Respuesta :
Answer:
A point is chosen at random inside the cylinder, then probability the point belongs outside of the spheres is 0.25
Step-by-step explanation:
Let's find volume of the cylinder and volume of the three spheres.
Volume of cylinder =[tex]\pi r^{2} h[/tex]
Volume of sphere =[tex]\frac{4}{3} \pi r^{3}[/tex]
Now, r =2.7
h=8.1
Find volume of each shape.
Volume of cylinder =[tex]\pi (2.7)^{2} (8.1)[/tex]
=59.049[tex]\pi[/tex]
Now, volume of 3 spheres =[tex]3*\frac{4}{3} *\pi *(2.7)^{3}[/tex]
=78.732[tex]\pi[/tex]
Now, p(E)=[tex]\frac{Outcomes favorable to event E}{Total number of outcomes}[/tex]
So, Probability of point is chosen at random inside the cylinder which belongs outside of the spheres is
[tex]\frac{78.732\pi -59.049\pi }{78.732\pi }[/tex]
Let's simplify them
0.25
A point is chosen at random inside the cylinder, then probability the point belongs outside of the spheres is 0.25
