X=1
x=2
x=3
2(1)-y=4
2-y=4
-2 both sides
-y=2
÷-1 both sides
y=-2
(1,-2)
2(2)-y=4
4-y=4
-4 both sides
-y=0
y=0
(2,0)
2(3)-y=4
6-y=4
-6 both sides
-y=-2
÷-1 both sides
y=2
(3,2)
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
Answer:
-true
Step-by-step explanation:
I'm correct if I'm rwong
Answer:
6=p
Step-by-step explanation:
We want to isolate p, so we can first multiply by p on both sides.
24/p=4
24=4p
Then, divide by 4.
6=p
A {1,2,3,4}<br>
B {2,4,9,16}<br>
C {1,2}<br>
D {1,2,3,4,9,16}
MaRussiya [10]
Answer:
B
Step-by-step explanation:
The range of the function consists of the set of output values ( on the right).
So its {2 4 9 16}
