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:
120
240
Step-by-step explanation:
We call the length of first part x
Length of second part = y
In the first scenario, it took the tortoise 110 sec to walk the first part and crawl the second.
So,
We have this equation,
x/4 + y/3 = 110
We take the LCM
(3x + 4y)/12 = 110
When we cross multiply
3x + 4y = 110x12
3x + 4y = 1320 ----- equation 1
For scenario 2
x/3 + y/4 = 100
When we take the LCM
(4x + 3y)/12 = 100
We cross multiply
4x + 3y = 100x12
4x + 3y = 1200 ------ equation 2
We now have two equations and we will solve for x and y using simultaneous linear equation.
3x + 4y = 1320 ----- 1
4x + 3y = 1200 ----- 2
We subtract equation 2 from 1 to get
- x + y = 120
We make y subject
y = x + 120 ----- 3
We put the value of y in equation 3 into equation 1
3x + 4(x + 120) = 1320
3x + 4x + 480 = 1320
7x + 480 = 1320
7x = 1320-480
7x = 840
We divide through by 7
x = 840/7
x = 120
We put value of x in equation 3
y = x + 120
y = 120 + 120
y = 240
120 and 240 are the lengths of the 2 parts of the journey.
Thanks
L=2W-4
PERIMETER=2L+2W
58=2(2W-4)+2W
58=4W-8+2W
58=6W-8
6W=58+8
6W=66
W=66/6
W=11 ANS. FOR THE WIDTH.
L=2*11-4
L=22-4
L=18 ANS. FOR THE LENGTH.
PROOF:
58=2*18=2*11
58=36+22
58=58
The answer is 8 it should be
