There are 10 balls in the Urn Total.
Red: 6
Green: 4
Question One: The probability that five red and two green is selected is likely. (as that is over half for both)
Question Two: Impossible. There is only 6 red balls, and 7 are taken from the urn. Thus it would at most be possible for 6 red and 1 green.
Question Three: At least four is likely, as there is more red then green in the Urn.
Hope I helped!
(Mark Brainliest if you can please!)
Answer:
Step-by-step explanation:
Given points are: (-4,2) and (5,6)

The markup percentage is 45.14%
Step-by-step explanation:
The given is:
- The selling price of a box of crackers is $1.75
- You mark the crackers up to $2.54
We need to find the markup percentage
The markup percentage =
× 100%
∵ The selling price of a box of crackers is $1.75
∴ Old = 1.75
∵ You mark the crackers up to $2.54
∴ New = 2.54
- Substitute these values in the rule above
∵ The markup percentage =
× 100%
∴ The markup percentage =
× 100%
∴ The markup percentage = 0.4514 × 100%
∴ The markup percentage = 45.14%
The markup percentage is 45.14%
Learn more:
You can learn more about percentage in brainly.com/question/1834017
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Answer: Man sorry I don’t know how to solve that go to a math solver or something
Step-by-step explanation:
<span>Let n = number
n = 32 + 8
n = 40</span>