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Line AD is observed in three sections, AB, BC, and CD, with lengths and standard deviations given. Part A If AB 302.97 ± 0.018 ft, BC = 498.67 ± 0.013 ft, and CD = 784.65±0.016 ft, what is the total length AD and its standard deviation? 1586.29 0.016 ft 1586.29 0.047 ft 1586.29 0.217 ft 1586.29 0.027 ft Submit Request Answer Part B If AB 254.061 ± 0.0060 m, BC = 486.578 ± 0.0056 m, CD=465.837 ± 0.0058 m, what is the total length AD and its standard deviation? 1206.476 0.1319 m 1206.476 0.0100 m 1206.476 0.0058 m 1206.476 ± 0.0174 m Submit Request Answer Provide Feedback Next>

Line AD is observed in three sections, AB, BC, and CD, with lengths and standard deviations given. Part A If AB 302.97 ± 0.018 ft, BC = 498.67 ± 0.013 ft, and CD = 784.65±0.016 ft, what is the total length AD and its standard deviation? 1586.29 0.016 ft 1586.29 0.047 ft 1586.29 0.217 ft 1586.29 0.027 ft Submit Request Answer Part B If AB 254.061 ± 0.0060 m, BC = 486.578 ± 0.0056 m, CD=465.837 ± 0.0058 m, what is the total length AD and its standard deviation? 1206.476 0.1319 m 1206.476 0.0100 m 1206.476 0.0058 m 1206.476 ± 0.0174 m Submit Request Answer Provide Feedback Next>

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Line AD is observed in three sections, AB, BC, and CD, with lengths and standard deviations given. Part A If AB 302.97 ± 0.018 ft, BC = 498.67 ± 0.013 ft, and CD = 784.65±0.016 ft, what is the total length AD and its standard deviation? 1586.29 0.016 ft 1586.29 0.047 ft 1586.29 0.217 ft 1586.29 0.027 ft Submit Request Answer Part B If AB 254.061 ± 0.0060 m, BC = 486.578 ± 0.0056 m, CD=465.837 ± 0.0058 m, what is the total length AD and its standard deviation? 1206.476 0.1319 m 1206.476 0.0100 m 1206.476 0.0058 m 1206.476 ± 0.0174 m Submit Request Answer Provide Feedback Next>
Transcribed Image Text:Line AD is observed in three sections, AB, BC, and CD, with lengths and standard deviations given. Part A If AB 302.97 ± 0.018 ft, BC = 498.67 ± 0.013 ft, and CD = 784.65±0.016 ft, what is the total length AD and its standard deviation? 1586.29 0.016 ft 1586.29 0.047 ft 1586.29 0.217 ft 1586.29 0.027 ft Submit Request Answer Part B If AB 254.061 ± 0.0060 m, BC = 486.578 ± 0.0056 m, CD=465.837 ± 0.0058 m, what is the total length AD and its standard deviation? 1206.476 0.1319 m 1206.476 0.0100 m 1206.476 0.0058 m 1206.476 ± 0.0174 m Submit Request Answer Provide Feedback Next>
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