Question 41749: The data shows the number of times a cricket chirped in 1 minute and the temperature at the time of the chirps:
Temp (ºC): 18 8 15 14 10 7 14 6 14 9 14 13 15
# Chirps: 20 17 18 18 16 16 17 15 16 15 17 16 17
A linear model that approximates this data is N = 0.281t + 13.4 where N represents the number of chirps and t represents the temperature. Plot the actual data and the model on the same graph. How closely does the model represent the data?
I really need some help
Answer by tutorcecilia(2152)   (Show Source):
You can put this solution on YOUR website! Set up a graph. The x-axis is the temperature. The y-axis are the chirps. Plot the graph according to the data:
(x-axis) Temp (ºC): 18 8 15 14 10 7 14 6 14 9 14 13 15
(y-axis)# Chirps: 20 17 18 18 16 16 17 15 16 15 17 16 17
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Plot these points: (18, 20), (8, 17), (15, 18), (14, 18), (10, 16), (7,16), (14, 17), (6, 15), etc.
Even though the points are somewhat scattered, draw a "best-fit" line through the cluster of points.
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Next, using the formula N = 0.281t + 13.4, plug in each value for the temperature (x)and see what you get for the value of the chirps. Plot these new points on the same graph (maybe in a different color ink). Compare both lines and see how close they are.
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Example: N = 0.281t + 13.4
Plug in the first temperature value of 18 taken from the old point of (18, 20):
N = 0.281(18) + 13.4
N = 18.458. The new point is (18, 18.458) which is pretty close to the old point of (18, 20)
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Take another old point and plug in the x-value into N = 0.281t + 13.4
Use the x value from (8, 17):
N = 0.281(8) + 13.4
N = 15.64.
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This new point of (8, 15.64) is pretty close to the old point of (8, 17). Plot (8, 15.64).
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Continue to plug in the temperature and plot the new points. Compare the old and the new plotted lines.
I hope this helps. It isn't as hard as it seems -- it is just a lot of work!!!
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