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# help

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Point $$(x, y)$$ is randomly picked from the rectangular region with vertices at $$(0,0),(2009,0),(2009,2010),$$ and $$(0,2010)$$. What is the probability that $$x > 7y$$?Express your answer as a common fraction.

Dec 8, 2018

#1
+2338
+2

I think the best thing to do here is to construct a diagram. The problem gives you vertices with coordinates from the Cartesian plane, and it even gives you an inequality. Here is the construction I created. I will reference it throughout the solving process:

I first located the given coordinates and connected them with vertical and horizontal lines. I then graphed the inequality$$x>7y$$, which can be rewritten as $$y<\frac{x}{7}$$ . I added a few points, as they will become relevant when I solve.

One way to solve this problem is to figure out the ratio of the area created by the inequality $$y<\frac{x}{7}$$ and the sides of the rectangle to the area of the entire rectangle. I will do this by first finding the coordinates of the point E. Point E is located on $$\overline{BC}$$ , so its x-coordinate is also 2009. If you know the x-coordinate, then the y-coordinate of any point on $$y=\frac{x}{7}$$ is $$\frac{x}{7}$$ . Therefore, the y-coordinate is $$\frac{2009}{7}$$

The formula for the area of the triangle is $$\frac{1}{2}bh$$$$\triangle ABE$$ is a right triangle, so the base and height are the side lengths that are not the hypotenuse. We know the length of the base, $$AB$$ , because it is in the diagram. The height is the length of $$BE$$ , or $$\frac{2009}{7}$$

$$A_{\triangle}=\frac{1}{2}*AB*BE=\frac{1}{2}*2009*\frac{2009}{7}=\frac{2009^2}{14}$$

We can also calculate the area of the rectangle

$$A_{\text{rect.}}=bh=AB*BC=2009*2010$$

As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas.

 $$P(x,y|x>7y)=\frac{A_{\triangle}}{A_{\text{rect.}}}$$ As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas. Substitute in the values and simplify. $$P(x,y|x>7y)=\frac{\frac{2009^2}{14}}{2009*2010}$$ Use algebraic manipulation to simplify this. $$P(x,y|x>7y)=\frac{2009^2}{14*2009*2010}$$ $$P(x,y|x>7y)=\frac{2009}{14*2010}$$ At this point, you must use a calculator. $$P(x,y|x>7y)=0.07139$$

Dec 8, 2018

#1
+2338
+2

I think the best thing to do here is to construct a diagram. The problem gives you vertices with coordinates from the Cartesian plane, and it even gives you an inequality. Here is the construction I created. I will reference it throughout the solving process:

I first located the given coordinates and connected them with vertical and horizontal lines. I then graphed the inequality$$x>7y$$, which can be rewritten as $$y<\frac{x}{7}$$ . I added a few points, as they will become relevant when I solve.

One way to solve this problem is to figure out the ratio of the area created by the inequality $$y<\frac{x}{7}$$ and the sides of the rectangle to the area of the entire rectangle. I will do this by first finding the coordinates of the point E. Point E is located on $$\overline{BC}$$ , so its x-coordinate is also 2009. If you know the x-coordinate, then the y-coordinate of any point on $$y=\frac{x}{7}$$ is $$\frac{x}{7}$$ . Therefore, the y-coordinate is $$\frac{2009}{7}$$

The formula for the area of the triangle is $$\frac{1}{2}bh$$$$\triangle ABE$$ is a right triangle, so the base and height are the side lengths that are not the hypotenuse. We know the length of the base, $$AB$$ , because it is in the diagram. The height is the length of $$BE$$ , or $$\frac{2009}{7}$$

$$A_{\triangle}=\frac{1}{2}*AB*BE=\frac{1}{2}*2009*\frac{2009}{7}=\frac{2009^2}{14}$$

We can also calculate the area of the rectangle

$$A_{\text{rect.}}=bh=AB*BC=2009*2010$$

As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas.

 $$P(x,y|x>7y)=\frac{A_{\triangle}}{A_{\text{rect.}}}$$ As aforementioned, the probability of (x,y) such that x>7y is just the ratio of these areas. Substitute in the values and simplify. $$P(x,y|x>7y)=\frac{\frac{2009^2}{14}}{2009*2010}$$ Use algebraic manipulation to simplify this. $$P(x,y|x>7y)=\frac{2009^2}{14*2009*2010}$$ $$P(x,y|x>7y)=\frac{2009}{14*2010}$$ At this point, you must use a calculator. $$P(x,y|x>7y)=0.07139$$

TheXSquaredFactor Dec 8, 2018