Question

39An amusement park issued a coupon to increase the number of visitors to the park each week. The function below representsthe number of visitors at the amusement park x weeks after the issuance of the couponVx) = 500(1.5)What is the approximate average rate of change over the interval [2,6]?OA 949 visitors per weekB 281 visitors per weekC1,143 visitors per weekD. 762 visitors per weekResetSubmitCrved12-39

Answer 1

The Solution.

Given the exponential function below:

`V(x)=500(1.5)^x`

The average rate of change over the interval [2,6] is given as below:

`\text{Average rate of change =}(V(6)-V(2))/(6-2)`

To find V(6):

`V(6)=500(1.5)^6=500*11.3906=5695.313`

To find V(2):

`V(2)=500(1.5)^2=500*2.25=1125`

So, substituting for the values of V(6) and V(2) in the above formula, we get

`\begin{gathered} \text{Average rate of change over \lbrack{}2,6\rbrack =}(5695.313-1125)/(6-2) \\ \\ \text{ = }(4570313)/(4)=1142.578\approx1143\text{ visitors per week} \end{gathered}`

Thus, the correct answer is 1143 visitors p