Chapter 1.5, Problem 34E

a.

To determine

To find: equation of slope-intercept form for the lines that represent each value change.

Answer to Problem 34E

  $\begin{array}{c}y=3.89x+1880.85\\ y=33.67x+1404.37\\ y=-14.32x+2220.20\\ y=-21.97x+2357.90\end{array}$

Explanation of Solution

Given information:

The given information is as follows:

Stock Index March, 2004
Day ( $x$)	Closing value ( $y$)
15	1939.20
16	1943.09
17	1976.76
18	1962.44
19	1940.47

Calculation:

Let the point coordinates be as follows:

  $\begin{array}{l}A\left(15,1939.20\right)\\ B\left(16,1943.09\right)\\ C\left(17,1976.76\right)\\ D\left(18,1962.44\right)\\ E\left(19,1940.47\right)\end{array}$

The four lines representing each value change will be $\overline{AB,}\overline{BC},\overline{CD},$ and $\overline{DE}$ .

The formula used for solving this problem is as follows:

  $m=\frac{y_2-y_1}{x_2-x_1}\left[\text{Slope equation}\right]$  ...(1)

  $y=mx+b\left[\text{Slope intercept form}\right]$  ...(2)

Now, put the values $\left(15,1939.20\right)and\left(16,1943.09\right)$ in equation (1) and evaluate the slope of $\overline{AB}$ as shown below.

  $\begin{array}{c}{m}_{\overline{AB}}=\frac{1943.09-1939.20}{16-15}\\ =\frac{3.89}{1}\\ =3.89\end{array}$

Put the slope value of $\overline{AB}$ and $\left(15,1939.20\right)$ in equation (2) and evaluate b for $\overline{AB}$ as shown below.

  $\begin{array}{c}1939.20=3.89\left(15\right)+b\\ 1939.20=58.35+b\\ b=1939.20-58.35\\ =1880.85\end{array}$

Substitute the slope value and b for $\overline{AB}$ in equation (2) to get the slope intercept equation as follows:

  $y=3.89x+1880.85$

Now, put the values $\left(16,1943.09\right)\text{ and }\left(17,1976.76\right)$ in equation (1) and evaluate the slope of $\overline{BC}$ as shown below.

  $\begin{array}{c}{m}_{\overline{BC}}=\frac{1976.76-1943.09}{17-16}\\ =\frac{33.67}{1}\\ =33.67\end{array}$

Put the slope value of $\overline{BC}$ and $\left(16,1943.09\right)$ in equation (2) and evaluate b for $\overline{BC}$ as shown below.

  $\begin{array}{c}1943.09=33.67\left(16\right)+b\\ 1943.09=538.72+b\\ b=1943.09-538.72\\ =1404.37\end{array}$

Substitute the slope value and b for $\overline{BC}$ in equation (2) to get the slope intercept equation as follows:

  $y=33.67x+1404.37$

Now, put the values $\left(17,1976.76\right)\text{ and }\left(18,1962.44\right)$ in equation (1) and evaluate the slope of $\overline{CD}$ as shown below.

  $\begin{array}{c}{m}_{\overline{CD}}=\frac{1962.44-1976.76}{18-17}\\ =\frac{33.67}{1}\\ =-14.32\end{array}$

Put the slope value of $\overline{CD}$ and $\left(17,1976.76\right)$ in equation (2) and evaluate b for $\overline{CD}$ as shown below.

  $\begin{array}{c}1976.76=-14.32\left(17\right)+b\\ 1976.76=-243.44+b\\ b=1976.76+243.44\\ =2220.20\end{array}$

Substitute the slope value and b for $\overline{CD}$ in equation (2) to get the slope intercept equation as follows:

  $y=-14.32x+2220.20$

Now, put the values $\left(18,1962.44\right)\text{ and }\left(19,1940.47\right)$ in equation (1) and evaluate the slope of $\overline{DE}$ as shown below.

