Chapter 9.1, Problem 98E

To determine

To write:The terms of the sequence $a_n=-0.0126n^3+0.391n^2-4.21n+48.5$

Answer to Problem 98E

  $-\frac{51}{32}$

Explanation of Solution

Given:The numbers (in thousands) of AIDS cases reported from 2003 though 2019 can be approximated by $a_n=-0.0126n^3+0.391n^2-4.21n+48.5,n=3,4,\mathrm{....},10$ where $n$ is the year, with $n=3$ corresponding to 2003.

First write the finite term of the sequence.

For $n=3,4,5,6,\mathrm{....10}$

  $\begin{array}{l}{a}_3=-0.0126{\left(3\right)}^3+0.391{\left(3\right)}^2-4.21\left(3\right)+48.5=39.049\\ a_4=-0.0126{\left(4\right)}^3+0.391{\left(4\right)}^2-4.21\left(4\right)+48.5=37.109\\ a_5=-0.0126{\left(5\right)}^3+0.391{\left(5\right)}^2-4.21\left(5\right)+48.5=35.65\\ a_6=-0.0126{\left(6\right)}^3+0.391{\left(6\right)}^2-4.21\left(6\right)+48.5=34.594\\ a_7=-0.0126{\left(7\right)}^3+0.391{\left(7\right)}^2-4.21\left(7\right)+48.5=33.867\\ a_8=-0.0126{\left(8\right)}^3+0.391{\left(8\right)}^2-4.21\left(8\right)+48.5=33.393\\ a_9=-0.0126{\left(9\right)}^3+0.391{\left(9\right)}^2-4.21\left(9\right)+48.5=33.096\\ a_{10}=-0.0126{\left(10\right)}^3+0.391{\left(10\right)}^2-4.21\left(10\right)+48.5=32.9\end{array}$

Now, we plot the points on graph and make histogram.

  

To determine

The number of casesreported in 7 years using graph.

Answer to Problem 98E

Decreases

Explanation of Solution

Given: The numbers (in thousands) of AIDS cases reported from 2003 though 2019 can be approximated by $a_n=-0.0126n^3+0.391n^2-4.21n+48.5,n=3,4,\mathrm{....},10$ where $n$ is the year, with $n=3$ corresponding to 2003.

The number of cased decreased from 2003 to 2010. As the height of histogram in graph decreases.

Using graph, we can say the number of cases of AIDS decreases since 2003.