Chapter 9.4, Problem 75E

(a)

To determine

To Find: The first difference of the data shown in Table 1 then find the linear model that approximates the data.

Answer to Problem 75E

The required linear model is $a_n=15n+4636$ .

Explanation of Solution

Given:

The value $n=10$ corresponds to 2010.

The given table is shown in table 1

Table 1

Year	Number of Residents   $a_g$
2010	4785
2011	4801
2012	4816
2013	4831
2014	4846
2015	4859

Calculation:

Consider the difference from the given table as,

  $\begin{array}{l}4801-478=16\\ 4816-4801=15\\ 4831-4816=15\\ 4846-4831=15\end{array}$

Consider the points $\left(11,4801\right),\left(14,4846\right)$ . Then, the slope by the point slope form is obtained as,

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ a_n-4801=\frac{4846-4801}{14-11}\left(n-11\right)\\ a_n=15n+4636\end{array}$

(b)

To determine

To Find: The graphing utility to find the number of linear model for the data and compare the model of part (a),

Answer to Problem 75E

The linear model is $a_n=14.9n+4637$ and it is similar to the model of part (a).

Explanation of Solution

Input the n values on L1 and $a_n$ on L2 as shown in Figure 1

  

Figure 1

From the above results the linear model is $a_n=14.9n+4637$ .

(c)

To determine

To Find: The prediction for the number of residents in 2021 and determine how the values compare.

Answer to Problem 75E

The number of residents is $4951$ and the results are similar for both part (a) and part (c).

Explanation of Solution

From part (b) use $n=21$ in the linear model as,

  $\begin{array}{c}{a}_{21}=14.9\left(21\right)+4637\\ =4949.9\end{array}$

From part (b) use $n=21$ in the linear model as,

  $\begin{array}{c}{a}_{21}=15\left(21\right)+4636\\ =4951\end{array}$

Thus, the values are similar of part (a) and part (b).