The answer is 6.75 and the answer is a decimal
the parabola has maximum at 9, meaning is a vertical parabola and it opens downwards.
it has a symmetry at x = -5, namely its vertex's x-coordinate is -5.
check the picture below.
so then, we can pretty much tell its vertex is at (-5 , 9), and we also know it passes through (-7, 1)
![\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} y=a(x- h)^2+ k\qquad \leftarrow \textit{using this one}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-5\\ k=9 \end{cases}\implies y=a[x-(-5)]^2+9\implies y=a(x+5)^2+9](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20y%3Da%28x-%20h%29%5E2%2B%20k%5Cqquad%20%5Cleftarrow%20%5Ctextit%7Busing%20this%20one%7D%5C%5C%5C%5C%20x%3Da%28y-%20k%29%5E2%2B%20h%20%5Cend%7Barray%7D%20%5Cqquad%5Cqquad%20vertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D-5%5C%5C%20k%3D9%20%5Cend%7Bcases%7D%5Cimplies%20y%3Da%5Bx-%28-5%29%5D%5E2%2B9%5Cimplies%20y%3Da%28x%2B5%29%5E2%2B9)
![\bf \textit{we also know that } \begin{cases} x=-7\\ y=1 \end{cases}\implies 1=a(-7+5)^2+9 \\\\\\ -8=a(-2)^2\implies -8=4a\implies \cfrac{-8}{4}=a\implies -2=a \\\\[-0.35em] ~\dotfill\\\\ ~\hfill y=-2(x+5)^2+9~\hfill](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bwe%20also%20know%20that%20%7D%20%5Cbegin%7Bcases%7D%20x%3D-7%5C%5C%20y%3D1%20%5Cend%7Bcases%7D%5Cimplies%201%3Da%28-7%2B5%29%5E2%2B9%20%5C%5C%5C%5C%5C%5C%20-8%3Da%28-2%29%5E2%5Cimplies%20-8%3D4a%5Cimplies%20%5Ccfrac%7B-8%7D%7B4%7D%3Da%5Cimplies%20-2%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20y%3D-2%28x%2B5%29%5E2%2B9~%5Chfill)
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:
6. 4
7. 40
8. 10
9. 4
10. 21
Step-by-step explanation:
