A point $(x,y)$ is randomly selected such that $0 \le x \le 8$ and $0 \le y \le 4$. What is the probability that $x+y \le 4$? Express your answer as a common fraction.
+128460 I believe we have these boundaries:
0 ≤ x ≤ 8 and 0 ≤ y ≤ 4
And we want to know the probability that
x + y ≤ 4 , given the above bounds
See the graph of these regions here......https://www.desmos.com/calculator/dz5d9hoagt
The region bounded by the lines x =0, y = 0, x = 8 and y = 4 has an area of (8)(4) = 32 sq units
And the region bounded by x=0, y=0 and x + y ≤ 4 has an area of (1/2)(4)(4) = (1/2)(16) = 8 units
So the probability that x + y ≤ 4, given the stated bounds = 8/32 = 1/4
+128460 I believe we have these boundaries:
0 ≤ x ≤ 8 and 0 ≤ y ≤ 4
And we want to know the probability that
x + y ≤ 4 , given the above bounds
See the graph of these regions here......https://www.desmos.com/calculator/dz5d9hoagt
The region bounded by the lines x =0, y = 0, x = 8 and y = 4 has an area of (8)(4) = 32 sq units
And the region bounded by x=0, y=0 and x + y ≤ 4 has an area of (1/2)(4)(4) = (1/2)(16) = 8 units
So the probability that x + y ≤ 4, given the stated bounds = 8/32 = 1/4
