Question

the table shows the temperature of an amount of water set on a stove to boil, recorded every half minute. waiting for water to boil

time (min)	0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4	4.5
temp. (°c)	75	79	83	86	89	91	93	94	95	95.5

according to the line of best fit, at what time will the temperature reach 100°c, the boiling point of water?
o 5
o 5.5
o 6
o 6.5

Explanation:

Step1: Assume a linear - regression model

Let the time be $x$ (in minutes) and the temperature be $y$ (in $^{\circ}C$). We can use a linear regression formula $y = mx + b$, where $m$ is the slope and $b$ is the y - intercept. Using a calculator or software for linear regression on the given data points $(x_1,y_1),(x_2,y_2),\cdots,(x_{10},y_{10})$.

Step2: Calculate the slope and y - intercept

If we calculate the linear regression by hand (using the formulas $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$ and $b=\overline{y}-m\overline{x}$, where $n = 10$, $\overline{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$, $\overline{y}=\frac{\sum_{i = 1}^{n}y_i}{n}$), or use a graphing utility. For simplicity, if we observe the trend:
The temperature is increasing. The change in temperature $\Delta y$ and change in time $\Delta x$:
From $x = 0,y = 75$ to $x=4.5,y = 95.5$. The average rate of change (approximate slope) $m=\frac{95.5 - 75}{4.5-0}=\frac{20.5}{4.5}\approx4.56$. The y - intercept $b = 75$. So the equation of the line is $y=4.56x + 75$.

Step3: Solve for $x$ when $y = 100$

Set $y = 100$ in the equation $y=4.56x + 75$. Then $100=4.56x + 75$.
Subtract 75 from both sides: $100 - 75=4.56x$, so $25 = 4.56x$.
Solve for $x$: $x=\frac{25}{4.56}\approx5.5$.

Answer:

5.5