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9966 [12]
3 years ago
14

Caterpillar nitrogen assimilation and body mass. we explored the relationship between nitrogen assimilation and body mass (both

on log scales) for data on a sample of caterpillars in Caterpillars. The Instar variable in the data codes different stages (1 to 5) of caterpillar development.
a. Fit a model to predict log nitrogen assimilation (LogNassim) using log body mass (LogMass). Report the value of for this model.
b. Fit a model to predict LogNassim using appropriate indicators for the categories of Instar. Report the value of for this model and compare it to the model based on LogMass.
c. Give an interpretation (in context) for the first two coefficients of the fitted model in (b).
d. Fit a model to predict LogNassim using LogMass and appropriate indicators for Instar. Report the value of and compare it to the earlier models.
e. Is the LogMass variable really needed in the model of part (d). Does the linear trend appear to be better when the caterpillars are in a free growth period? (Again, we are not ready to fit more complicated models, but we are looking at the plot for linear trend in the two groups.)
Mathematics
1 answer:
Lady_Fox [76]3 years ago
3 0
A. Produce a scatterplot for predicting nitrogen assimilation (Nassim) based on Mass. Comment on any patterns.

b B.Produce a similar plot using the log (base 10) transformed variables, LogNassim versus Log Mass. Again, comment on any patterns.

C. Would you prefer the plot in part (a) or part (b) to predict the nitrogen assimilation of caterpillars with a linear model? Fit a linear regression model for the plot you chose and write down the prediction equation.
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3 years ago
Which expression is equivalent to 18x + 45
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Transform the given function f(x) as described.
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2 years ago
On Novermber 27, 1993, the New York times reported that wildlife biologists have found a direct ink between the increase in the
Korolek [52]

Answer:

a.  1 263 888

b. 130 701

c. 72 years

Step-by-step explanation:

a. The differential equation applies here.

Let the quantity increase for a certain time be given by Q(t)

Every unity of time, the quantity increases by 1+\frac{r}{100} so that after the time t, the quantity remaining will be given by:

Q(t) = (1+ \frac{r}{100} )^{t}

In a similar manner, the quantity R(t) decreases at a rate given by the following expression:

1-\frac{r}{100} and after the time , t the quantity of R remaining will be given by:

R(t) = (1-\frac{r}{100} )^{t}

a. To find the population of humans in 1953

Q(t) = (1+ \frac{r}{100} )^{t}

1993 - 1953 = 40 years = t

Q(40) = Q×1.06^{40}

Q = 1 263 888.44

    ≈ 1 263 888

b. For bear population in 1993:

R(t) = (1-\frac{r}{100} )^{t}

t = 40

R(40) = b 0.94^{40} = 11 000

b = 130 700. 889

    ≈130 701

c. time taken for black bear population number less than 100 is given by:

130 = 11000×0.94^{t}

solving using natural logarithms gives t = 72.72666

                                                                  =   72 years Ans

8 0
3 years ago
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