Let's factorise it :
![\: {\qquad \dashrightarrow \sf {x}^{3} (x + 3) + [-5(x + 3)] }](https://tex.z-dn.net/?f=%5C%3A%20%7B%5Cqquad%20%20%5Cdashrightarrow%20%5Csf%20%20%20%20%7Bx%7D%5E%7B3%7D%20%28x%20%2B%203%29%20%2B%20%5B-5%28x%20%2B%203%29%5D%20%20%7D)
Using Distributive property we get :
⠀
Therefore,
The answer is b I believe
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.
D = 2r => 12.6 = 2*r => r=6.3
A = π*r^2 = (3.14) * (6.3)^2 = (3.14)*(39.69) = 124.6266 which is approximately 124.63
Hope this helps!
<h2>y = -0.25x + 2</h2><h3></h3><h3><em>Please let me know if I am wrong.</em></h3>
