Q/9 = whole number
Q/12 = whole number
Q /(3*3) = whole number
Q/(2*2*3) = whole number
Q at least 3*3*2*2 = 9*4 =36
of course 72 would also work :)
1) If 9 and 12 each divide Q without remainder, which of the following must Q divide without remainder?
A)1
B)3
C)36
D)72
E)The answer can't be determined from the given info
^I got B
2) If mn does not equal 0, then 1/n^2 times (m^5n^3/m^3)^2
A)mn^4
B)m^4n^2
C)m^4n^3
D)m^4n^4
E)m^4n^5
^I got D
3) Out of a group of 3 female and 3 males, 3 people at random enter a room. What is the probability that there are exactly 2 males in the room?
A)1/4
B)3/8
C)9/20
D)2/3
E)5/6
^I got D
10 answers
2. so did I
wrong on the third.
p(male) = .5
p(not male) = .5
P(2 out of 3) = C(3,2) .5^2 *.5^1
= 3!/[(1!)2!]*.5^3
=3 *.5^3 = .375 = 1/[2 2/3) = 1/ (8/3) = 3/8
p(not male) = .5
P(2 out of 3) = C(3,2) .5^2 *.5^1
= 3!/[(1!)2!]*.5^3
=3 *.5^3 = .375 = 1/[2 2/3) = 1/ (8/3) = 3/8
For number 1, wouldn't it be none of them?
And if number 4 is wrong, then can you please explain?
And if number 4 is wrong, then can you please explain?
Opps! sorry thank you!!
Did I not do so ?
And I said two of them work for number 1, neither of which you picked.
And I said two of them work for number 1, neither of which you picked.
You are welcome.
Changed my ind about #4
Not binomial because p is not constant
try MMF , MFM , FMM
MMF = 3/6 * 2/5 * 3/4 = 3/20
MFM = 3/6 * 3/5 * 2/4 = 3/20
FMM = 3/6 * 3/5 * 2/4 = 3/20
sum = 9/20
Not binomial because p is not constant
try MMF , MFM , FMM
MMF = 3/6 * 2/5 * 3/4 = 3/20
MFM = 3/6 * 3/5 * 2/4 = 3/20
FMM = 3/6 * 3/5 * 2/4 = 3/20
sum = 9/20
ohh Okay