pg97 #26
sports major league baseball, or mlb, rules state that baseballs must have a circumference of 9 inches. the national softball association, or nsa, rules state that softballs must have a circumference not exceeding 12 inches.

a. find the ratio of the circumference of an mlb baseball to the circumference of a 12 - inch nsa softball.

b. find the ratio of the volume of the mlb baseball to the volume of the nsa softball. round your answer to the nearest tenth.

Question

pg97 #26
sports major league baseball, or mlb, rules state that baseballs must have a circumference of 9 inches. the national softball association, or nsa, rules state that softballs must have a circumference not exceeding 12 inches.

a. find the ratio of the circumference of an mlb baseball to the circumference of a 12 - inch nsa softball.

b. find the ratio of the volume of the mlb baseball to the volume of the nsa softball. round your answer to the nearest tenth.

Accepted Answer

### Part a
# Explanation:
## Step1: Recall the formula for the circumference of a circle, $ C = 2\pi r $ or $ C=\pi d $ (where $ d $ is the diameter). The ratio of two circumferences $ C_1 $ and $ C_2 $ is $ \frac{C_1}{C_2}=\frac{\pi d_1}{\pi d_2}=\frac{d_1}{d_2} $ (since $ \pi $ cancels out). For a sphere (baseball/softball), the circumference is related to the diameter. Let the circumference of MLB baseball be $ C_{mlb}=9 $ inches and NSA softball be $ C_{nsa}=12 $ inches.
## Step2: Calculate the ratio $ \frac{C_{mlb}}{C_{nsa}}=\frac{9}{12} $
Simplify $ \frac{9}{12}=\frac{3}{4} $
# Answer:
$ \frac{3}{4} $

### Part b
# Explanation:
## Step1: Recall the formula for the volume of a sphere, $ V = \frac{4}{3}\pi r^3 $. First, find the radii of the MLB baseball and NSA softball. From circumference $ C = 2\pi r $, so $ r=\frac{C}{2\pi} $. For MLB baseball, $ C_{mlb}=9 $, so $ r_{mlb}=\frac{9}{2\pi} $. For NSA softball, $ C_{nsa}=12 $, so $ r_{nsa}=\frac{12}{2\pi}=\frac{6}{\pi} $.
## Step2: Calculate the volume of MLB baseball: $ V_{mlb}=\frac{4}{3}\pi (\frac{9}{2\pi})^3=\frac{4}{3}\pi\times\frac{729}{8\pi^3}=\frac{4\times729}{3\times8\pi^2}=\frac{2916}{24\pi^2}=\frac{243}{2\pi^2} $
## Step3: Calculate the volume of NSA softball: $ V_{nsa}=\frac{4}{3}\pi (\frac{6}{\pi})^3=\frac{4}{3}\pi\times\frac{216}{\pi^3}=\frac{4\times216}{3\pi^2}=\frac{864}{3\pi^2}=\frac{288}{\pi^2} $
## Step4: Find the ratio $ \frac{V_{mlb}}{V_{nsa}}=\frac{\frac{243}{2\pi^2}}{\frac{288}{\pi^2}}=\frac{243}{2\pi^2}\times\frac{\pi^2}{288}=\frac{243}{2\times288}=\frac{243}{576} $
Simplify $ \frac{243}{576}=\frac{27}{64}\approx0.4 $ (rounded to nearest tenth)
# Answer:
$ 0.4 $ (or $ \frac{27}{64} $ if not rounded)

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Question

Explanation:

Part a

Step2: Calculate the ratio \( \frac{C_{mlb}}{C_{nsa}}=\frac{9}{12} \)

Step2: Calculate the volume of MLB baseball: \( V_{mlb}=\frac{4}{3}\pi (\frac{9}{2\pi})^3=\frac{4}{3}\pi\times\frac{729}{8\pi^3}=\frac{4\times729}{3\times8\pi^2}=\frac{2916}{24\pi^2}=\frac{243}{2\pi^2} \)

Step3: Calculate the volume of NSA softball: \( V_{nsa}=\frac{4}{3}\pi (\frac{6}{\pi})^3=\frac{4}{3}\pi\times\frac{216}{\pi^3}=\frac{4\times216}{3\pi^2}=\frac{864}{3\pi^2}=\frac{288}{\pi^2} \)

Step4: Find the ratio \( \frac{V_{mlb}}{V_{nsa}}=\frac{\frac{243}{2\pi^2}}{\frac{288}{\pi^2}}=\frac{243}{2\pi^2}\times\frac{\pi^2}{288}=\frac{243}{2\times288}=\frac{243}{576} \)

Answer:

Part b