A rack of 15 billiard balls is shown. If one ball is selected at random, determine the odds against it containing a number grea
1 answer:
Using the probability and odds concepts, it is found that the odds against it containing a number greater than or equal to 7 is .
- A probability is the <u>number of desired outcomes divided by the number of total outcomes</u>.
- An odd is the <u>number of desired outcomes divided by the number of non-desired outcomes</u>.
In the rack, there are 15 balls, numbered from 1 to 15. Of those, <u>6 are less than 7</u>(against it containing a number greater than or equal to 7 is equivalent to it containing a number less than 7), thus:
- There are 6 desired outcomes.
- There are 9 non-desired outcomes.
The odd is:
The odds against it containing a number greater than or equal to 7 is .
A similar problem is given at brainly.com/question/21094006
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Answer:
The image triangle ΔD'L'W' after the dilation by a scale factor of 1/2 will have the vertices:
- D' (0, 1)
- L' (1/2, 5/2) or L'(0.5, 2.5)
- W' (5/2, 2) or W'(2.5, 2)
`The figure of the dilated image is attached below.
Step-by-step explanation:
Given the triangle ΔDLW withe the vertices
- D(0, 2)
- L(1, 5)
- W(5, 4)
As the scale factor 1/2 < 0, thus the image triangle will be smaller than the original triangle.
The rule of dilation is given by
Dilation: (x, y) → (1/2x, 1/2y), centered at (0, 0)
D(0, 2) → D' (1/2x, 1/2y) → D' (1/2(0), 1/2(2)) = D' (0, 1)
L(1, 5) → L'(1/2x, 1/2y) → L'(1/2(1), 1/2(5)) = L' (1/2, 5/2)
W(5, 4) → W'(1/2x, 1/2y) → W'(1/2(5), 1/2(4)) = W' (5/2, 2)
Therefore, the image triangle ΔD'L'W' after the dilation by a scale factor of 1/2 will have the vertices:
- D' (0, 1)
- L' (1/2, 5/2) or L'(0.5, 2.5)
- W' (5/2, 2) or W'(2.5, 2)
`The figure of the dilated image is attached below.
