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PSYCHO15rus [73]
3 years ago
15

A ball is made of a plastic sphere covered with a layer of rubber. The plastic sphere has a radius of 18 in., and the ball itsel

f has a radius of 22 in.
What is the volume of the rubber layer? Use 3.14 to approximate pi and round the answer to two decimal places.

___in3
Mathematics
1 answer:
VikaD [51]3 years ago
3 0
Rl - rubber layer, s - sphere, b - ball

V_{rl}=V_b-V_s\\
V=\dfrac{4}{3}\pi R^3\\\\\\
V_{rl}=\dfrac{4}{3}\cdot3.14\cdot22^3-\dfrac{4}{3}\cdot3.14\cdot18^3\\
V_{rl}=\dfrac{4}{3}\cdot3.14(10648-5832)\\
V_{rl}=\dfrac{4}{3}\cdot3.14\cdot4816\\
V_{rl}=20162.99\text { in}^3

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earnstyle [38]

Answer:

(a) The probability that at least one of these components will need repair within 1 year is 0.0278.

(b) The probability that exactly one of these component will need repair within 1 year is 0.0277.

Step-by-step explanation:

Denote the events as follows:

<em>A</em> = video components need repair within 1 year

<em>B</em> = electronic components need repair within 1 year

<em>C</em> = audio components need repair within 1 year

The information provided is:

P (A) = 0.02

P (B) = 0.007

P (C) = 0.001

The events <em>A</em>, <em>B</em> and <em>C</em> are independent.

(a)

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P (At least 1 component needs repair)

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Thus, the probability that at least one of these components will need repair within 1 year is 0.0278.

(b)

Compute the probability that exactly one of these component will need repair within 1 year as follows:

P (Exactly 1 component needs repair)

= P (A or B or C)

=P(A\cap B^{c}\cap C^{c})+P(A^{c}\cap B\cap C^{c})+P(A^{c}\cap B^{c}\cap C)\\=[0.02\times (1-0.007)\times (1-0.001)]+[(1-0.02)\times 0.007\times (1-0.001)]\\+[(1-0.02)\times (1-0.007)\times 0.001]\\=0.02766642\\\approx 0.0277

Thus, the probability that exactly one of these component will need repair within 1 year is 0.0277.

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