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In the coordinate plane, rectangular region R has vertices at
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08 Jun 2026, 05:21
08 Jun 2026, 05:21
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Question Stats:
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60% (01:42) wrong based on 5 sessions
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In the coordinate plane, rectangular region R has vertices at \((0,0),(0,3),(4,3)\) , and \((4,0)\) . If a point in region R is randomly selected, what is the probability that the point's y coordinate will be greater than its x -coordinate?
\(\begin{aligned}
& A. 7 / 12 \\
& B. 5 / 12 \\
& C. 3 / 8 \\
& D. 1 / 3 \\
& E. 1 / 4
\end{aligned}\)
\(\begin{aligned}
& A. 7 / 12 \\
& B. 5 / 12 \\
& C. 3 / 8 \\
& D. 1 / 3 \\
& E. 1 / 4
\end{aligned}\)
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In the coordinate plane, rectangular region R has vertices at
[#permalink]
11 Jun 2026, 22:55
11 Jun 2026, 22:55
Expert Reply
Explanation
The rectangle R is bounded by
\(0≤x≤4,0≤y≤3.\)
We want the probability that
\(y>x.\)
Since points are chosen uniformly from the rectangle, the probability equals
area where \(y>x\)/area of \(R\)
Area of the rectangle
Area(R)\(=4*3=12\).
Find the area where \(y>x\)
The line \(y=x\) intersects the rectangle at (0,0) and (3,3).
Within the rectangle, the region where \(y>x\) is the triangle with vertices
\((0,0), (0,3), (3,3).\)
Its area is
\(\frac{1}{2}(3)(3)=\frac{9}{2}.\)
Compute the probability
P(y>x)=(9/2)/12=\(\frac{9}{24}=\frac{3}{8}\).
\(\frac{3}{8}\)
Answer: C
The rectangle R is bounded by
\(0≤x≤4,0≤y≤3.\)
We want the probability that
\(y>x.\)
Since points are chosen uniformly from the rectangle, the probability equals
area where \(y>x\)/area of \(R\)
Area of the rectangle
Area(R)\(=4*3=12\).
Find the area where \(y>x\)
The line \(y=x\) intersects the rectangle at (0,0) and (3,3).
Within the rectangle, the region where \(y>x\) is the triangle with vertices
\((0,0), (0,3), (3,3).\)
Its area is
\(\frac{1}{2}(3)(3)=\frac{9}{2}.\)
Compute the probability
P(y>x)=(9/2)/12=\(\frac{9}{24}=\frac{3}{8}\).
\(\frac{3}{8}\)
Answer: C
gmatclubot
In the coordinate plane, rectangular region R has vertices at [#permalink]
11 Jun 2026, 22:55
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