-- The first urn has 9 balls in it all together, and 2 of them are white.
If you don't peek, then the prob of pulling out a white ball is 2/9 .
-- The second urn has 13 balls in it all together, and 3 of them are white.
If you don't peek, then the prob of pulling out a white ball is 3/13 .
-- The probability of being successful BOTH times is
(2/9) x (3/13) = ( 6/117 ) = about 0.0513 or 5.13% (rounded)
We have that AB || DC.
By a similar argument used to prove that AEB ≅ CED,we can show that (AED) ≅ CEB by (SAS) . So, ∠CAD ≅ ∠ (ACB) by CPCTC. Therefore, AD || BC by the converse of the (
ALTERNATE INTERIOR ANGLES) theorem. Since both pair of opposite sides are parallel, quadrilateral ABCD is a parallelogram
1. AED
2. SAS
3. ACB
4. ALTERNATE INTERIOR ANGLES
<h2>follow me please and be my friend..please request</h2>
When we Simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3], the result obtained is (1/18)x^2
<h3>Data obtained from the question</h3>
- [(x^2)^3 × 5x] / [6x^2 × 15x^3]
- Simplification =?
<h3>How to simplify [(x^2)^3 × 5x] / [6x^2 × 15x^3]</h3>
[(x^2)^3 × 5x] / [6x^2 × 15x^3]
Recall
(M^a)^b = M^ab
Thus,
(x^2)^3 = x^6
- [(x^2)^3 × 5x] / [6x^2 × 15x^3] = [x^6 × 5x] / [6x^2 × 15x^3]
Recall
M^a × M^b = M^(a+b)
Thus,
x^6 × 5x = 5x^(6 + 1) = 5x^7
6x^2 × 15x^3] = (6×15)x^(2 + 3) = 90x^5
- [x^6 × 5x] / [6x^2 × 15x^3] = 5x^7 / 90x^5
Recall
M^a ÷ M^b = M^(a - b)
Thus,
5x^7 ÷ 90x^5 = (5÷90)x^(7 - 5) = (1/18)x^2
Therefore,
- [(x^2)^3 × 5x] / [6x^2 × 15x^3] = (1/18)x^2
Learn more about algebra:
brainly.com/question/2768008
#SPJ1
