A bag contains 1 green, 4 red and 5 yellow balls. Two balls are selected at random. Find the probability if there is no replacement of P(1 green and 1
yellow).

The probability is 1/18.

Step-by-step explanation:

There are 10 balls in total. Let's start with the probability of just the green. There is 1 green ball.out of 10 balls total, so the probability of pickin just the green one would be 1/10. Since we aren't replacing the balls after we choose them, the total number of balls is now 9. now the probability of picking one yellow ball out of the nine balls left is 5/9 because there are 5 yellow balls. To get the probability of the green and yellow balls picked, we multiply the individual probabilities to get the actual probability of choosinf one green and one yellow ball without replacement. So multiply 1/10×5/9= 5/90, simplified to 1/18.

38.5

step-by-step explanation:

lets say, first number = r

second number = 3r+7

total = 49

(r)+(3r+7)= 49

4r+7 = 49

4r = 42

r = 42/4

r= 10.5

3r+7 = 3(10.5)+7 = 38.5

hope this

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0