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Pages:
2 pages/≈550 words
Sources:
Level:
MLA
Subject:
Mathematics & Economics
Type:
Lab Report
Language:
English (U.S.)
Document:
MS Word
Date:
Total cost:
$ 8.64
Topic:
Nonlinear Quadratic Model (Lab Report Sample)
Instructions:
This is a laboratory report that requires the development of a nonlinear quadratic model.
source..Content:
Insert Name
Professor’s name
Course/class
Date
Nonlinear Quadratic Model
QR-1: Scatter plot for the data
The 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 1
Time of day and temperature figures
Time of day (hours)Temperature (degrees F.)735950115613591461176220592344
Fig. SEQ Figure \* ARABIC 1. Scatter plot diagram for temperature against elapsed time
QR-2: Quadratic polynomial of best fit
The 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 = -23.078
QR-3: r2 value and significance
The 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 changes 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 hours
y = -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 = 58.23
y = 58.2
Temperature estimate = 58.2 degrees F.
QR 5: Maximum temperature and time when it occurred
The value of x when the temperature is at its peak value is determined by the equation x = -b/2a
Thus, x = -10.948/2(-0.3476) = 15.75 or 15¾ hours
The 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...
Professor’s name
Course/class
Date
Nonlinear Quadratic Model
QR-1: Scatter plot for the data
The 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 1
Time of day and temperature figures
Time of day (hours)Temperature (degrees F.)735950115613591461176220592344
Fig. SEQ Figure \* ARABIC 1. Scatter plot diagram for temperature against elapsed time
QR-2: Quadratic polynomial of best fit
The 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 = -23.078
QR-3: r2 value and significance
The 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 changes 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 hours
y = -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 = 58.23
y = 58.2
Temperature estimate = 58.2 degrees F.
QR 5: Maximum temperature and time when it occurred
The value of x when the temperature is at its peak value is determined by the equation x = -b/2a
Thus, x = -10.948/2(-0.3476) = 15.75 or 15¾ hours
The 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...
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