All these are true when X and Y are independent events:
1) The occurrence or lack of event X does not influence the occurrence of event Y, and the occurrence or lack of event Y does not influence the occurrence of event X.
2) P(X∩Y) = P(X) * P(Y)
3) P(X | Y) = P(X).... this is the probability of X given Y is equal to the probability of X.
4) P (Y | X) = P(Y)
5) P(A∪B) = P(A) + P(B) - P(A∩B)
Answer:
The answer in the procedure
Step-by-step explanation:
Let
A1 ------> the area of the first square painting
A2 ----> the area of the second square painting
D -----> the difference of the areas
we have
case 1) The area of the second square painting is greater than the area of the first square painting
The difference of the area of the paintings is equal to subtract the area of the first square painting from the area of the second square painting
D=A2-A1
case 2) The area of the first square painting is greater than the area of the second square painting
The difference of the area of the paintings is equal to subtract the area of the second square painting from the area of the first square painting
D=A1-A2
Step-by-step explanation:
a.) To model this scenario
Let the height of ball = y
The height of 1st= 0.5y
2nd =0.5(0.5y)
3rd = 0.5*(0.5(0.5y))
Hence the height of nth bounce can be modeled as
Height of nth bounce =(0.5ⁿ-1)*y
The exponential equation is
hn= (0.5ⁿ-1)*y
b.) if the ball is dropped from 9ft above the ground
y= 9ft
On the 4th bounce
n=4
Substituting in the exponential equation we have
h4=(0.5^4-1)*9
h4=0.5³*9
h4= 0.125*9
h4= 1.125ft
On the 4th bounce, the ball will reach a height of 1.125ft
Let the distance of the first part of the race be x, and that of the second part, 15 - x, then
x/8 + (15 - x)/20 = 1.125
5x + 2(15 - x) = 40 x 1.125
5x + 30 - 2x = 45
3x = 45 - 30 = 15
x = 15/3 = 5
Therefore, the distance of the first part of the race is 5 miles and the time is 5/8 = 0.625 hours or 37.5 minutes
The distance of the second part of the race is 15 - 5 = 10 miles and the time is 1.125 - 0.625 = 0.5 hours or 30 minutes.
