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The oxygen equivalence number of a weld is a number that can be used to predict properties such as hardness, strength, and ductility. The article “Advances in Oxygen Equivalence Equations for Predicting the Properties of Titanium Welds" (D. Harwig, W. Ittiwattana, and H. Castner, The Welding Journal, 2001:126s–136s) presents several equations for computing the oxygen equivalence number of a weld. An equation designed to predict the strength of a weld is X = 1.12C + 2.69N + O – 0.21 Fe, where X is the oxygen equivalence, and C, N, O, and Fe are the amounts of carbon, nitrogen, oxygen, and iron, respectively, in 0.0247, HN weight percent, in the weld. Suppose that for welds of a certain type, Hc = 0.0255, Ho = 0.0340, and OFe = 0.0597, ởC 0.0131, ƠN = 0.0194, oo 0.1668, µFe 0.0413. Furthermore assume that correlations are given by PC.N = -0.44, PC.0 = 0.58, PC,Fe = 0.39, PN,0 = -0.35. = -0.32, PN.Fe = 0.09, and PO,Fe Find µx- a. Find Cov(C, H), Cov(C, O), Cov(C, Fe), Cov(N, 0), Cov(N, Fe), and Cov(O, Fe). Find Ox. b. C.

The oxygen equivalence number of a weld is a number that can be used to predict properties such as hardness, strength, and ductility. The article “Advances in Oxygen Equivalence Equations for Predicting the Properties of Titanium Welds" (D. Harwig, W. Ittiwattana, and H. Castner, The Welding Journal, 2001:126s–136s) presents several equations for computing the oxygen equivalence number of a weld. An equation designed to predict the strength of a weld is X = 1.12C + 2.69N + O – 0.21 Fe, where X is the oxygen equivalence, and C, N, O, and Fe are the amounts of carbon, nitrogen, oxygen, and iron, respectively, in 0.0247, HN weight percent, in the weld. Suppose that for welds of a certain type, Hc = 0.0255, Ho = 0.0340, and OFe = 0.0597, ởC 0.0131, ƠN = 0.0194, oo 0.1668, µFe 0.0413. Furthermore assume that correlations are given by PC.N = -0.44, PC.0 = 0.58, PC,Fe = 0.39, PN,0 = -0.35. = -0.32, PN.Fe = 0.09, and PO,Fe Find µx- a. Find Cov(C, H), Cov(C, O), Cov(C, Fe), Cov(N, 0), Cov(N, Fe), and Cov(O, Fe). Find Ox. b. C.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.7: Distinguishable Permutations And Combinations
Problem 30E
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The oxygen equivalence number of a weld is a number that can be used to predict properties such as hardness, strength, and ductility. The article “Advances in Oxygen Equivalence Equations for Predicting the Properties of Titanium Welds" (D. Harwig, W. Ittiwattana, and H. Castner, The Welding Journal, 2001:126s–136s) presents several equations for computing the oxygen equivalence number of a weld. An equation designed to predict the strength of a weld is X = 1.12C + 2.69N + O – 0.21 Fe, where X is the oxygen equivalence, and C, N, O, and Fe are the amounts of carbon, nitrogen, oxygen, and iron, respectively, in 0.0247, HN weight percent, in the weld. Suppose that for welds of a certain type, Hc = 0.0255, Ho = 0.0340, and OFe = 0.0597, ởC 0.0131, ƠN = 0.0194, oo 0.1668, µFe 0.0413. Furthermore assume that correlations are given by PC.N = -0.44, PC.0 = 0.58, PC,Fe = 0.39, PN,0 = -0.35. = -0.32, PN.Fe = 0.09, and PO,Fe Find µx- a. Find Cov(C, H), Cov(C, O), Cov(C, Fe), Cov(N, 0), Cov(N, Fe), and Cov(O, Fe). Find Ox. b. C.
Transcribed Image Text:The oxygen equivalence number of a weld is a number that can be used to predict properties such as hardness, strength, and ductility. The article “Advances in Oxygen Equivalence Equations for Predicting the Properties of Titanium Welds" (D. Harwig, W. Ittiwattana, and H. Castner, The Welding Journal, 2001:126s–136s) presents several equations for computing the oxygen equivalence number of a weld. An equation designed to predict the strength of a weld is X = 1.12C + 2.69N + O – 0.21 Fe, where X is the oxygen equivalence, and C, N, O, and Fe are the amounts of carbon, nitrogen, oxygen, and iron, respectively, in 0.0247, HN weight percent, in the weld. Suppose that for welds of a certain type, Hc = 0.0255, Ho = 0.0340, and OFe = 0.0597, ởC 0.0131, ƠN = 0.0194, oo 0.1668, µFe 0.0413. Furthermore assume that correlations are given by PC.N = -0.44, PC.0 = 0.58, PC,Fe = 0.39, PN,0 = -0.35. = -0.32, PN.Fe = 0.09, and PO,Fe Find µx- a. Find Cov(C, H), Cov(C, O), Cov(C, Fe), Cov(N, 0), Cov(N, Fe), and Cov(O, Fe). Find Ox. b. C.
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