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puteri [66]
3 years ago
11

Benny sees an inflatable beach ball for sale at a pool store. The packaging says that the ball has a diameter of 20 inches. Benn

y purchases and inflates the
ball. What is its volume?
O A 400x
3
in
OB. 1,0007 in
0 C 4,000
in3
D
32,000
3
in?
Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
7 0

Answer:

Volume of beach ball = 4176.2 inch³ (Approx.)

Step-by-step explanation:

Given:

Diameter of beach ball = 20 inches

Find:

Volume of beach ball

Computation:

Radius of beach ball = Diameter / 2

Radius of beach ball = 20 / 2

Radius of beach ball = 10 inches

We know that beach ball is a sphere shape

So,

Volume of beach ball = Volume of sphere

Volume of sphere = [4/3][π][r³]

Volume of beach ball = [4/3][π][r³]

Volume of beach ball = [4/3][22/7][10³]

Volume of beach ball = [1.33][3.14][1,000]

Volume of beach ball = 4176.2 inch³ (Approx.)

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3 0
3 years ago
15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.
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Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = 30C5 = \frac{30 !}{(30-5)!5!} \\

30C5 = 142506 ways

Number of ways of selecting 5 good bulbs  from 26 bulbs = 26C5 = \frac{26 !}{(26-5)!5!} \\

26C5 = 65780 ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = \frac{26C4  * 4C1}{30C5}

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) =  \frac{26C3  * 4C2}{30C5}

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) +  Pr(2 defective) +  Pr(3 defective) +  Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) =  \frac{26C2  * 4C3}{30C5}

Pr(3 defective) = 0.009

Pr(4 defective) =  \frac{26C1  * 4C4}{30C5}

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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