A microscopic worm is eating its way around the inside of a spherical apple of radium 6 cm. What is the probability that the worm is with 1 cm if the surface of the apple? (Hint: V=(4/3)Pir^3)
volume of whole apple
= (4/3)π(6^3) = 288π cm^3
volume of inner part of apple with radius 5 cm
= (4/3)π(5^3) = 500π/3 cm^3
part of apple within 1 cm of outside
= 288π - 500π/3 = 364π/3
prob(being in outer part of apple)
= (364π/3) / (288π)
= 91/216
or
appr .421
To determine the probability that the worm is within 1 cm of the surface of the apple, we need to compare the volume of the 1 cm layer on the surface of the apple with the total volume of the apple.
First, let's find the volume of the apple using the given formula V = (4/3) * π * r^3, where r is the radius of the apple (6 cm).
V = (4/3) * π * (6 cm)^3
V = (4/3) * 3.14 * (6 cm)^3
V ≈ 904.32 cm^3
Now, let's find the volume of the 1 cm layer on the surface of the apple. This can be calculated by subtracting the volume of the apple with a radius of 5 cm from the volume of the entire apple.
V_layer = V - V_inner_apple
V_inner_apple = (4/3) * π * (5 cm)^3
V_inner_apple ≈ 523.60 cm^3
V_layer = 904.32 cm^3 - 523.60 cm^3
V_layer ≈ 380.72 cm^3
Finally, we can calculate the probability by dividing the volume of the 1 cm layer by the total volume of the apple:
Probability = V_layer / V
Probability ≈ 380.72 cm^3 / 904.32 cm^3
Probability ≈ 0.4206 or approximately 42.1%
Therefore, the probability that the worm is within 1 cm of the surface of the apple is approximately 42.1%.