Lily R.
asked • 08/31/21help asap please!!! :) A raw potato was found to…
A raw potato was found to have a mass of 330 g. The potato was sliced and the slices were set in a single layer on a piece of screen wire to dry in the sun. The potato was measured after 7, 14, and 21 days and found to be 212 g, 148 g, and 126 g respectively.
what is the total mass of the potato likely to be on day 23?
then, determine when the potatoes mass was 170 g.
and finally, what was the total mass of the potato slices after three days of drying time
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2 Answers By Expert Tutors
William W. answered • 08/31/21
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Experienced Tutor and Retired Engineer
If you have a TI-84 calculator (or other similar), you can enter the data in a list and then do regression analysis on it to determine if it fits some standard patterns (linear, quadratic, etc). Doing so in this case (assuming the numbers are correctly typed) I found that a cubic regression fits perfectly.
This calculates the values of a, b, c, and d that fit y = ax3 + bx2 + cx + d and then you can use that function to find the answers to your question. If you need help using your calculator this way, let me know.
After the fact addition: Despite a perfect fit for the cubic regression model, it does not match the real world conditions outside of the 0 to 21 day range. Please look at the comment below for more.
Dayv O.
even if a<0 the way the data tapers as t increases does not intuitively fit cubic, maybe the calculator means recipocal of a cubic.
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08/31/21
William W.
Cubic functions generally have both increasing and decreasing intervals. This just happens to be on a decreasing interval. The value of r^2 for this fit is 1.0 in other words the function perfectly fits the data. A person may choose to reject this regression model based on the fact that outside of this range, the function does NOT behave as you would expect the real world problem to behave (it increases right after x = 21). This is the problem with using a regression outside the range of the data (or extrapolating). I have 100% confidence that the model will accurately give a projection of the day that the potato was 170 g. I do NOT have good confidence to use the model to project what the potato will weight at day 23 (which I did not notice until you questioned it). So in this case, at least for the projection of the day 23 weight, it would be better to use an exponential decay model even though the r^2 value you got from the cubic was a perfect fit. I used y1 ~ a*b^(x1 - h) + k in desmos and got an excellent fit (r^2 = 0.9991) AND it fits what you would expect the real life situation to be. Thanks for questioning my work.
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08/31/21
Dayv O.
how thin were the slices of potato??,,I am sure desmos is great and everything. But just first principles are for exponential functions;;; y(t)=Ab^t, here A=330 at time t=0.,,,,averaging I calculate b as .944 given t is in days. The data at day 21 is outlier so I ignore. ,,,, At day 3, y= 277.6 ,,, at day 23 y=87.7. ,,, 11.5 days is (ln(170/330))/ln(.944) or when weight is 170. ,,,, day 7 y=220.5,,, day 14 y=147.3,,, day 21 y=98.4.,, Is the r^2=.999 y1-a*e^(x1-h)+k much diffferent?,,,Thank you too. update--- lets say final potato weight is 100 grams after many days, then first principles model would be y=230*.89^t+100
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09/01/21
William W.
When doing regression, you cannot assume A = 330, despite the fact that it is the "normal" way to consider an exponential function. Doing so forces the function to fit the first data point which may or may not be true. Regression is all about taking the minimum residual for all data points (none more important than the other). Anyway, here is the result I got: The function is y(t) = 8.4842*0.900286^(t - 31.4945) + 98.5089 y(0) = 330.4 y(7) = 209.7 y(11.2) = 170 y(14) = 151.8 y(21) = 124.1 y(23) = 119.2
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09/01/21
Dayv O. answered • 09/01/21
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first principles are for exponential functions;;; y(t)=Ab^t,
here A=330 at time t=0.
averaging I calculate b as .944 given t is in days.
y=330*(.944)t
update -- find b in this case = (y(14)/y(7))(1/7),,, it is agebra called geometric (growth, or decline dependin on b>1 or b<1)
The data at day 21 is outlier so I ignore.
At day 3, y= 277.6
at day 23 y=87.7
11.5 days is (ln(170/330))/ln(.944) or when weight is 170
my estimated model for grams is y=330*(.944)t
day 7 y=220.5,,, day 14 y=147.3,,, day 21 y=98.4.,,
so it is not exact but might be close,
depending on "how thin were the slices of potato?"
update--- let's say final potato weight is 100 grams after many days, then first principles model would be ;;; y(t)=230*.89^t+100.,,,,y(0)=330,,,y(7)=201.7,,,y(14)=145.0,,,y(21)=119.9,,,y(23)=115.8,,,y(3)262.1,,,and y=170 in 10.2 days;;;;;;;;;;;;;;;;;;;not bad for the basic way and thinking, hey the potato weight doesn't go to near zero after 50 days.
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Dayv O.
the value at 21 days, if it was 106, would indicate exponential decay - are you sure 126 is correct? Are you studying exponential functions or is William W. on the right track and you are studying regression?08/31/21