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

Explanation:

Step1: Assume linear - relationship

Let the time be $x$ (in minutes) and the temperature be $y$ (in $^{\circ}C$). We can use the least - squares method to find the line of best fit $y = mx + b$. First, we calculate the means of $x$ and $y$ values. The mean of $x$ values: $\bar{x}=\frac{0 + 0.5+1.0 + 1.5+2.0+2.5+3.0+3.5+4+4.5}{10}=2.25$ The mean of $y$ values: $\bar{y}=\frac{75 + 79+83+86+89+91+93+94+95+95.5}{10}=87.05$

Step2: Calculate slope $m$

We use the formula $m=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}$. $\sum_{i = 1}^{10}(x_{i}-\bar{x})(y_{i}-\bar{y})=(0 - 2.25)(75 - 87.05)+(0.5 - 2.25)(79 - 87.05)+(1.0 - 2.25)(83 - 87.05)+(1.5 - 2.25)(86 - 87.05)+(2.0 - 2.25)(89 - 87.05)+(2.5 - 2.25)(91 - 87.05)+(3.0 - 2.25)(93 - 87.05)+(3.5 - 2.25)(94 - 87.05)+(4 - 2.25)(95 - 87.05)+(4.5 - 2.25)(95.5 - 87.05)$ $=(- 2.25)\times(-12.05)+(-1.75)\times(-8.05)+(-1.25)\times(-4.05)+(-0.75)\times(-1.05)+(-0.25)\times1.95 + 0.25\times3.95+0.75\times5.95+1.25\times6.95+1.75\times7.95+2.25\times8.45$ $=27.1125 + 14.0875+5.0625 + 0.7875-0.4875+0.9875+4.4625+8.6875+13.9125+19.0125$ $=96.6$

$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=(0 - 2.25)^{2}+(0.5 - 2.25)^{2}+(1.0 - 2.25)^{2}+(1.5 - 2.25)^{2}+(2.0 - 2.25)^{2}+(2.5 - 2.25)^{2}+(3.0 - 2.25)^{2}+(3.5 - 2.25)^{2}+(4 - 2.25)^{2}+(4.5 - 2.25)^{2}$ $=5.0625+3.0625 + 1.5625+0.5625+0.0625+0.0625+0.5625+1.5625+3.0625+5.0625$ $=20.625$

$m=\frac{96.6}{20.625}\approx4.68$

Step3: Calculate intercept $b$

We know that $\bar{y}=m\bar{x}+b$, so $b=\bar{y}-m\bar{x}$. $b = 87.05-4.68\times2.25$ $=87.05 - 10.53$ $=76.52$

The line of best fit is $y = 4.68x+76.52$.

Step4: Solve for $x$ when $y = 100$

Set $y = 100$ in the equation $y = 4.68x+76.52$. $100=4.68x+76.52$ $4.68x=100 - 76.52$ $4.68x=23.48$ $x=\frac{23.48}{4.68}\approx5$

Answer:

5