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

the table shows the temperature of an amount of water set on a stove to boil, recorded every half - minute.\nwaiting for water to boil\n| time (min) | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4 | 4.5 |\n| temp. (°c) | 75 | 79 | 83 | 86 | 89 | 91 | 93 | 94 | 95 | 95.5 |\naccording to the line of best fit, at what time will the temperature reach 100°c, the boiling point of water?\no 5\no 5.5\no 6\no 6.5

the table shows the temperature of an amount of water set on a stove to boil, recorded every half - minute.\nwaiting for water to boil\n| time (min) | 0 | 0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 | 3.5 | 4 | 4.5 |\n| temp. (°c) | 75 | 79 | 83 | 86 | 89 | 91 | 93 | 94 | 95 | 95.5 |\naccording to the line of best fit, at what time will the temperature reach 100°c, the boiling point of water?\no 5\no 5.5\no 6\no 6.5

Answer

Explanation:

Step1: Assume linear - regression model

Let the time be $x$ (in minutes) and the temperature be $y$ (in $^{\circ}C$). The general form of a linear equation is $y = mx + b$. We can use two - point form to find the equation of the line of best - fit. Let's take two points $(x_1,y_1)=(0,75)$ and $(x_2,y_2)=(4.5,95.5)$. First, find the slope $m$. $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{95.5 - 75}{4.5-0}=\frac{20.5}{4.5}=\frac{41}{9}\approx4.56$

Step2: Find the y - intercept

Since the line passes through $(0,75)$, the y - intercept $b = 75$. So the equation of the line is $y=\frac{41}{9}x + 75$.

Step3: Solve for x when y = 100

Set $y = 100$ in the equation $y=\frac{41}{9}x + 75$. $100=\frac{41}{9}x+75$ Subtract 75 from both sides: $100 - 75=\frac{41}{9}x$ $25=\frac{41}{9}x$ Multiply both sides by $\frac{9}{41}$ to solve for $x$: $x=\frac{25\times9}{41}=\frac{225}{41}\approx5.5$

Answer:

5.5