Question

The following data represent (hypothetical) energy consumption normalized to the year 1900. Plot the data. Test the model Q = ae^bx by plotting the transformed data. Estimate the parameters of the model graphically. x
0 10 20 30 40 50 60 70 80 90 100 Year
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 Consumption Q
1.00 2.01 4.06 8.17 16.44 33.12 66.69 134.29 270.43 544.57 1096.63 Make an appropriate trunsformation to fit the model P aeh using Equation (3.4). Estimate a and h

Answer 1

The correct value of a or initial amount in exponential growth $Q = ae^b^x$ is 789.

Quantity rises over time through a process called exponential growth and it happens when the derivative, or instantaneous rate of change, of a quantity with respect to time is proportionate to the original quantity.

Given that, hypothetical energy consumption normalized to the year 1990 we have to estimate a and h

$Q = ae^b^x$

$ln Q= ln a+bx$

For the year 1990:
$x = 0$     $lnQ =0$

For the year 1910:

$x= 10$    $lnQ= 0.698$

For the year 1920:

$x=20$    $lnQ= 1.4012$

For the year 1930:

$x=30$     $lnQ=2.1$

For the year 1940:

$x=40$     $lnQ=2.8$

For the year 1950:

$x=50$     $lnQ=3.5$

For the year 1960:

$x=60$     $lnQ=4.2$

For the year 1970:

$x=70$     $lnQ=4.9$

For the year 1980:

$x=80$     $lnQ=5.6$

For the year 1990:

$x=90$     $lnQ=6.3$

For the year 2000:

$x= 100$     $lnQ= 7$

Here, estimate the parameters of the model graphically, the slope of line is approximated as follows:

a = $\frac{1096.63-544.57}{7-6.3}$

a = $\frac{552.06}{0.7}$

a = 788.657

a ≈ 789

Hence, a or initial amount in exponential growth $Q = ae^b^x$ is 789.

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