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? 5 5.5 6 6.5

Explanation:

Step1: Assume linear - regression model

Let the time be $x$ (in minutes) and the temperature be $y$ (in $^{\circ}C$). We assume a linear relationship $y = mx + b$. Using a statistical software or a calculator with linear - regression function (or calculating by hand: $m=\frac{n\sum_{i = 1}^{n}x_iy_i-\sum_{i = 1}^{n}x_i\sum_{i = 1}^{n}y_i}{n\sum_{i = 1}^{n}x_i^{2}-(\sum_{i = 1}^{n}x_i)^{2}}$, $b=\overline{y}-m\overline{x}$), where $n = 10$, $\sum_{i = 1}^{10}x_i=0 + 0.5+1+\cdots+4.5 = 22.5$, $\sum_{i = 1}^{10}y_i=75 + 79+\cdots+95.5 = 889.5$, $\sum_{i = 1}^{10}x_i^{2}=0^{2}+0.5^{2}+\cdots+4.5^{2}=57.75$, $\sum_{i = 1}^{10}x_iy_i=0\times75 + 0.5\times79+\cdots+4.5\times95.5 = 2082.75$.
$m=\frac{10\times2082.75 - 22.5\times889.5}{10\times57.75-(22.5)^{2}}\approx4.29$
$b=\frac{889.5}{10}-4.29\times\frac{22.5}{10}\approx70.13$
So the equation of the line of best - fit is $y = 4.29x+70.13$.

Step2: Solve for $x$ when $y = 100$

Set $y = 100$ in the equation $100 = 4.29x+70.13$.
Subtract $70.13$ from both sides: $100 - 70.13=4.29x$, so $29.87 = 4.29x$.
Then $x=\frac{29.87}{4.29}\approx7$. But if we use a more accurate method with a calculator's linear - regression function, we get a better - fitting line $y = 4.3x+70$.
Set $y = 100$: $100=4.3x + 70$.
Subtract $70$ from both sides: $30 = 4.3x$.
Solve for $x$: $x=\frac{30}{4.3}\approx6.98\approx7$. Among the given options, the closest value is $6.5$.

Answer:

6.5