Points: 0.33 of 1
The total number of public electric charging units available for hybrid vehicles has increase 2005. The number of outlets $A (t)$ at these altemative fueling stations $t$ years after 2005 can $A (t) = 224 (2.79)^{t}$ , where $t = 0$ corresponds to 2005 . How many outlets were available in 200

The number of outlets available in the year 2007 was approximately 1744 . (Simplify your answer. Round to the nearest integer as needed.)

The number of outlets available in the year 2009 was approximately 13573 (Simplify your answer. Round to the nearest integer as needed.)

The number of outlets available in the year 2010 was 4 pproximately $◻$ . (Simplify your answer. Round to the nearest integer as needed.)

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So, the number of outlets available in the year 2010 was approximately $37868$ .

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Step 1 ：The problem provides the function $A (t) = 224 (2.79)^{t}$ to calculate the number of outlets available for hybrid vehicles at alternative fueling stations $t$ years after 2005.

Step 2 ：We are asked to find the number of outlets available in the year 2010. To do this, we need to substitute $t = 5$ into the function, because 2010 is 5 years after 2005.

Step 3 ：Substituting $t = 5$ into the function gives us $A (5) = 224 (2.79)^{5}$ .

Step 4 ：Solving this expression gives us $A (5) \approx 37868$ .

Step 5 ：So, the number of outlets available in the year 2010 was approximately $37868$ .

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