm/yr) 0.48 0.56 0.68 0.85 1.05 0.99 1.26 1.13 1.45 1.58

Answer 1

solution - Ten at various measured steel specimens were temperature and the submerged corrosion in seawater rates were Temperature (°C) 4 ( 2 ) 7.5 8.2 comosion (mmlyer (Yi) 0.48 0.56 0.68 0.85 1.05 8.7 15.8 0.99 1.26 3.6 I 56.25 6.592 67.24 S.916 75.69 13.43 249.64 11.865 l 127.69 14.75) 1 222.01 28.602 515.29 30.962 750176 39. 15 I 729 42,028 707.56 194.896 3501.13 this quantity and 11.3 14.9 22.7 27.4 0 27. 26.6 170. What is correlation 0, 2304 0.3136 0.4624 0.7225 1.1025 0.9801 1.5876 1,2769 2.1025 2.4964 11.2749 the 1.4 1-13 1 .45 L 1.58 10.03 the relation between coefficient. 0 Sample mean (a) = Eni - 17001 - 12.01 © sample mean (T) = 2y n a 10.03 10 ? 1.003 G standard deviation (6x) = Eriz - (87)2 (17.01) 3501.13 ✓ 10 = 7.7956 *

② standard devication (by) = / Eyr2 - (9)2 11.2749 - wazung - (108) (1.003)² Joy = 0.3485 / ♡ cov (May) - Eniyi - (T) (9) - (17.01) (1.003). = 194.896 10 = 2.4285 COV(214) (1 comecetion coefficient (o)= Concury) (6x) (64) (7.2956) (0.3485) r = 0.8938| 6 The least square regression line is Ij = atbai where, bacov (219) 612 and a = 9 bał Then b = b = 2.4285 (7.7956) 2 16= 0.0399

and a=ýba a = (1.003) - (0.0399) (17.01) a= 0,3243 Then least square regression line is y=atby ie. y = 0.3243 + (0,0399) op To predict the corrosion of 30°C ? predict the strength ie. X = 30°c. predicted value Then of y is y = 0.3243 + (010399 ) (30) y = 1.5213 Therefore the predicted strength is 1.3213