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Bumek [7]
2 years ago
10

A large shipment of batteries (several million) contains 4% that are defecttive.It 60 batteries are selected randomly,What is th

e probability that at most two are defective?
Mathematics
1 answer:
Ksenya-84 [330]2 years ago
7 0

Answer:

0.5676

Step-by-step explanation:

We have been given these informations from this question ,

probability of 4% = 0.04

q = 1 - p = 1 - 0.04 = 0.96

Sample size n = 60

We solve further by Using binomial probability formula,

P(X = x) = (nCx)*px*(1 - p)^n - x

We have:

(60C0)x 0.04⁰x(0.96⁶⁰+(60C1)x0.04¹x(0.96)⁵⁹+(60C2)x0.04²x (0.96)⁵⁸

From this we would get the probability to be = 0.5676

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The complete question is:

Find the missing x- and y-values and Pythagorean triples using the identity given:

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Step-by-step explanation:

Pythagorean triples are three numbers p, q, r, that satisfy Pythagoras' theorem.

That is, they are numbers such that

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We can say tha

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Y Value: 3

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r = 3² + 4² = 25

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X Value: 5

Y Value: ? = (4)

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y = 40/10 = 4

Pythagorean Triples: (9,40,41)

X Value: ? = (6)

Y Value: 3

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x = 36/6 = 6

Pythagorean Triples: (27,36,45)

X Value: 7

Y Value: 5

p = 7² - 5² = 24

q = 2×7×5 = 70

r = 7² + 5² = 74

Pythagorean Triples: ? = (24,70,74)

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3 years ago
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In this case the answer would be 269.6.

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