Question

Candlepin bowling balls have no holes in them and are smaller than the bowling balls used
in tenpin bowling. The regulations size is 4.5 in in diameter, and their density is 0.05
Ibs/in^3. What is the regulation weight of the candlepin bowling ball? Round your answer
to the nearest tenth of a pound.

Answer 1

Answer:

2.4 lbs

Step-by-step explanation:

Firstly, we need to identify that a bowling ball is spherical in shape.

Hence, to answer this question at first, we need to know the volume of this bowling ball.

Since it is a sphere, the volume  V = 4/3π $r^{3}$

From the question, we are made to know that D = 4.5 inches, this means that r = D/2 = 4.5/2 = 2.25 inches

Substituting this into the equation of the volume, we have;

V = 4/3 ×π ×$2.25^{3}$  = 47.71 $inch^{3}$

Now, we proceed to calculate the weight

Mathematically,

Weight = volume × density

Hence weight = 47.71 × 0.05 = 2.3855 lbs = 2.4 lbs to the nearest tenth