We are given the wind chill temperature formula:

$y = 0.0055x^2 - 0.67x + 15.4$

where $y$ is the wind chill temperature, and $x$ is the wind speed in mph. We want to determine for what values of $x$ (the wind speed) between 0 and 90 mph, the wind chill temperature is $-3^\circ F$ or below.

So, we need to solve the inequality:

$0.0055x^2 - 0.67x + 15.4 \leq -3$

Step 1: Set up the equation

First, set the equation equal to $-3$:

$0.0055x^2 - 0.67x + 15.4 = -3$

Step 2: Move all terms to one side of the equation

$0.0055x^2 - 0.67x + 15.4 + 3 = 0$ $0.0055x^2 - 0.67x + 18.4 = 0$

Step 3: Solve the quadratic equation

We can now solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation:

• $a = 0.0055$
• $b = -0.67$
• $c = 18.4$

Let's plug these values into the quadratic formula to find the wind speeds.

I'll calculate this next.The wind speeds at which the wind chill temperature is exactly $-3^\circ F$ are approximately 41.82 mph and 80 mph.

Step 4: Interpretation

Since the quadratic equation gives us these two wind speeds, the wind chill temperature will be $-3^\circ F$ or below for wind speeds between 41.82 mph and 80 mph.

Final Answer:

For wind speeds between approximately 41.82 mph and 80 mph, the wind chill temperature is $-3^\circ F$ or below.

Do you want more details on the steps or explanations?

Here are 5 related questions to expand on:

1. How does the wind chill temperature change with increasing wind speeds?
2. What happens to the wind chill temperature when the wind speed exceeds 90 mph?
3. How would the wind chill equation change if the outside temperature is different?
4. Can you use this quadratic equation to find the wind chill at specific wind speeds?
5. What are the physical factors that affect wind chill and how it is calculated?

Tip: The discriminant in the quadratic formula tells you how many real solutions exist: if it's negative, there are no real solutions!