某种放射性物质的质量为 $m$ 毫克（mg），随时间 $t$ 小时（hours）递减。 已知 $m$ 和 $t$ 满足方程 $m = M e^{-kt}$，其中 $M$ 和 $k$ 是常数。 下表显示了 $m$ 和 $t$ 的测量值。

t (小时)	1	2	3	4	5
m (毫克)	6.55	5.36	4.40	3.60	2.94

1. Calculate the values of $\ln m$ for each corresponding value of $t$, rounding to 2 decimal places for plotting.

For $t = 1$, $\ln 6.55 \approx 1.88$
For $t = 2$, $\ln 5.36 \approx 1.68$
For $t = 3$, $\ln 4.40 \approx 1.48$
For $t = 4$, $\ln 3.60 \approx 1.28$
For $t = 5$, $\ln 2.94 \approx 1.08$

2. Plot the points on the graph paper and draw a straight line of best fit.

Plot the coordinates $(1, 1.88)$, $(2, 1.68)$, $(3, 1.48)$, $(4, 1.28)$, and $(5, 1.08)$ on the grid with $t$ on the horizontal axis and $\ln m$ on the vertical axis. Draw a straight line passing through these points.

1. Identify the relationship between the initial mass and the graph's y-intercept

Taking the natural logarithm of both sides of $m = M e^{-kt}$ gives:
$\ln m = \ln M - kt$
This is in the linear form $Y = c + mX$, where $Y = \ln m$, $X = t$, gradient $= -k$, and the vertical intercept is $\ln M$.
The initial mass is $M$.

2. Read the y-intercept from the graph and calculate the initial mass

From the plotted graph of $\ln m$ against $t$, the line intersects the vertical axis ($t = 0$) at approximately $2.08$.
$\ln M = 2.08$
$M = e^{2.08} \approx 8.00$

1. Relate the constant $k$ to the gradient of the straight line graph

From $m = M e^{-kt}$, taking natural logarithms gives $\ln m = \ln M - kt$.
The graph of $\ln m$ against $t$ is a straight line with gradient $-k$.

2. Calculate the gradient using points from the line of best fit

Using the points $(1, \ln 6.55)$ and $(5, \ln 2.94)$ from the line:
$\text{Gradient} = \frac{\ln 2.94 - \ln 6.55}{5 - 1}$
$-k = \frac{1.0784 - 1.8795}{4}$
$-k = -0.200$
$k = 0.200$ (to 3 s.f.)