Question:
In this sample, are the events "222 doors" and "444 doors" mutually exclusive?
Answer to question above: Yes
Answer to big question:
P( 2 doors OR 4 doors) = 0.77
Step-by-step explanation:
14 + 18 + 20 + 25 + 0 + 0 = 77 of the 100 cars that had two doors or four doors.
Y= -2(x²-x)+2= -2(x-1/2)²+2+1/2= -2(x-1/2)²+5/2
<span>x=6, y= -2(11/2)²+5/2= -121/2+5/2= -58
I hope this helps</span>
∠BDC=30, so ∠CBD + ∠BCD=180-30=150
ΔDBC is isosceles, so ∠CBD=∠BCD=half of 150=75
∠ABC=∠ABD-∠CBD=155-75=80
ΔABC is isosceles, so ∠ABC=∠ACB=80
∠BAC=180-80-80=20
You are going to use the equation y2-y1/x2-x1, and you will get negtive 4
