Answer:
The answer is r < or equal to 15.5.
Step-by-step explanation:
I will be including both an basic explanation of what it is and its proof.
I'm guessing you are either learning about conditional probability at school or preparing for competitions.
Baye's theorem states:

That is the theorem itself and it means that the probability that event A happens given B is true equals the probability event B happens given A is true times the probability event A happens divided by the probability B happens.
That was the basic of the theorem and the proof of this is basically just testing how well you understand what conditional probability is.


Now we know that the probably that A and B both happens is the same as the probably that B and A both happens.
Therefore P(A|B) can be seen as P(B|A) multiplied by P(A) and then divided by P(B) which gives the right hand side of the first equation. And this is basically the theorem.

**Note P(B) have to be not equal to 0 because having a 0 in the denominator would make this equation undefined.
If you have any questions or need further explanations please ask me in the comments of the answer, I hope this helped!
Answer:
D.) 10 2/3 square feet
Step-by-step explanation:
The area will change by
multiply the area by 4/9 and you get 32/3 = 10 and 2/3 (because 10 = 30/3)
The answer is D. 9(y + 3) and 27 + 9y because 9(y + 3) when expanded is 9y + 27. I hope this helps!
Hello,
Let's place the last digit: it must be 2 or 4 or 8 (3 possibilities)
It remainds 4 digits and the number of permutations fo 4 numbers is 4!=4*3*2*1=24
Thus there are 3*24=72 possibilities.
Answer A
If you do'nt believe run this programm
DIM n(5) AS INTEGER, i1 AS INTEGER, i2 AS INTEGER, i3 AS INTEGER, i4 AS INTEGER, i5 AS INTEGER, nb AS LONG, tot AS LONG
tot = 0
n(1) = 1
n(2) = 2
n(3) = 4
n(4) = 7
n(5) = 8
FOR i1 = 1 TO 5
FOR i2 = 1 TO 5
IF i2 <> i1 THEN
FOR i3 = 1 TO 5
IF i3 <> i2 AND i3 <> i1 THEN
FOR i4 = 1 TO 5
IF i4 <> i3 AND i4 <> i2 AND i4 <> i1 THEN
FOR i5 = 1 TO 5
IF i5 <> i4 AND i5 <> i3 AND i5 <> i2 AND i5 <> i1 THEN
nb = ((((n(i1) * 10) + n(i2)) * 10 + n(i3)) * 10 + n(i4)) * 10 + n(i5)
IF nb MOD 2 = 0 THEN
tot = tot + 1
END IF
END IF
NEXT i5
END IF
NEXT i4
END IF
NEXT i3
END IF
NEXT i2
NEXT i1
PRINT "tot="; tot
END