Question 1181509
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Since the length of piece 1 measured 1.3 meters, we know that the true measure of that length, *[tex \Large x_1] is in the range *[tex \Large 1.25\,\leq\,x_1\,<\,1.35].  Similarly, *[tex \Large 0.935\,\leq\,x_2\,<\,0.945].  So the greatest that that sum of the measures of the two pieces could be would be *[tex \Large 1.35\,+\,0.945\ =\ 2.295] meters and the shortest the sum could be is *[tex \Large 1.25\,+\,0.935\ =\ 2.185].  However, the original piece of pipe could be actually in the range *[tex \Large 2.95\,\leq\,l\,<\,3.05].  So the shortest the third length could be is the smallest the original pipe could be minus the longest the sum of the first two pieces could be, namely *[tex \Large 2.95\ -\ 2.295\ =\ 0.655] meters and the longest the third piece could be is the longest the original pipe could be minus the shortest the sum of the first two pieces could be, namely *[tex \Large 3.05\ -\ 2.185\ =\ 0.815] meters.


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.655\ \leq\ x_3\ <\ 0.815]


																
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
*[illustration darwinfish.jpg]

From <https://www.algebra.com/cgi-bin/upload-illustration.mpl> 
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