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The oxygen equivalence number of a weld is a number that

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 14E Chapter 2.5

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 14E

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. One equation, designed to predict the hardness of a weld, is X = 0 + 2N + (2/3)C, where X is the oxygen equivalence, and O, N, and C are the amounts of oxygen, nitrogen, and carbon, respectively, in weight percent, in the weld. Suppose that for welds of a certain type, μO = 0.1668, μN = 0.0255, μC = 0.0247, σO= 0.0340, σN = 0.0194. and σC = 0.0131.

a. Find μx

b. Suppose the weight percents of O, N, and C are independent. Find σx.

Step-by-Step Solution:

Step 1</p>

The equation which used to predict the hardness of  a weld is X= 0+ 2N+ (2/3) C , where X is the oxygen equivalence , O, N, C are the amounts of oxygen, nitrogen and carbon.

We have to find

  if the weight percentage of O, N, and C are independent.

Step 2</p>

It is given that the weight of oxygen, nitrogen and carbon for wields of a certain type are

0.1668 ,  , , , 0.0194 and .

        The expected amount of oxygen equivalence :

         The given equation  X = O + 2 N + (2/3) C

        Since E(ax + b ) = a E(x) + b

                E(x) = E(O) + 2 E(N) + (2/3) E(C)

                           =  0.1668 + 2 (0.0255) + (2/3)

                                =   0.2342

     Therefore the expected amount of  oxygen equivalence , = 0.2342 .

Step 3 of 3

Chapter 2.5, Problem 14E is Solved
Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

This full solution covers the following key subjects: oxygen, weld, equivalence, weight, suppose. This expansive textbook survival guide covers 153 chapters, and 2440 solutions. The answer to “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. One equation, designed to predict the hardness of a weld, is X = 0 + 2N + (2/3)C, where X is the oxygen equivalence, and O, N, and C are the amounts of oxygen, nitrogen, and carbon, respectively, in weight percent, in the weld. Suppose that for welds of a certain type, ?O = 0.1668, ?N = 0.0255, ?C = 0.0247, ?O= 0.0340, ?N = 0.0194. and ?C = 0.0131.a. Find ?x________________b. Suppose the weight percents of O, N, and C are independent. Find ?x.” is broken down into a number of easy to follow steps, and 145 words. Statistics for Engineers and Scientists was written by and is associated to the ISBN: 9780073401331. This textbook survival guide was created for the textbook: Statistics for Engineers and Scientists , edition: 4. The full step-by-step solution to problem: 14E from chapter: 2.5 was answered by , our top Statistics solution expert on 06/28/17, 11:15AM. Since the solution to 14E from 2.5 chapter was answered, more than 762 students have viewed the full step-by-step answer.

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