The correct value of <em>a </em>or initial amount in exponential growth
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 <em>a</em> and <em>h</em>
![Q = ae^b^x](https://tex.z-dn.net/?f=Q%20%3D%20ae%5Eb%5Ex)
![ln Q= ln a+bx](https://tex.z-dn.net/?f=ln%20Q%3D%20ln%20a%2Bbx)
For the year <em>1990:</em>
![lnQ =0](https://tex.z-dn.net/?f=lnQ%20%3D0)
For the year <em>1910:</em>
![lnQ= 0.698](https://tex.z-dn.net/?f=lnQ%3D%200.698)
For the year <em>1920:</em>
![lnQ= 1.4012](https://tex.z-dn.net/?f=lnQ%3D%201.4012)
For the year <em>1930:</em>
![lnQ=2.1](https://tex.z-dn.net/?f=lnQ%3D2.1)
For the year <em>1940:</em>
![lnQ=2.8](https://tex.z-dn.net/?f=lnQ%3D2.8)
For the year <em>1950:</em>
![lnQ=3.5](https://tex.z-dn.net/?f=lnQ%3D3.5)
For the year <em>1960:</em>
![lnQ=4.2](https://tex.z-dn.net/?f=lnQ%3D4.2)
For the year <em>1970:</em>
![lnQ=4.9](https://tex.z-dn.net/?f=lnQ%3D4.9)
For the year <em>1980:</em>
![lnQ=5.6](https://tex.z-dn.net/?f=lnQ%3D5.6)
For the year <em>1990:</em>
![lnQ=6.3](https://tex.z-dn.net/?f=lnQ%3D6.3)
For the year <em>2000:</em>
![lnQ= 7](https://tex.z-dn.net/?f=lnQ%3D%207)
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}](https://tex.z-dn.net/?f=%5Cfrac%7B1096.63-544.57%7D%7B7-6.3%7D)
a = ![\frac{552.06}{0.7}](https://tex.z-dn.net/?f=%5Cfrac%7B552.06%7D%7B0.7%7D)
a = 788.657
a ≈ 789
Hence, <em>a </em>or initial amount in exponential growth
is 789.
To know more about 'exponential growth' related questions
visit- brainly.com/question/12490064
#SPJ4