Since x³ = x^(1/3) you can say that
(x¹⁰)^(1/3) = x^10/3 = x^(3 + 1/3) =
x³ ·∛x
The one and the negative one
Step-by-step explanation:
1. P(light OR domestic) = P(light) + P(domestic) − P(light AND domestic)
P(light OR domestic) = 0.62 + 0.70 − 0.55
P(light OR domestic) = 0.77
2. P(light AND not domestic) = P(light) − P(light AND domestic)
P(light AND not domestic) = 0.62 − 0.55
P(light AND not domestic) = 0.07
3. P(light GIVEN not domestic) = P(light AND not domestic) / P(not domestic)
P(light GIVEN not domestic) = 0.07 / (1 − 0.70)
P(light GIVEN not domestic) = 0.233
4. Two events are independent if P(A) × P(B) = P(A and B).
P(light) × P(domestic) = 0.62 × 0.70 = 0.434
P(light AND domestic) = 0.55
Therefore, the type and location are not independent.
5X^4 + 2X^2 – 8
Its quadratic form is 5(x²)² +2x² -8
<span>Let be X²=p, so the final expression is 5p²+2p-8</span>
