Question

HELP ASAP! What is the probability to the nearest hundredth that a point chosen randomly inside the rectangle is in the triangle?

Answer 1

Answer: C .08

Explanation:

Probability is the part out of the whole. you first need to find the part, which is the triangle

In other words find out how much of the triangle is taking up the rectangle, which is done through area

A(triangle)=1/2 b h where b=6 h=9

A(triangle)=1/2 (6)(9) = 27

A(rectangle)= Lw where L=24 h=15

A(rectangle)= (24)(15) = 360

P(triangle in rectangle)= 27/360

=.075 rounded = .08

Answer 2

Explanation:

To answer this question, we need to know the dimensions of the rectangle and the triangle. Assuming that the triangle is right-angled and has one side coinciding with one side of the rectangle, we can use the formula for the area of a triangle: \ ( \frac {1} {2} \) × base × height1. The base and height of the triangle are the same as the length and width of the rectangle. The area of the rectangle is length × width. The probability that a point chosen randomly inside the rectangle is in the triangle is then the ratio of the area of the triangle to that of the rectangle, which is: \ ( \frac {\frac {1} {2} \times length \times width} {length \times width} \) = \ ( \frac {1} {2} \). So, the probability is 0.5 or 50% to the nearest hundredth2.