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Nikitich [7]
3 years ago
14

On a particular production line, the likelihood that a light bulb is defective is 10%. seven light bulbs are randomly selected.

What is the probability that at most 4 of the light bulbs will be defective
Mathematics
1 answer:
amid [387]3 years ago
6 0

Answer:

0.9995

Step-by-step explanation:

10% = 0.10

1 - 0.10 = 0.9

n = number of light bulbs = 7

we calculate this using binomial distribution.

p(x) = nCx × p^x(1-p)^n-x

our question says at most 4 is defective

= (7C0 × 0.1⁰ × 0.9⁷) + (7C1 × 0.1¹ × 0.9⁶) + (7C2 × 0.1² × 0.9⁵) + (7C3 × 0.1³ × 0.9⁴) + (7C4 × 0.1⁴ × 0.9³)

= 0.478 + 0.372 + 0.1239 + 0.023 + 0.0026

= 0.9995

we have 0.9995 probability that at most 4 light bulbs are defective.

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Genevieve wants to verify that negative x is the simplified expression of one-fifth (5 x minus 20) minus one-half (4 x minus 8).
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Answer: -x

Apply Multiplicative Distribution Law:- 1/5 * 5x - 1/5 * 20 - 1/2 * 4x - 1/2 *(-8)

Determine the sign for multiplication or division: 1/5 * 5x - 1/5 * 20 - 1/2 * 4x + 1/2 * 8

Cross out the common factor: x - 1/5 * 20 - 1/2 * 4x + 1/2 * 8

Cross out the common factor: x - 4 - 1/2 * 4x + 1/2 * 8 - 4 - 2x + 1/2 * 8

Cross out the common factor: x - 4 - 2x + 1/2 * 8

Cross out the common factor:  x - 4 - 2x + 4

Reorder and gather like terms: (x - 2x) + (- 4 + 4)

Collect coefficients for the like terms: (1 - 2) * x + (- 4 + 4)

Calculate the sum or difference :  - x + (- 4 + 4)

The sum of two opposites equals 0: -x

Answer: -x

Learn more about Multiplicative Distribution Law here:brainly.com/question/2898526

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