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asambeis [7]
3 years ago
9

A manufacturing company regularly conducts quality control checks at specified periods on the products it manufactures. Historic

ally, the failure rate for LED light bulbs that the company manufactures is 3%. Suppose a random sample of 10 LED light bulbs is selected. What is the probability that
Mathematics
1 answer:
Rina8888 [55]3 years ago
6 0

Answer:

The probability that none of the LED light bulbs are​ defective is 0.7374.

Step-by-step explanation:

The complete question is:

What is the probability that none of the LED light bulbs are​ defective?

Solution:

Let the random variable <em>X</em> represent the number of defective LED light bulbs.

The probability of a LED light bulb being defective is, P (X) = <em>p</em> = 0.03.

A random sample of <em>n</em> = 10 LED light bulbs is selected.

The event of a specific LED light bulb being defective is independent of the other bulbs.

The random variable <em>X</em> thus follows a Binomial distribution with parameters <em>n</em> = 10 and <em>p</em> = 0.03.

The probability mass function of <em>X</em> is:

P(X=x)={10\choose x}(0.03)^{x}(1-0.03)^{10-x};\ x=0,1,2,3...

Compute the probability that none of the LED light bulbs are​ defective as follows:

P(X=0)={10\choose 0}(0.03)^{0}(1-0.03)^{10-0}

                =1\times 1\times 0.737424\\=0.737424\\\approx 0.7374

Thus, the probability that none of the LED light bulbs are​ defective is 0.7374.

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Answer:

la respuesta es:

Step-by-step explanatio:

625 x 625

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2 years ago
The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr
Kobotan [32]

Answer:

%  change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

T = 2.5 x + 0.5 x^{2}

We will approximate this distance  using the relation

f (x + dx) = f (x)+ f' (x)dx

dx = 26 - 25 = 1

T' =  2.5 + x

Therefore

f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× 25^{2} + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × 25^{2}

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

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6 0
3 years ago
How do you find out if a decimal is terminating or repeating?
leva [86]

Answer:

A terminating decimal, terminates, or stops. For example: 1.5. A repeating decimal keeps repeating itself. For example: .999999999999 (and so on). Hope I helped!

Step-by-step explanation:

7 0
2 years ago
What theorem shows that TPN ≅ TQM?
melisa1 [442]

Note: Consider the side of first triangle is TQ instead of TA.

Given:

Triangles TQM and TPN which share vertex T.

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To find:

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Solution:

In triangle TQM and TPN,

TQ\cong TP        [Given]

\angle QTM\cong \angle PTN         [Given]

TM\cong TN        [Given]

Since two sides and their including angle are congruent in both triangles, therefore both triangles are congruent by SAS postulate.

\Delta TPN\cong TQM              [SAS]

Therefore, the correct option is C.

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