Question

this table shows the height, in inches, of a candle x minutes after it has been lit.
minutes (x) | 0 | 2 | 3 | 4
height (y) | 12 | 11.3 | 10.5 | 9.8
based on the best - fit linear model, what is the height of the candle 10 minutes after it has been lit?
a 7.3 inches
b 6.6 inches
c 5.2 inches
d 3.9 inches

Explanation:

Step1: Calculate the slope

Slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{11.3 - 12}{2 - 0} = -0.35$

Step2: Determine the linear equation

Intercept $b = 12$ (when $x=0$), so $y = -0.35x + 12$

Step3: Predict height at x=10

$y = -0.35(10) + 12 = 8.5$ (Note: The closest option is B. 6.6 inches? Correction: Recheck slope with more points: $\frac{10.5-11.3}{3-2}=-0.8$, $\frac{9.8-10.5}{4-3}=-0.7$. Average slope ≈ -0.75. Equation: $y=12-0.75x$. At x=10: $12-7.5=4.5$? No, original table: x=0→12, x=2→11.3 (Δx=2, Δy=-0.7→slope=-0.35); x=3→10.5 (Δx=1, Δy=-0.8→slope=-0.8); x=4→9.8 (Δx=1, Δy=-0.7→slope=-0.7). Best-fit slope calculation: $\bar{x}=2.25$, $\bar{y}=10.9$. Covariance: $\sum(x_i-\bar{x})(y_i-\bar{y})=(0-2.25)(12-10.9)+(2-2.25)(11.3-10.9)+(3-2.25)(10.5-10.9)+(4-2.25)(9.8-10.9)=(-2.25)(1.1)+(-0.25)(0.4)+(0.75)(-0.4)+(1.75)(-1.1)=-2.475-0.1-0.3-1.925=-4.8$. Variance of x: $\sum(x_i-\bar{x})^2=(0-2.25)^2+(2-2.25)^2+(3-2.25)^2+(4-2.25)^2=5.0625+0.0625+0.5625+3.0625=8.75$. Slope $m=-4.8/8.75≈-0.5486$. Equation: $y-10.9=-0.5486(x-2.25)→y≈-0.5486x+1.234+10.9≈-0.5486x+12.134$. At x=10: $y≈-5.486+12.134≈6.648$.

Answer:

B. 6.6 inches