1. **State the problem:** We have data for mass $y$ (grams) at time $t$ (minutes) and want to check if $y = at^n$ fits the data. 2. **Show the formula and transform it:** Starting with $y = at^n$, take logarithms of both sides: $$\log y = \log (at^n) = \log a + \log t^n = \log a + n \log t$$ This shows the model can be written as: $$\log y = n \log t + \log a$$ 3. **Explain the graph:** If the model is valid, plotting $\log y$ (vertical axis) against $\log t$ (horizontal axis) will produce a straight line with slope $n$ and intercept $\log a$. 4. **Calculate $\log t$ and $\log y$ values:** | $t$ | 4 | 9 | 14 | 19 | 24 | 29 | |---|---|---|----|----|----|----| | $y$ | 3.00 | 4.50 | 5.61 | 6.54 | 7.35 | 8.08 | Calculate $\log t$ and $\log y$ (base 10): $\log 4 = 0.6021$, $\log 9 = 0.9542$, $\log 14 = 1.1461$, $\log 19 = 1.2788$, $\log 24 = 1.3802$, $\log 29 = 1.4624$ $\log 3.00 = 0.4771$, $\log 4.50 = 0.6532$, $\log 5.61 = 0.7497$, $\log 6.54 = 0.8153$, $\log 7.35 = 0.8665$, $\log 8.08 = 0.9076$ 5. **Plot and find the line of best fit:** Using the points $(\log t, \log y)$, calculate slope $n$ and intercept $\log a$ by linear regression. Calculate slope $n$: $$n = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{\sum (x_i - \bar{x})^2}$$ where $x_i = \log t$, $y_i = \log y$. Calculate means: $$\bar{x} = \frac{0.6021 + 0.9542 + 1.1461 + 1.2788 + 1.3802 + 1.4624}{6} = 1.1206$$ $$\bar{y} = \frac{0.4771 + 0.6532 + 0.7497 + 0.8153 + 0.8665 + 0.9076}{6} = 0.7449$$ Calculate numerator and denominator: $$\sum (x_i - \bar{x})(y_i - \bar{y}) = (0.6021-1.1206)(0.4771-0.7449) + ... + (1.4624-1.1206)(0.9076-0.7449) = 0.3883$$ $$\sum (x_i - \bar{x})^2 = (0.6021-1.1206)^2 + ... + (1.4624-1.1206)^2 = 0.4413$$ So, $$n = \frac{0.3883}{0.4413} = 0.8799 \approx 0.88$$ Calculate intercept $\log a$: $$\log a = \bar{y} - n \bar{x} = 0.7449 - 0.88 \times 1.1206 = -0.2361$$ 6. **Find $a$:** $$a = 10^{\log a} = 10^{-0.2361} = 0.579$$ 7. **Write the model:** $$y = 0.579 t^{0.88}$$ 8. **Find time for $y=10$ g:** $$10 = 0.579 t^{0.88} \Rightarrow t^{0.88} = \frac{10}{0.579} = 17.27$$ Raise both sides to power $\frac{1}{0.88}$: $$t = 17.27^{\frac{1}{0.88}} = 22.3 \text{ minutes (approx)}$$ **Final answers:** - Model: $y = 0.579 t^{0.88}$ - Time for 10 g: approximately 22.3 minutes