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Ugo [173]
3 years ago
5

The life of light bulbs is distributed normally. The standard deviation of the lifetime is 20 hours and the mean lifetime of a b

ulb is 500 hours. Find the probability of a bulb lasting for between 480 and 526 hours.
Mathematics
1 answer:
FinnZ [79.3K]3 years ago
5 0

Answer:

The probability of a bulb lasting for between 480 and 526 hours=0.74454

Step-by-step explanation:

We are given that

Standard deviation of the lifetime,\sigma=20hours

Mean, \mu=500hours

We have to find the probability of a bulb lasting for between 480 and 526 hours.

P(480

P(480

P(480

P(480

P(480

Hence, the probability of a bulb lasting for between 480 and 526 hours=0.74454

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BlackZzzverrR [31]

Answer: The correct statements are

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

David applied the distributive property.

Step-by-step explanation:

GCF = Greatest common factor

1) GCF of coefficients : (80,32,48)

80 = 2 × 2 × 2 × 2 × 5

32 = 2 × 2 × 2 × 2 × 2

48 = 2 × 2 × 2 × 2 × 3

GCF of coefficients : (80,32,48) is 16.

2) GCF of variables :(b^4,b^2,b^4)

b^4= b × b × b × b

b^2 = b × b

b^4 =b × b × b × b

GCF of variables :(b^4,b^2,b^4) is b^2

3) GCF of c^3 and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.

4) 80b^4-32b^2c^3+48b^4c

=16b^2(5b^2-2c^3+3b^2c)

David applied the distributive property.

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Solution

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For this case we can do this:

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Problem 7

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The answers are MK and PY and AH.
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The probabilities of all possible outcomes in a distribution must sum to 1.

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Luba_88 [7]

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