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Karo-lina-s [1.5K]
4 years ago
10

What is the exponential regression equation that fits these data

Mathematics
2 answers:
s2008m [1.1K]4 years ago
7 0

Answer:

Correct choice is B.

Step-by-step explanation:

Plot the data from the given table on the coordinate plane.

Note that option C and D represent the quadratic and linear regressions (not exponential), thus, these options are false.

Plot both graphs from options A and B. The exponential regression from option A is shown by red curve (see attached diagram) and the the exponential regression from option B is shown by blue curve. As you can see the green points better suit blue curve, thus, option B is correct.

<u>Remark:</u>

Note that

\dfrac{570}{240}, \dfrac{240}{85}, \dfrac{85}{27}, \dfrac{27}{8}, \dfrac{8}{3} are greater than 1.03, then option A is false.

weqwewe [10]4 years ago
6 0
C and D are not exponential functions, so eliminate C and D.
Between A and B, B is the better fit. 
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Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

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                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

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                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

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