Question

1. this scatter plot shows a linear relationship between \\(x\\) and \\(y\\) in a small class of students. a trend line passes through \\((3.2, 10.8)\\) and \\((8, 7.2)\\).

(a) find the equation of the trend line.

(b) find the equation to estimate \\(y\\) when \\(x = 13\\).

(c) find the equation to estimate \\(x\\) when \\(y = 13\\).

(d) find the equation to estimate \\(x\\) when \\(y = 2\\).

1. this scatterplot shows the growth of a seedling over time. find the equation of the trend line, and use this equation to predict the height of the seedling after 25 days.

Explanation:

Calculate the slope of the trend line

We find the slope \(m\) using the points \((3.2, 10.8)\) and \((8, 7.2)\).

$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7.2 - 10.8}{8 - 3.2} = \frac{-3.6}{4.8} = -0.75 $$

Find the equation of the trend line

We use the point-slope form with \((8, 7.2)\) to find the \(y\)-intercept.

$$ y - 7.2 = -0.75(x - 8) \implies y = -0.75x + 6 + 7.2 \implies y = -0.75x + 13.2 $$

Solve for y when x is 13

We substitute \(x = 13\) into our trend line equation.

$$ y = -0.75(13) + 13.2 = -9.75 + 13.2 = 3.45 $$

The equation to estimate \(y\) is \(y = -0.75x + 13.2\).

Solve for x when y is 13

We rearrange the equation to solve for \(x\) in terms of \(y\).

$$ y = -0.75x + 13.2 \implies 0.75x = 13.2 - y \implies x = \frac{13.2 - y}{0.75} $$

Substituting \(y = 13\):

$$ x = \frac{13.2 - 13}{0.75} = \frac{0.2}{0.75} = \frac{4}{15} \approx 0.27 $$

Solve for x when y is 2

We use the rearranged equation to estimate \(x\) when \(y = 2\).

$$ x = \frac{13.2 - 2}{0.75} = \frac{11.2}{0.75} = \frac{112}{7.5} = \frac{224}{15} \approx 14.93 $$

Answer:

Question 2(a)

\(y = -0.75x + 13.2\)

Question 2(b)

\(y = -0.75x + 13.2\) (Substituting \(x = 13\) gives \(y = 3.45\))

Question 2(c)

\(x = \frac{13.2 - y}{0.75}\) (Substituting \(y = 13\) gives \(x \approx 0.27\))

Question 2(d)

\(x = \frac{13.2 - y}{0.75}\) (Substituting \(y = 2\) gives \(x \approx 14.93\))