Answer:
It is B
Step-by-step explanation:
I think it’s 1.3, if it’s not than I’m so sorry
Answer:
Step-by-step explanation:
1 In general, given a{x}^{2}+bx+cax
2
+bx+c, the factored form is:
a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})(x-\frac{-b-\sqrt{{b}^{2}-4ac}}{2a
2a
−b+√
b
2
−4ac
)(x−
2a
−b−√
b
2
−4ac
)
2 In this case, a=1a=1, b=-2b=−2 and c=-2c=−2.
(x-\frac{2+\sqrt{{(-2)}^{2}-4\times -2}}{2})(x-\frac{2-\sqrt{{(-2)}^{2}-4\times -2}}{2})(x−
2
2+√
(−2)
2
−4×−2
)(x−
2
2−√
(−2)
2
−4×−2
)
3 Simplify.
(x-\frac{2+2\sqrt{3}}{2})(x-\frac{2-2\sqrt{3}}{2})(x−
2
2+2√
3
)(x−
2
2−2√
3
)
4 Factor out the common term 22.
(x-\frac{2(1+\sqrt{3})}{2})(x-\frac{2-2\sqrt{3}}{2})(x−
2
2(1+√
3
)
)(x−
2
2−2√
3
)
5 Cancel 22.
(x-(1+\sqrt{3}))(x-\frac{2-2\sqrt{3}}{2})(x−(1+√
3
))(x−
2
2−2√
3
)
6 Simplify brackets.
(x-1-\sqrt{3})(x-\frac{2-2\sqrt{3}}{2})(x−1−√
3
)(x−
2
2−2√
3
)
7 Factor out the common term 22.
(x-1-\sqrt{3})(x-\frac{2(1-\sqrt{3})}{2})(x−1−√
3
)(x−
2
2(1−√
3
)
)
8 Cancel 22.
(x-1-\sqrt{3})(x-(1-\sqrt{3}))(x−1−√
3
)(x−(1−√
3
))
9 Simplify brackets.
(x-1-\sqrt{3})(x-1+\sqrt{3})(x−1−√
3
)(x−1+√
3
)
Answer:
7920
Step-by-step explanation:
12C3 × 9C2
= 220×36
= 7920
Complete question:
Consider 3 boxes, each of which contains 4 balls in particular, box 1 contains 4 white balls, box 2 contains 3 white balls and 1 red ball, box 3 contains 2 white balls and 2 red balls. Frank chooses a box at random. If boxes 1 and 3 are chosen, Frank extracts 2 balls without replacement. If box 2 is chosen Frank extracts 2 balls with replacement. USE ONLY THE CONDITIONAL PROBABILITY FORMALISM and derive an expression for the probability that the 2 extracted balls are red.
Answer:
11/288
Step-by-step explanation:
We are given:
Box 1: ( 4White, ORed)
Box 2: (3White, 1Red)
Box 3: (2White, 2 Red)
We are told that if Frank choose from Box 1 and Box 3, 2 balls are extracted without replacement.
Since there is no red ball in Box 1, there is no way 2 red balls will come out from Box1.
Our Event, E = getting 2 red balls.
Now Box 1 is ruled out, we have:
P[E(B1)]= 0
P[E/B3)] = (2 2) / (4 2)
= 1/6
If box 2 is chosen, 2 balls are extracted with replacement. Therefore for Box 2:
P(E/B2) = (1/4) *(1/4)
= 1/16
Therefore probability that 2 balls extracted are red, we have:
P(E)=P(E/B1) P(B1) + P(E/B2) P(B2)+P(E/B3) P(B3)

= 11/288