A microscopic worm is eating its way around the inside of a spherical apple of a radius of 6 cm. What is the probability that the worm is within 1 cm of the surface of the apple? ( hint: v=4/3pier^3 where r is the radius)
To solve this problem, we need to find the volume of the region within 1 cm of the surface of the apple and divide it by the total volume of the apple.
Given the radius of the apple, r = 6 cm, we can use the formula for the volume of a sphere:
V = (4/3) * pi * r^3.
Substituting the value of the radius, we get:
V = (4/3) * pi * 6^3 = (4/3) * pi * 216 = 288π cm^3.
Now, let's find the volume within 1 cm of the surface. Since the worm is moving around the inside of the apple, we need to subtract the volume of the sphere with a radius of 5 cm from the volume of the apple.
The volume of the smaller sphere is given by:
V_inner = (4/3) * pi * (5^3) = (4/3) * pi * 125 = 500π cm^3.
To find the probability, we divide the volume within 1 cm of the surface by the total volume of the apple:
P = V_inner / V = (500π) / (288π) ≈ 1.736.
So, the probability that the worm is within 1 cm of the surface of the apple is approximately 1.736 or 173.6% (as a decimal percentage).