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10. if a point is chosen at random in the solid below, what is the prob…

Question

  1. if a point is chosen at random in the solid below, what is the probability that it lies within the hemisphere? 11. if a point in the cylinder below is chosen at random, what is the probability that it lies within the shaded region?

Explanation:

Step1: Calculate the volume of the hemisphere

The formula for the volume of a hemisphere is \(V_{hemisphere}=\frac{2}{3}\pi r^{3}\). Given \(r = 7m\), we have \(V_{hemisphere}=\frac{2}{3}\pi(7)^{3}=\frac{2}{3}\pi\times343=\frac{686\pi}{3}m^{3}\).

Step2: Calculate the volume of the cone

The formula for the volume of a cone is \(V_{cone}=\frac{1}{3}\pi r^{2}h\). Given \(r = 7m\) and \(h=12m\), we have \(V_{cone}=\frac{1}{3}\pi(7)^{2}\times12=\frac{1}{3}\pi\times49\times12 = 196\pi m^{3}\).

Step3: Calculate the total volume of the solid

\(V_{total}=V_{hemisphere}+V_{cone}=\frac{686\pi}{3}+196\pi=\frac{686\pi + 588\pi}{3}=\frac{1274\pi}{3}m^{3}\).

Step4: Calculate the probability

The probability \(P\) that a point lies within the hemisphere is \(P=\frac{V_{hemisphere}}{V_{total}}=\frac{\frac{686\pi}{3}}{\frac{1274\pi}{3}}=\frac{686}{1274}=\frac{49}{91}=\frac{7}{13}\).

Answer:

\(\frac{7}{13}\)