Chapter 11.7, Problem 6CFU

(a)

To determine

To find:

A regression equation for the amount $y$ of lead as a function of time $x$ .

Answer to Problem 6CFU

  $y=\left(10.0170\right){\left(0.9703\right)}^x$

Explanation of Solution

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $y=a\times b^x$

Calculation:

The table is

Time (min)	$0$	$5$	$10$	$15$	$20$
U- $239$ percent(g)	$10$	$8.6$	$7.5$	$6.3$	$5.5$

  $y=a\times b^x\mathrm{....................}\left(1\right)$

  $\begin{array}{l}a=10.0170\\ b=0.9703\end{array}$

From equation (1)

  $\begin{array}{l}y=a\times b^x\\ y=\left(10.0170\right){\left(0.9703\right)}^x\end{array}$

(b)

To determine

To find:

The equation in terms of base $e$ .

Answer to Problem 6CFU

  $N=10.0170\left(e^{-0.0301}\right)x$

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{kt}$

Calculation:

The table is

Time (min)	$0$	$5$	$10$	$15$	$20$
U- $239$ percent(g)	$10$	$8.6$	$7.5$	$6.3$	$5.5$

  $y=a\times b^x\mathrm{....................}\left(1\right)$

  $\begin{array}{l}a=10.0170\\ b=0.9703\end{array}$

From equation (1)

  $\begin{array}{l}y=a\times b^x\\ y=\left(10.0170\right){\left(0.9703\right)}^x\end{array}$

  $\begin{array}{l}N=N_0{e}^{kt}\\ N=10.0170{\left(e^{\ln 0.9703}\right)}^x\\ N=10.0170\left(e^{\ln 0.9703}\right)x\\ N=10.0170\left(e^{-0.0301}\right)x\end{array}$

(c)

To find:

The time required for an amount is half-life at the given model.

  $x\approx 23.08$

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{rt}$

Calculation:

The table is

Time (min)	$0$	$5$	$10$	$15$	$20$
U- $239$ percent(g)	$10$	$8.6$	$7.5$	$6.3$	$5.5$

  $N=N_0{e}^{rt}\mathrm{....................}\left(1\right)$

The time needed to quadruple an investment replace $N$ by $\frac{1}{2}{N}_0$ if

From equation (1)

  $\begin{array}{l}\frac{1}{2}{N}_0=N_0{e}^{rt}\\ 5\approx 10.0170e^{-0.0301x}\\ -0.0301x\approx \ln \frac{5}{10.0170}\\ x=\frac{\ln \frac{5}{10.0170}}{-0.0301}\\ x\approx 23.08\end{array}$

Explanation of Solution

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{kt}$

Calculation:

The table is

Time (min)	$0$	$5$	$10$	$15$	$20$
U- $239$ percent(g)	$10$	$8.6$	$7.5$	$6.3$	$5.5$

  $y=a\times b^x\mathrm{....................}\left(1\right)$

  $\begin{array}{l}a=10.0170\\ b=0.9703\end{array}$

From equation (1)

  $\begin{array}{l}y=a\times b^x\\ y=\left(10.0170\right){\left(0.9703\right)}^x\end{array}$

  $\begin{array}{l}N=N_0{e}^{kt}\\ N=10.0170{\left(e^{\ln 0.9703}\right)}^x\\ N=10.0170\left(e^{\ln 0.9703}\right)x\\ N=10.0170\left(e^{-0.0301}\right)x\end{array}$

(c)

To determine

To find:

The time required for an amount is half-life at the given model.

Answer to Problem 6CFU

  $x\approx 23.08$

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{rt}$

Calculation:

The table is

Time (min)	$0$	$5$	$10$	$15$	$20$
U- $239$ percent(g)	$10$	$8.6$	$7.5$	$6.3$	$5.5$

  $N=N_0{e}^{rt}\mathrm{....................}\left(1\right)$

The time needed to quadruple an investment replace $N$ by $\frac{1}{2}{N}_0$ if

From equation (1)

  $\begin{array}{l}\frac{1}{2}{N}_0=N_0{e}^{rt}\\ 5\approx 10.0170e^{-0.0301x}\\ -0.0301x\approx \ln \frac{5}{10.0170}\\ x=\frac{\ln \frac{5}{10.0170}}{-0.0301}\\ x\approx 23.08\end{array}$

Explanation of Solution

Given:

The data of number of the amount.

Concept used:

Formula:

The equation is

  $N=N_0{e}^{rt}$

Calculation:

The table is

Time (min)	$0$	$5$	$10$	$15$	$20$
U- $239$ percent(g)	$10$	$8.6$	$7.5$	$6.3$	$5.5$

  $N=N_0{e}^{rt}\mathrm{....................}\left(1\right)$

The time needed to quadruple an investment replace $N$ by $\frac{1}{2}{N}_0$ if

From equation (1)

  $\begin{array}{l}\frac{1}{2}{N}_0=N_0{e}^{rt}\\ 5\approx 10.0170e^{-0.0301x}\\ -0.0301x\approx \ln \frac{5}{10.0170}\\ x=\frac{\ln \frac{5}{10.0170}}{-0.0301}\\ x\approx 23.08\end{array}$