Question

The total number of public electric charging units available for hybrid vehicles has increased exponentially since 2005. The number of outlets​ A(t) at these alternative fueling stations t years after 2005 can be approximated by A(t)=233(2.08)t​, where t=0 corresponds to 2005. How many outlets were available in 2007​, in 2009​, in 2010​?

Answer 1

a.

The number of outlets​ in 2007 are 1008 outlets

Since  the number of outlets​ A(t) at these alternative fueling stations t years after 2005 is $A(t) = 233(2.08)^{t}$ with t = 0 corresponding to 2005.

To find the number of outlets in 2007, we substitute t = 2 into the equation.

So,

$A(t) = 233(2.08)^{t}\\A(2) = 233(2.08)^{2}\\A(2) = 233(4.3264)\\A(2) =1008.0512$

A(2) ≅ 1008 outlets (we round down to the nearest whole number since we can only have a whole number of outlets).

So, the number of outlets​ in 2007 are 1008 outlets

b.

The number of outlets in 2009 are 4361 outlets

To find the number of outlets in 2009, we substitute t = 4 into the equation.

So,

$A(t) = 233(2.08)^{t}\\A(4) = 233(2.08)^{4}\\A(4) = 233(18.72)\\A(4) =4361.23$

A(4) ≅ 4361 outlets (we round down to the nearest whole number since we can only have a whole number of outlets).

So, the number of outlets in 2009 are 4361 outlets

c.

The number of outlets in 2010 are 9071 outlets

To find the number of outlets in 2010, we substitute t = 5 into the equation.

So,

$A(t) = 233(2.08)^{t}\\A(5) = 233(2.08)^{5}\\A(5) = 233(38.93)\\A(5) = 9071.36$

A(5) ≅ 9071 outlets (we round down to the nearest whole number since we can only have a whole number of outlets).

So, the number of outlets in 2010 are 9071 outlets

Learn more about exponential increase here:

https://brainly.com/question/18261561