Question

the scatter plot shows the average monthly temperature, x, and a familys monthly heating cost, y, for 25 different months. use the scatter plot to answer the parts below. (note that you can use the graphing tools to help you approximate the line.) (a) write an approximate equation of the line of best fit. round the coefficients to the nearest hundredth. y = \square (b) using your equation from part (a), predict the monthly heating cost for a month with an average temperature of 45 °f. round your prediction to the nearest hundredth. $\square$

Explanation:

Step1: Identify two points on the line

Looking at the scatter plot, we can approximate two points on the line of best fit. Let's take (10, 90) and (70, 10) as approximate points (these are estimates from the plot).

Step2: Calculate the slope (m)

The slope formula is \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Plugging in the points: \( m = \frac{10 - 90}{70 - 10} = \frac{-80}{60} \approx -1.33 \) (rounded to nearest hundredth).

Step3: Find the y-intercept (b)

Using the point-slope form \( y - y_1 = m(x - x_1) \) with (10, 90) and \( m \approx -1.33 \):
\( y - 90 = -1.33(x - 10) \)
\( y - 90 = -1.33x + 13.3 \)
\( y = -1.33x + 103.3 \) (approximate, may vary slightly based on point selection; another common estimate could be using (20, 80) and (60, 30): \( m = \frac{30 - 80}{60 - 20} = \frac{-50}{40} = -1.25 \), then \( y - 80 = -1.25(x - 20) \) → \( y = -1.25x + 25 + 80 = -1.25x + 105 \); for simplicity, let's use a more accurate approximation. Let's take two points: (10, 95) and (70, 10). Then \( m = \frac{10 - 95}{70 - 10} = \frac{-85}{60} \approx -1.42 \), then \( y - 95 = -1.42(x - 10) \) → \( y = -1.42x + 14.2 + 95 = -1.42x + 109.2 \). However, a common approximate line from such plots is \( y = -1.25x + 105 \) or \( y = -1.30x + 103 \). Let's use \( y = -1.33x + 103.33 \) (from first calculation).

Step4: Part (b) - Predict for x = 45

Substitute \( x = 45 \) into the equation from part (a). Using \( y = -1.33x + 103.33 \):
\( y = -1.33(45) + 103.33 = -59.85 + 103.33 = 43.48 \). If we used \( y = -1.25x + 105 \), then \( y = -1.25(45) + 105 = -56.25 + 105 = 48.75 \). The exact value depends on the line approximation, but let's proceed with the slope calculation from (10, 90) and (70, 10): \( m = -80/60 ≈ -1.33 \), b = 103.33. So for part (a), approximate equation: \( y = -1.33x + 103.33 \) (or similar). For part (b), plug x=45: \( y = -1.33(45) + 103.33 = -59.85 + 103.33 = 43.48 \), or with a different line, say \( y = -1.25x + 105 \), then \( y = -56.25 + 105 = 48.75 \).

Answer:

(Part a):
\( y = -1.33x + 103.33 \) (or similar, e.g., \( y = -1.25x + 105 \))