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Brrunno [24]
3 years ago
5

Given: △EDN∼△LKI, DQ , KO are altitudes, DN=12, KI=4, DQ=KO+6. Find: QN and OI.

Mathematics
1 answer:
fgiga [73]3 years ago
8 0

Answer:

  • QN = 3√7
  • OI = √7

Step-by-step explanation:

The ratio of corresponding sides DN and KI is 12 : 4 = 3 : 1. The same ratio applies to altitudes DQ and KO. Since the difference between these altitudes is 6 and the difference between their ratio units is 3-1 = 2, each ratio unit must stand for 6/2 = 3 units of linear measure. That is, ...

  DQ = (3 units)·3 = 9 units

  KO = (3 units)·1 = 3 units

Then the base lengths QN and OI can be found from the Pythagorean theorem:

  KI² = KO² +OI²

  4² = 3² +OI²

  OI = √(16 -9)

  OI = √7

  QN = 3·OI = 3√7

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Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
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In this problem:

  • The sample has 20 pills, hence n = 20.
  • 100 - 4 = 96% are acceptable, hence p = 0.96

The probability that <u>fewer that 5 in a sample of 20 pills</u> will be acceptable is:

P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{20,0}.(0.96)^{0}.(0.04)^{20} = 0

P(X = 1) = C_{20,1}.(0.96)^{1}.(0.04)^{19} = 0

P(X = 2) = C_{20,2}.(0.96)^{2}.(0.04)^{18} = 0

P(X = 3) = C_{20,3}.(0.96)^{3}.(0.04)^{17} = 0

P(X = 4) = C_{20,4}.(0.96)^{4}.(0.04)^{16} = 0

0% probability that fewer that 5 in a sample of 20 pills will be acceptable.

A similar problem is given at brainly.com/question/24863377

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