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:
The solution to this question is 39
Step-by-step explanation:
Given that :
√(3x+7)+2√(x-8)=0.
√(3x+7)=-2√(x-8) (i)
square of the equation (i)
Then
(√(3x+7))^2=(-2√(x-8))^2
we know that √a*√a=(√a)^2=a So,
(3x+7)=4(x-8)
3x+7=4x-32 //multiply by 4
32+7=4x-3x exchange the value.
39=1x
x=39.
5y = 135....divide both sides by 5
y = 135 / 5
y = 27 <==
To answer this question, we just need to subtract the saved cloth from the requested cloth:

Since the fractions don't have the same denominator, we have to find the least common denominator. Luckily, since 6 goes into 12 twice, we can convert

into

:

Now that the fractions have a common denominator, we can subtract them:

The difference between the requested cloth and the saved cloth is 5/12 of a foot, or 0.4166'.