Chapter 4, Problem 4P

a.

To determine

Make a scatter plot of the data.

Answer to Problem 4P

  

Explanation of Solution

Given information:

A student is trying to determine the half-life of radioactive iodine- $131$ . He measures the amount of iodine- $131$ in a sample solution every $8$ hour. His data are shown in the table in the margin.

  

Make a scatter plot of the data.

Calculation:

Consider the following data,

  

Draw a scatter diagram by using $TI-84$ plus.

Press $STAT$ and $ENTER$ to enter the stat edit mode.

  

Now enter the data in $L_1$ press $\boxed{⊳}$ to get into $L_2$ then enter the values.

The display is as shown.

  

Now press $2nd\boxed{Y=}$ and set plot on,

  

Press $zoom9$ to view the scatter plot,

  

Hence, the solution is scatter plot.

b.

To determine

To find an exponential model.

Answer to Problem 4P

  $f\left(x\right)=4.792{\left(0.996\right)}^x$

Explanation of Solution

Given information:

A student is trying to determine the half-life of radioactive iodine- $131$ . He measures the amount of iodine- $131$ in a sample solution every $8$ hour. His data are shown in the table in the margin.

  

Use a calculator to find an exponential model.

Calculation:

To find an exponential model.

Draw a scatter diagram by using $TI-84$ plus.

Press $STAT$ key and choose CALC zero.

  

Now press enter to produce the regression results.

The display is as shown.

  

The required exponential model is, $f\left(x\right)=4.792{\left(0.996\right)}^x$

Hence, the solution is $f\left(x\right)=4.792{\left(0.996\right)}^x$

c.

To determine

To find the half-life of iodine- $131$ .

Answer to Problem 4P

  $\boxed{4.71}$

Explanation of Solution

Given information:

A student is trying to determine the half-life of radioactive iodine- $131$ . He measures the amount of iodine- $131$ in a sample solution every $8$ hour. His data are shown in the table in the margin.

  

Use your model to find the half-life of iodine- $131$ .

Calculation:

To find the half-life of iodine- $131$ .

Put $x=4$ in expression $f\left(x\right)=4.792{\left(0.996\right)}^x$ .

  $\begin{array}{l}f\left(4\right)=4.792{\left(0.996\right)}^4\\ f\left(4\right)\approx 4.7157\end{array}$

Hence, the solution is $\boxed{4.71}$