The volume of a sphere, V = (4/3)(PI)(R^3)
Let k = (4/3)(PI)
Therefore, V = k (R^3)
Let R’ = new radius = 2R
V’ =k (R’^3)
= k (2R)^3
= 8 k R^3
= 8 V
The volume would be eight time the original volume.
Answer: B
just make the included side equal
5 children so you have 2^5=32 possibilities to "assign" genders
P(3 girls):
how many possibilities are there to "assign" the 3 girl-genders to the 5 children? the first girl has 5 possibilities then the next 4, 3 -> 5*4*3=60
but these possibilities include orders of assigned genders, while children 1-5 might differ the gender "girl" is always the same so we have the remove the orderings of the 3 girl-gender assignments which is 3*2*1=6
if we divide 60/6 we get 10 possibilities to have 3 girls, so what is the resulting chance? the 10 possibilities divided by the total 32 possibilities: 10/32=5/16=P(3 girls)=P(2 boys)
this is a bit of lengthy way of saying "use the binomial coefficient" equation/explaining it a bit which is (n!)/(k!(n-k)!) with n=5, k=3:
5*4*3*2*1/((3*2*1)*(2*1))=
5*4*3*2/(3*2*2)=
5*4*3*2/(3*4)=
5*2=
10 possibilities again
P(girls>=4)=P(boys<=1)=P(boys=1)+P(boys=0)
(or P(girls=4)+P(girls=5))
P(boys=0) is the easy case: simply multiply the chance of getting a girl 5 times: (1/2)^5=1/32
P(boys=1)= again the binomial coefficient with n=5 and k=1:
5*4*3*2*1/((1)*(4*3*2*1))=
5*4*3*2/(4*3*2)=
5 possibilities
so the P(boys=0)=1 possibility + P(boys=1)=5 possibilities totals to 6 possibilities
again the chance is the 6 possibilities divided by all 32 possibilities: 6/32=3/16
P(alternate gender starting with boy): when thinking about the possibilities then there is only a single way to build that order: bgbgb, so one possibility
knowing there is only one way we already know P(alternate...)=1/32 by again dividing by the total amount of possibilities
the alternative way would be to multiply P(boy)*P(girl)*P(boy)*P(girl)*P(boy)=(1/2)^5= 1/32 again
D = -(8-5) = -3
a = 47
an = a + (n-1)d = 47 + (-3) = 47 - 3 = 44
