V(t), left parenthesis, t, right parenthesis models the number of visitors in a park as a function of the outside temperature t (in degrees Celsius).

Look at picture posted below.
t 18 20 37
------------------------
v(t) 10 18 46

When does the number of visitors in the park increase faster?
Choose 1 answer:
(A). Between outside temperatures of 181818 degrees Celsius and 202020 degrees Celsius
(B). Between outside temperatures of 202020 degrees Celsius and 272727 degrees Celsius
(C).The number of visitors increases at the same rate over both intervals

Answer:

The number of visitors increases at the same rate over both intervals

Step-by-step explanation:

The unit rate at which the number of visitors in the park increases over a given temperature interval is called the average rate of change, or ARCARCA, R, C.

To find the average rate of change of a function over an interval, we need to take the total change in the function value over the interval and divide it by the length of the interval.

Hint #22 / 3

We are asked to compare the rates at which the number of visitors increases over the interval between an outside temperature of 181818 degrees Celsius and 202020 degrees Celsius, and over the interval between an outside temperature of 202020 degrees Celsius and 272727 degrees Celsius. These correspond to the domain intervals [18,20][18,20]open bracket, 18, comma, 20, close bracket and [20,27][20,27]open bracket, 20, comma, 27, close bracket.

Let's calculate the average rate of change of VVV over those intervals:

ARC_{[18,20]}ARC

[18,20]

​

A, R, C, start subscript, open bracket, 18, comma, 20, close bracket, end subscript ARC_{[20,27]}ARC

[20,27]

​

A, R, C, start subscript, open bracket, 20, comma, 27, close bracket, end subscript

\begin{aligned} \dfrac{V(20)-V(18)}{20-18}&=\dfrac{18-10}{2}\\\\&=\dfrac{8}{2}\\\\&=4\end{aligned}\quad

20−18

V(20)−V(18)

​

​

=

2

18−10

​

=

2

8

​

=4

​

 \begin{aligned} \dfrac{V(27)-V(20)}{27-20}&=\dfrac{46-18}{7}\\\\&=\dfrac{28}{7}\\\\&=4\end{aligned}

27−20

V(27)−V(20)

​

​

=

7

46−18

​

=

7

28

​

=4

​

Hint #33 / 3

The average rate of change over the interval [18,20][18,20]open bracket, 18, comma, 20, close bracket is the same as the average rate of change over the interval [20,27][20,27]open bracket, 20, comma, 27, close bracket.

Therefore, the number of visitors increases at the same rate over both intervals.

Using the average rate of change, it is found that the correct option is:

(A) Between outside temperatures of 18 degrees Celsius and 20 degrees Celsius.

The rate of change of a function is given by the change in the output divided by the change in the input.

When the temperature changes from 18 to 20 degrees, the number of visitors increase from 10 to 18, thus, the rate of change is:

[tex]r = \frac{18 - 10}{20 - 18} = \frac{8}{2} = 4[/tex]

When the temperature changes from 20 to 37 degrees, the number of visitors increase from 18 to 46, thus, the rate of change is:

[tex]r = \frac{46 - 18}{37 - 20} = \frac{28}{17} = 1.65[/tex]

First case has higher rate of change, thus, option A is correct.

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