The equation of line $l$ is $y = -x + 6$. The equation of line $m$ is $y = -x + 1$. What is the probability that a point randomly selected in the first quadrant and below $l$ will fall between $l$ and $m$? Express your answer as a decimal to the nearest hundredth.

Guest Aug 9, 2023

#1**-1 **

Below is an image that representst this situation well.

We can find the area of the triangle created by \(y = -x + 6\). This is just the area of a right triangle. \(A_{\triangle 1}= \frac{1}{2} * 6 * 6 = 18\). We can then find the area underneath the equation \(y = -x + 1\). In this case, \(A_{\triangle 2} = \frac{1}{2} * 1 * 1 = \frac{1}{2}\).

The probability that a point lies between \(y = -x + 6 \) and \(y = -x + 1\) is essentially finding the ratio of the part and the whole. The whole is the area underneath the triangle created by y = -x + 6 and the boundaries of the first quadrant. The part is the area between y = -x + 6 and y = -x + 1 in the first quadrant.

\(\frac{\text{part}}{\text{whole}} = \frac{18 - \frac{1}{2}}{18} = \frac{35}{36} \approx 0.97\). This is rounded to the nearest hundreds place, as suggested.

The3Mathketeers Aug 9, 2023

#1**-1 **

Best Answer

Below is an image that representst this situation well.

We can find the area of the triangle created by \(y = -x + 6\). This is just the area of a right triangle. \(A_{\triangle 1}= \frac{1}{2} * 6 * 6 = 18\). We can then find the area underneath the equation \(y = -x + 1\). In this case, \(A_{\triangle 2} = \frac{1}{2} * 1 * 1 = \frac{1}{2}\).

The probability that a point lies between \(y = -x + 6 \) and \(y = -x + 1\) is essentially finding the ratio of the part and the whole. The whole is the area underneath the triangle created by y = -x + 6 and the boundaries of the first quadrant. The part is the area between y = -x + 6 and y = -x + 1 in the first quadrant.

\(\frac{\text{part}}{\text{whole}} = \frac{18 - \frac{1}{2}}{18} = \frac{35}{36} \approx 0.97\). This is rounded to the nearest hundreds place, as suggested.

The3Mathketeers Aug 9, 2023