Chapter 6.5, Problem 50E

a.

To determine

To find: When the events will have about the same attendance.

Answer to Problem 50E

In 2007 they have same attendance.

Explanation of Solution

Given information: It is given with graph for the two equations.

  $\begin{array}{}$ y=4000{(1.25)}^x$$ y=12000{(0.87)}^x$\end{array}$

  

Calculation:

Read the graph where they intersect.

Approximately when x =3, they meet.

It means in 2007 they have same attendance. Because 0 represents 2004.

Conclusion: In 2007 they have same attendance.

b.

To determine

To Explain: About to verify the answer for part a.

Explanation of Solution

Given information: It is given two equations.

  $\begin{array}{}$ y=4000{(1.25)}^x$$ y=12000{(0.87)}^x$\end{array}$

Calculation:

Verify by plugging in x as 3 into the equations.

When $x=3$

  $y=4000{(1.25)}^3$

  $y=7812.5$

Now, in second equation,

  $y=12000{(0.87)}^3$

  $y=7902.036$

They are approximately same. So, they are verified using equations.