Question

1) Please write clearly.

The total number of public electric charging units available for hybrid vehicles has increased exponentially since 2006. The number of outlets A(t) at these alternative fueling stations t years after 2006 can be approximated by A(t) = 233(2.01)", where t = 0 corresponds to 2006. How many outlets were available in 2008, in 2010, in 2011? The number of outlets available in the year 2008 was approximately (Simplify your answer. Round to the nearest integer as needed.) The number of outlets available in the year 2010 was approximately (Simplify your answer. Round to the nearest integer as needed.) The number of outlets available in the year 2011 was approximately (Simplify your answer. Round to the nearest integer as needed.)

Answer 1

Given that, the total number of public electric charging units available for hybrid vehicles has increased exponentially since '2006'. Since it is increasing exponentially, the function should be in the exponential form i.e., "f(x) = a.b^x ".

The number of outlets 'A(t)' at these alternative fueling stations 't' years after '2006' can be approximated by $"\underline{A(t) = 233(2.01)^t}"$ , where 't = 0' corresponds to 2006.

That means, 't' value increases by '1' for each successive year of 2006. In other words, 't' is simply the difference between the required year we want to calculate and 2006.

We have to determine the number of outlets available in the years 2008, 2010 & 2011.

In order to do so, we have to find 't' value for each year i.e., the difference between that year and 2006 and then substitute it in the given exponential model.

i) In the year 2008:

The year 2008 means, it is '2' years after 2006. So, 't = 2' ($\because$2008 - 2006 = 2).

Now, put 't = 2' in the given model to calculate the available number of outlets.

Alt) = 2332.01)

  $\Rightarrow A(t) = 233\;(2.01)^2$

At) = 233 x (4.0401

  $\Rightarrow \mathbf{A(t) = 941.3433 \simeq \underline{941}}$

ii) In the year 2010:

The year 2010 means, it is '4' years after 2006. So, 't = 4' ($\because$2010 - 2006 = 4).

Now, put 't = 4' in the given model to calculate the available number of outlets.

Alt) = 2332.01)

  $\Rightarrow A(t) = 233\;(2.01)^4$

  $\Rightarrow A(t) = 233\times (16.32241)$

  $\Rightarrow \mathbf{A(t) = 3803.1211 \simeq \underline{3803}}$

iii) In the year 2011:

The year 2011 means, it is '5' years after 2006. So, 't = 5' ($\because$2011 - 2006 = 5).

Now, put 't = 5' in the given model to calculate the available number of outlets.

Alt) = 2332.01)

  $\Rightarrow A(t) = 233\;(2.01)^5$

  $\Rightarrow A(t) = 233\times (32.80804)$

  $\Rightarrow \mathbf{A(t) = 7644.2733 \simeq \underline{7644}}$

Therefore,

The number of outlets available in the year 2008 was approximately $\mathbf{\underline{"941"}}$ .

The number of outlets available in the year 2010 was approximately $\mathbf{\underline{"3803"}}$ .

The number of outlets available in the year 2011 was approximately $\mathbf{\underline{"7644"}}$ .