P(arrives on time given leaves on time)=
P(B | A)= P(B n A) / P(A)
= P(A n B) / P(A)
= 0.36 / 0.9
= 0.4
X=15
Hope this helps I had the same problem. That’s what I got.
Answer:
C. (-3,11)
Step-by-step explanation:
Tp is (-3,6) implies the quadratic could have been
f(x) = (x+3)²+6
(2/3)f(x) = (2/3)[(x+3)²+6]
= (2/3)(x+3)²+4
(2/3)f(x)+3 = (2/3)(x+3)²+4+3
= (2/3)(x+3)²+7
Tp at (-3,7)
Alternately,
No change in domain so x remains-3
(2/3)f(x) changes y from 6 to 4 (6×2/3)
+3 increases the y by 3
i.e 4+3 = 7
So, (-3,7)
Answer:
just do 20 percent of 1200 then add it to 1200
ABCEG. Not D because non terminating numbers are irrational, same goes for Pi
