First two rolls have to be 1-4 that is 2/3 chance twice and the third can be 4or 5
2/3*2/3*1/3 + the chance that the fourth is the 5 or 6.
2/3*2/3*2/3*1/3
So the solution is : P=2/3*2/3*1/3 + 2/3*2/3*2/3*1/3
A(n,s)=(ns^2)/(4tan(180/n)), n=number of sides, s=side length
A(8,4.6)=(8*4.6^2)/(4tan22.5)
A(8,4.6)=42.32/tan22.5 m (exact)
A(8, 4.6)≈102.17 m^2 (to nearest hundredth of a square meter)
V(cylinder)= πr²h
d=2r,
d is diameter, r is radius.
When diameter is tripled, a new diameter D=3d,
new radius R=3d/2=(3*2r)/2=3r
R=3r
V(new cylinder)= πR²h = π(3r)²h=9πr²h
V(new cylinder)/V(cylinder)=9πr²h/πr²h=9
Volume new cylinder 9 times more the volume of old cylinder.
