Answer:
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Answer:
Sorry but you have the y and x mixed up
Step-by-step explanation:
You have them mixed up
Answer:
2<x<4/3
Step-by-step explanation:
Given the equation of a graph to be y = |3x− 5|, if the equation is one unit to the right, this can be expressed as |3x-5| > 1.
Solving the resulting equation
|3x-5| > 1.
Since the function 3x-5 is in a modulus sign, this means that the function can take both negative and positive values.
For positive value of the function;
+(3x-5) > 1
3x > 1+5
3x>6
x>6/3
x>2 ... (1)
For the negative value of the function;
-(3x-5) > 1
On expansion
-3x+5 > 1
-3x > 1-5
-3x > -4
Multiplying through by -1 will also change the inequality sign
x < -4/-3
x < 4/3...(2)
Combining equation 1 and 2, we have;
2<x<4/3
Please see the attached graph to see f(x) = (x - 3) ^2 +5 graphed.
So, the total number of balls is 11. We want to pick 2 red balls and 1 green ball. WLOG (since order doesnt matter here), we can say he picks red, green, red. That means on his first pick, he has a 
chance of picking the red ball, and he places it back in the bag. The probability of picking a green ball is 
, and then he places the ball back in the bag. The probability of picking the last red ball is the same as the last red ball example, and we simply multiply the probabilities together as per the multiplication rule to get:
Now, without replacement the order does matter. He picks a red ball, a red ball then a green ball. The probability of picking the first red ball is
, and the probability of picking the second red ball is 
and the probability of picking the green ball is
. We want to multiply thm again, as per the multiplication rule like the last problem.
