Question

1. ice rink maury has an ice rink in his back yard. the scatter plot shows the thickness of the ice relative to the temperature.

a. write an equation in slope - intercept form for the line that is drawn.
b. use the equation to make a conjecture about the \\(\frac{1}{3}\\) inches.

Explanation:

Step1: Identify two points on the line

Assume from the scatter plot (since it's not shown, we'll use a common example for ice thickness vs temperature). Let's say two points are (0, 4) and (20, 0) (common for such a negative slope relationship, temperature on x - axis, thickness on y - axis).

Step2: Calculate the slope (m)

The slope formula is $m=\frac{y_2 - y_1}{x_2 - x_1}$. Using the points (0, 4) and (20, 0):
$m=\frac{0 - 4}{20 - 0}=\frac{-4}{20}=-\frac{1}{5}$

Step3: Determine the y - intercept (b)

The y - intercept is the value of y when x = 0. From the point (0, 4), we see that when x = 0, y = 4. So, b = 4.

Step4: Write the slope - intercept form equation

The slope - intercept form is $y=mx + b$. Substituting m = $-\frac{1}{5}$ and b = 4, we get $y =-\frac{1}{5}x+4$.

(Note: Since the scatter plot is not provided, this is a general example. If the actual points from the plot are different, the calculation will change accordingly. For part b, assuming we have the equation $y =-\frac{1}{5}x + 4$, if we want to find the temperature when thickness $y=\frac{1}{3}$, we set up $\frac{1}{3}=-\frac{1}{5}x + 4$. Then, $-\frac{1}{5}x=\frac{1}{3}-4=\frac{1 - 12}{3}=-\frac{11}{3}$. Multiply both sides by - 5: $x=\frac{55}{3}\approx18.33$ degrees. But this is based on the assumed equation. The key for part a is to use the two - point formula to find slope and y - intercept.)

Answer:

a. (Assuming the points as above) The equation in slope - intercept form is $y =-\frac{1}{5}x + 4$.
b. (Based on the above equation) If the thickness $y=\frac{1}{3}$ inches, then $\frac{1}{3}=-\frac{1}{5}x + 4$. Solving for x:
$\frac{1}{5}x=4-\frac{1}{3}=\frac{12 - 1}{3}=\frac{11}{3}$
$x=\frac{55}{3}\approx18.33$ degrees. So the temperature would be approximately $\frac{55}{3}$ degrees (or about 18.33 degrees) when the ice thickness is $\frac{1}{3}$ inches.