Answer:
A) The best way to picture this problem is with a probability tree, with two steps.
The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.
The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.
If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.
If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.
B) So a person can have a red marble and a blue marble in two ways:
1) Picking the red first and the blue last
2) Picking the blue first and the red last
C) P(R&B) = 3/5 = 60%
Step-by-step explanation:
C) P(R&B) = P(RB) + P(BR) = (2/5)*(3/4) + (3/5)*(2/4) = 3/10 + 3/10 = 3/5
Answer:
c. 5x + 25
Step-by-step explanation:
5( x + 5)
5×x + 5×5
5x + 25
Answer:
(-3, -3) ?
Step-by-step explanation:
it looks like you already answered your question
Answer:
0.4
Step-by-step explanation:
Round 0.600 - 0.200 = 0.4
Answer:
- y = -6
- x=2 and x=6
- Greatest value of y is y=2 and it occurs when x = 4
- For x between x = 2 and x = 6, y > 0
Step-by-step explanation:
<u>Definition</u>
- A parabola is a curve where any point is at an equal distance from:
a fixed point (the focus ), and
a fixed straight line (the directrix )
From the graph we can see that this is indeed so. We can even calculate the parabola equation from the given graph but since that is not required, I am not illustrating the steps here to do that
- The y-intercept is the value of y where the parabola cuts the y axis and from the graph we see that this occurs at y = -6
- The x-intercepts are the x-values where the parabola crosses the x-axis and we can see that this occurs at x = 2 and x = 6
- The greatest value of y occurs at the vertex of the parabola and we see that the vertex is at (4,2) ie greatest value of y is at y = 2 and occurs at x=4
- Between x =2 and x = 6 we see that the y values greater than 0 ie y > 0
(Note on last question: If you exclude these two points then y > 0 between x=2 and x=6.Specifically it is 0 at x =2 and x=6 and > 0. So if you include these two points then y ≥ 0. I have taken it as excluding the two points, x = 2 and x =6)
