Question

jace gathered the data in the table. he found the approximate line of best fit to be $y = -0.7x + 2.36$.

$x$	$y$
1	1
4	0
5	-2
7	-2

what is the residual value when $x = 5$?

• $-3.14$
• $-0.86$
• $0.86$
• $3.14$

Explanation:

Step1: Recall residual formula

Residual = Observed \( y \) - Predicted \( y \).

Step2: Find predicted \( y \) at \( x = 5 \)

Use the line of best fit \( y = -0.7x + 2.36 \). Substitute \( x = 5 \):
\( y_{\text{predicted}} = -0.7(5) + 2.36 = -3.5 + 2.36 = -1.14 \).

Step3: Find observed \( y \) at \( x = 5 \)

From the table, when \( x = 5 \), observed \( y = -2 \).

Step4: Calculate residual

Residual = \( -2 - (-1.14) = -2 + 1.14 = -0.86 \)? Wait, no—wait, residual is observed - predicted. Wait, observed \( y = -2 \), predicted \( y = -1.14 \). So residual = \( -2 - (-1.14) = -0.86 \)? Wait, no, wait: Wait, maybe I messed up. Wait, the line is \( y = -0.7x + 2.36 \). Let's recalculate predicted \( y \) at \( x = 5 \):
\( -0.7 \times 5 = -3.5 \); \( -3.5 + 2.36 = -1.14 \). Observed \( y \) is \( -2 \). So residual = observed - predicted = \( -2 - (-1.14) = -0.86 \)? Wait, but wait, maybe I made a mistake. Wait, no—wait, the options include -0.86. Wait, but let me check again. Wait, \( -0.7 \times 5 = -3.5 \), plus 2.36 is -1.14. Observed \( y \) is -2. So residual is \( -2 - (-1.14) = -0.86 \). Yes, that matches the option.

Answer:

\(-0.86\) (corresponding to the option with -0.86)