...42, -193.. thats the 4th and 5th Hope this helps
Answer:
I believe the answer would be C (sorry i messed uo the first time)
Strictly speaking, x^2 + 2x + 4 doesn't have solutions; if you want solutions, you must equate <span>x^2 + 2x + 4 to zero:
</span>x^2 + 2x + 4= 0. "Completing the square" seems to be the easiest way to go here:
rewrite x^2 + 2x + 4 as x^2 + 2x + 1^2 - 1^2 = -4, or
(x+1)^2 = -3
or x+1 =i*(plus or minus sqrt(3))
or x = -1 plus or minus i*sqrt(3)
This problem, like any other quadratic equation, has two roots. Note that the fourth possible answer constitutes one part of the two part solution found above.
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.
