Answer:
(Hope this helps can I pls have brainlist (crown) ☺️)
Step-by-step explanation:
In pic
(credit: Symbolab)
Bet imma go measure it rn!! Be back to you very soon xoxo
Im doin the same stuff and can’t figure it out
Remember the chain rule.
L(x)=f(g(x))
L'(x)=f'(g(x))g'(x)
take the derivative of f(g(x)). just treat them like they are variables. so you get:
h'=f'(g(x))g'(x)
now plug in your x value and evaluate:
h'(1)=f'(g(1))(g'(1))
substitute in values that you know and evaluate again
h'(1)=f'(3)(-3)
h'(1)=(-5)(-3)=15
Answer:
0.79
Step-by-step explanation:
Here,
Let X be the event that the flights depart on time
Let Y be the event that flights arrive on time
So,
X∩Y will denote the event that the flights departing on time also arrive on time.
Let P be the probability
P(X∩Y)=0.65
And
P(X)=0.82
We have to find P((Y│X)
We know that
P((Y│X)=P(X∩Y)/P(X) )
=0.65/0.82
=0.79
So the probability that a flight that departs on schedule also arrives on schedule is: 0.79
