Wednesday, June 24, 2020

Nonlinear Quadratic Model Lab Report - 550 Words

Nonlinear Quadratic Model (Lab Report Sample) Content: Insert NameProfessors nameCourse/classDateNonlinear Quadratic ModelQR-1: Scatter plot for the dataThe data used to create the scatter plot is in the table below (see table 1) and the graph is the subsequent figure (see fig. 1)Table  SEQ Table \* ARABIC 1Time of day and temperature figuresTime of day (hours)Temperature (degrees F.)735950115613591461176220592344Fig.  SEQ Figure \* ARABIC 1. Scatter plot diagram for temperature against elapsed time QR-2: Quadratic polynomial of best fitThe quadratic polynomial of best fit, presented as y = ax2+bx+c, is:y = -0.3476x2 + 10.948x - 23.078, Where a = -0.3476 b = 10.948 c = -23.078QR-3: r2 value and significanceThe r2 value for the quadratic polynomial of best fit is 0.9699. In this case, the r2 value is the coefficient of determination and is used to predict data modeling results and infer data points based on the model and collected data. In this respect, the assumption is that all data chan ges will conform to the statistical model. For that matter, if r2 is 1.0, then any forecasts will have a 0.0% margin of error, while if r2 is 0.0, then any predictions will have a 100% margin of error. Considering that the r2 value 0.9699 is closer to 1.0 than 0.0, any predictions will have a 2.92% margin of error. Thus, the parabola is a good fit for the data since margin of error is small. QR-4: Determine the temperature estimate given that the time is 19.5 hoursy = -0.3476x2 + 10.948x - 23.078, x is the time in hours and y is the temperature estimate.y = -0.3476 (19.5)2 + 10.948(19.5) 23.078 = 58.23y = 58.2Temperature estimate = 58.2 degrees F.QR 5: Maximum temperature and time when it occurredThe value of x when the temperature is at its peak value is determined by the equation x = -b/2aThus, x = -10.948/2(-0.3476) = 15.75 or 15Â ¾ hoursThe time at maximum temperature is 15.75 hours or 3:45 pm.Substituting for x in the quadratic equation gives us the maximum temperature y = -0.347...