Question

a chemist has a sample of radioactive material. he records the amount of radioactive material every week for 6 weeks and obtains the following data: use the exponential regression equation (rounded to the nearest thousandth) to determine when there will be only 1 gram of the material left (rounded to the nearest week).

Explanation:

Step1: Recall exponential regression form

Exponential regression equation: $y = ab^x$, where $x$=week, $y$=weight.

Step2: Calculate $a$ and $b$

Using data: $a \approx 100.000$ (week 0 value), $b \approx \frac{88.3}{100} \approx 0.883$ (check consistency: $\frac{75.9}{88.3}\approx0.859$, $\frac{69.4}{75.9}\approx0.914$; better via calculator: $a\approx100.0$, $b\approx0.88$). Accurate regression gives $y = 100(0.88)^x$.

Step3: Set $y=1$ and solve for $x$

$$1 = 100(0.88)^x$$
$$0.01 = (0.88)^x$$
Take natural log: $\ln(0.01) = x\ln(0.88)$
$$x = \frac{\ln(0.01)}{\ln(0.88)} \approx \frac{-4.605}{-0.128} \approx 36.0$$

Answer:

36