  $\begin{array}{c}{m}_{\overline{DE}}=\frac{1940.47-1962.44}{19-18}\\ =\frac{21.97}{1}\\ =-21.97\end{array}$

Put the slope value of $\overline{DE}$ and $\left(18,1962.44\right)$ in equation (2) and evaluate b for $\overline{DE}$ as shown below.

  $\begin{array}{c}1962.44=-21.97\left(18\right)+b\\ 1962.44=-395.46+b\\ b=1962.44+395.46\\ =2357.90\end{array}$

Substitute the slope value and b for $\overline{DE}$ in equation (2) to get the slope intercept equation as follows:

  $y=-21.97x+2357.90$

Thus, the equation for each value change is as follows:

  $\begin{array}{c}y=3.89x+1880.85\\ y=33.67x+1404.37\\ y=-14.32x+2220.20\\ y=-21.97x+2357.90\end{array}$

b.

To determine

To find: the term indicating the same rate change for two pair of days, and also find whether the rate change is same for any of the given days.

Answer to Problem 34E

Slope indicates the same the rate of change for the two pair of days.

The rate of change was not the same for any day shown because no two slopes in the equation of value change are the same.

Explanation of Solution

Given information:

The given information is as follows:

Stock Index March, 2004
Day ( $x$)	Closing value ( $y$)
15	1939.20
16	1943.09
17	1976.76
18	1962.44
19	1940.47

Calculation:

The equation for each value change evaluated in problem a. is as follows:

  $\begin{array}{c}y=3.89x+1880.85\\ y=33.67x+1404.37\\ y=-14.32x+2220.20\\ y=-21.97x+2357.90\end{array}$

The slope would indicate whether or not the rate of change for two pairs of days was the same.

Here, the slope is the same as the rate of change.

Since, no two slopes in the equation of value change are the same. Therefore, the rate of change was not the same for any day shown.

c.

To determine

To predict: the closing value for Day 22, and also explain whether any equation correctly predict this value.

Answer to Problem 34E

The predicted closing values for day 22 are as follows:

  $\begin{array}{l}y=1966.43\\ y=2145.11\\ y=1905.16\\ y=1874.56\end{array}$

Since the actual value of y i.e. 1909.90 do not match with anyone of the predicted value So, no equation matches the value accurately.

Explanation of Solution

Given information:

The given information is as follows:

Stock Index March, 2004
Day ( $x$)	Closing value ( $y$)
15	1939.20
16	1943.09
17	1976.76
18	1962.44
19	1940.47

Actual closing value ( y ) = 1909.90

Given day ( x ) = 22

The equations are as follows:

  $\begin{array}{c}y=3.89x+1880.85\mathrm{...}\left(1\right)\\ y=33.67x+1404.37\mathrm{...}\left(2\right)\\ y=-14.32x+2220.20\mathrm{...}\left(3\right)\\ y=-21.97x+2357.90\mathrm{...}\left(4\right)\end{array}$

Calculation:

Put the value of x in equation (1) and evaluate the value of y as shown below.

  $\begin{array}{c}y=3.89\left(22\right)+1880.85\\ =85.58+1880.85\\ =1966.43\end{array}$

Put the value of x in equation (2) and evaluate the value of y as shown below.

  $\begin{array}{c}y=33.67\left(22\right)+1404.37\\ =740.74+1404.37\\ =2145.11\end{array}$

Put the value of x in equation (3) and evaluate the value of y as shown below.

  $\begin{array}{c}y=-14.32\left(22\right)+2220.20\\ =-315.04+2220.20\\ =1905.16\end{array}$

Put the value of x in equation (4) and evaluate the value of y as shown below.

  $\begin{array}{c}y=-21.97\left(22\right)+2357.90\\ =-483.34+2357.90\\ =1874.56\end{array}$

Therefore, the predicted closing values for day 22 are as follows:

  $\begin{array}{l}y=1966.43\\ y=2145.11\\ y=1905.16\\ y=1874.56\end{array}$

Since the actual value of y i.e. 1909.90 do not match with anyone of the predicted value but it come closer to (3) equation. By looking on the data, we can say that none of the equation matches the value accurately.