91Ó°ÊÓ

Given data:

$\begin{array}{|llll|}\hline \text{Speed (mph)}& \text{Distance (ft)}& \text{Speed (mph)}& \text{Distance (ft)}\\ 6& 1.42& 32& 52.08\\ 9& 4.92& 40& 84\\ 19& 18& 48& 110.33\\ 30& 44.75& & \\ \hline\end{array}$

Log of given data:$\begin{array}{|llll|}\hline \text{Speed (mph)}& \text{Distance (ft)}& \log (\text{speed)}& \log (\text{distance)}\\ 6& 1.42& 0.7781513& 0.15228834\\ 9& 4.92& 0.9542425& 0.6919651\\ 19& 18& 1.2787536& 1.25527251\\ 30& 44.75& 1.4771213& 1.65079304\\ 32& 52.08& 1.50515& 1.71667098\\ 40& 84& 1.60206& 1.92427929\\ 48& 110.33& 1.6812412& 2.04269362\\ \hline\end{array}$

Making use of a Ti83/84 calculator

Step 1: Press STAT,

Step 2: select 1: EDIT.

Step 3: in list $L1$, type the data for logarithmic speed, and in list $L2$, type the data for logarithmic distance.

Step 4: hit STAT once more, select CALC, and then $LinReg(a+bx)$.

The result:

$y=a+bx$

$a=-1.3506$

$b=2.0361$

Substituting the value in $a$and $b$

$y=-1.3506+2.0361x$

The logarithm of speed is $x$, whereas the logarithm of distance is $y$.

localid="1654264833872" $\log y=-1.3506+2.0361\log x$

$x$denotes the speed and $y$denotes the distance.

Given data:

$\begin{array}{|llll|}\hline \text{Speed (mph)}& \text{Distance (ft)}& \text{Speed (mph)}& \text{Distance (ft)}\\ 6& 1.42& 32& 52.08\\ 9& 4.92& 40& 84\\ 19& 18& 48& 110.33\\ 30& 44.75& & \\ \hline\end{array}$

Substituting the value $x$by $48$:

$\log y=-1.3506+2.0361\log x$

$\log y=-1.3506+2.0361\log 48$

$\log y=2.0726$

Using the exponential function with a base of ten:

$y={10}^{\log y}$

$={10}^{2.0726}$

$=118.195$