PROBLEM NO.04. The length of a line was measured repeatedly on threedifferent occasions and probable error of each mean value was computed withthe following results: 1st set of measurement= 1202.50=0.03m, 2nd set ofmeasurement3 1202.75±0.05m, 3rd set of measurement%3 1202.85-0,08m.Determine the weighted mean of the three sets of measurements.

Answered: PROBLEM NO.04. The length of a line was…

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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The following values were determined in a series of a tape measurements of a line: 1000.58,1000.40, 1000.38,1000.48, 1000.40 and 1000.46 meters. Assume these values are equally reliable and their variations result only from accidental error. Which of the following most nearly gives the probable error of a single measurement and probable error of the mean?
The five observations of a distance are recorded as 422.16, 452.27, 382.25, and 472.28, 444.24 and given weights of 1, 2, 2, and 4, 5 respectively, by the surveyor. Determine the weighted mean.
Problem ABC. Five measurements were made to determine the lenght of a line and recorded as follows: 350.33, 350.22, 350.30, 350.27, and 350.30 meters. The most probable value of the lenght measured was. answer: 350.28 now, On Problem ABC. What will be the residual of the second measurement choices: a. -0.06, b. 0.06, c. -0.02, d. 0.02 -On Problem ABC. What will be the value of probable error of the mean choices: a. 0.08, b. 0.094, c. .085, d. 0.095
During a series of tape measurements of a line, following values were determined: 887.57 m, 887.42 m, 887.38 m, 887.24 m, 887.00 m, 887.46 m, 887.42 m, and 887.38 m. Compute for the: *Final expression of the most probable length
the following values were determined in a series of tape measurements of a line: 300.58,301.23, 299.25, 300.02, 301.08, and 301.2 meters. Determine the following: Most probable value measured length. Probable error of a single measurement and probable error of the mean.
A line 100m long was paced by a surveyor for four times with the following data, 142, 145, 145.5, and 146. Then another line was paced for four times again with the following results, 893, 893.5, 891, and 895.5. Determine the length of the line.
1. A pacer walks along a 90m course on fairly level ground for a survey party and counted 99.0, 100.0, 101.50, 103.0, 105.50 and 104.0 paces respectively. He then started walking an unknown distance OP which he recorded 174.0, 175.50, 176,0, and 177.0 paces. Determine the following: A] Unknown distance OP B] Relative Precision of the measurement if the taped length of OP is 155.0 meters.(Simplify to the nearest unit fraction) 2. A distance of 12.0m is measured with a tape. During measurement, the other end was discovered to be 2.19 m lower. Find the leveled distance.
Measurement of Horizontal Distance in Elementary Surveying: The following values of a line were determined in a series of tapemeasurements: 2050.67m, 2050.86m, 2050.41m, 2050.18m, 2050.68m,2050.03m. Determine the following:a. Most probable value.b. Probable error of a single measurement.c. Probable error of the mean.d. Final expression for the most probable length.e. Relative precision of the measurement.
The following values were determined in a series of tape measurements of a line: 1256.66, 1256.35, 1256.32, 1256.55, 1256.50, and 1256.72 meters. Determine the probable error of a single measurement. Determine the probable error of the mean.
A surveying instructor sent all the 40 students in his class out to measure a distance between two points marked ona runway. The students working in groups of four came up with 10 different measurements as follows: 922.11, 921.24, 921.86, 920.99, 922.59, 922.15, 922.33, 922.11, 920.22, and 920.01meters. Assuming these values are equally reliable and that variations result only from accidental errors, determine the relative precision of a single measurement and the relative precision of the mean.
with the given table below, find the following:- Linear Error of Closure- Relative Error of Closure- BSE Bearing Distance Lat Dep cLat cDep AdjLat AdjDep AB S 68 5 35 W 472.68 -176.36 -438.55 45.67633474 45.67926201 -130.68 -392.87 BC N 19 46 0 W 216.3 73.09 -203.40 20.88522093 20.8865594 93.98 -182.51 CD N 45 55 20 E 276.52 192.35 198.65 26.72086842 26.72258088 219.08 225.37 DA S 54 59 15 E 382.24 -219.31 313.06 36.93687525 36.93924243 -182.38 350.00 1347.57 ∈latitude=el=-130.22 ∈departure=ed=-130.23 130.2192993 130.2276447 0.00 0.00
Instruction: Show all solution. Problem: If for the particular measurement, the probable error of the mean is 0.075m and the most probable value of the measurement is 365.70 m, the relative precision would be expressed as ___. _ Thank you!

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