EGR 120Introduction to EngineeringFile: EXPREG.MCDMathCAD Example: Exponential RegressionProblem 3.18 - Graph Counts per second ,C, vs. plate thickness, W, for a Geiger counter on a semi-log graph and determine an equation representing the data.ORIGIN1N16..Note: There are 6 data points. Plate Thickness W, mm Geiger CounterC, Counts Per SecondW2.05.010.020.027.532.5C5500370025501300715470Geiger CounterCounts per second vs plate thicknessCNWN0 5 10 15 20 25 30 351001.1031.104Note: Double-click on the graph and change the y-axis to a log scale.Note: To determine the exponential equation representing the data we recall that the exponential formula can be expressed as: mxy = be or ln(y) = mx + ln(b)so the slope m = change in ln(y) and the intercept b = ln(b) change in xso we need to generate a new function ln(y) for finding the slope.Note: A new column has been generated for regression analysis:LNCNln CNLNC8.6138.2167.8447.176.5726.153=m slope W LNC,() m0.077=Note: form is slope(x,y)lnb intercept W LNC,()lnb8.675=Note: Intercept = ln(b), not bbelnbb5.856 103.=so mx -0.077WResult: y = be or C = 5856eNote: If we want to graph the results using the empirical formula just determined to generate a straight line, we should do the following: 1. Graph the original data showing points only 2. Form a new set of values, CNEW, using the empirical formula 3. Graph the new set of values showing only the lineCNEWNbemWN..CNEW5.018 103.3.982 103.2.707 103.1.252 103.701.755477.156=Geiger CounterCounts per second vs plate thicknessCNEWNCNWN0 5 10 15 20 25 30
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