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Numerade Educator

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Problem 8 Medium Difficulty

The table gives the number of yeast cell in a new laboratory culture.
(a) Plot the data and use the plot to estimate the carrying capacity for the yeast population.
(b) Use the data to estimate the initial relative growth rate.
(c) Find both an exponential model and a logistic model for these data.
(d) Compare the predicted values with the observed values, both in a table and with graphs. Comment on how well your models fit the data.
(e) Use your logistic model to estimate the number of yeast cells after 7 hours.

Answer

a) 680
b) $\simeq 0.5833$
c) $P(t)=\frac{680}{1+36.78 e^{-\frac{7 t}{12}}}$
d) Clearly the logistic model fits in better as compared to the exponential function.
e) $\simeq 420$ yeast cells

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Video Transcript

Hey, it's Clara. So when you married Hyun So we see that the curve has a carrying capacity to sticks 100 and really so the east population ISS 6 80 for part B, we have the initial relative growth rate to be won over P subzero, which is 18 times 39 minus 18. Which is the change in P over the change in time, which is to minus zero, which gives us 0.583 for part C. We have the exponential model, which is pft. Be equal to 18 which is our initial population e to the 70 over 12. We know that our K value is seven over 12. So when we calculate our a value, we get 680 minus 18 over 18 and we get around 36.78 So you get P. F T is equal to 6 80 all over one plus 36.78 he to the negative, 70 over 12 for party. We're going to use this table that I drew here, and when we make a graph reflecting it, we have our exponential moto that looks like this, and then our logistic model will look something like this. And when we compare it to part A this our graph, we see that the logistic model is a better fit for part E were using the equation we found in part. See? And we're finding p of tea when a T is equal to seven. So we just plug in seven to this equation when we get about 480 east cells.