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DerKrebs [107]
3 years ago
13

Explain why the greatest common factor of two numbers is sometimes 1.

Mathematics
2 answers:
Lady_Fox [76]3 years ago
8 0

Step-by-step explanation:

The greatest common factor of two numbers is sometimes 1 when they are coprime to each other.

Coprime numbers are those numbers who are relatively prime.

Now, prime numbers are those numbers who are divisible by 1 and itself only.

So, eventually when we going to find the greatest common factor of two prime numbers it comes out to be 1 only.

For example, we takes two numbers 2 and 5 ,

Factor of 2 = 1,2

Factor of 5= 1,5

So , 1 is only common factor between them.

rjkz [21]3 years ago
5 0
Because one of the numbers can be prime numbers, for example the greatest common factor of 12 and 13 it's 1 because 13 it's a prime number or because they don't have common prime factors.Prime numbers are the numbers who can be divided just by 1 and themselves. 
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DIstributive property to match equivalent expression 7(-4-x)
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Answer:

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Step-by-step explanation:

You mulitply 7 by -4 to get -28 and then mulitply 7 by -x to get -7x

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Answer:

(a) P (X < 109.78) = 0.9484.

(b) P (X < 109.78) = 0.9484.

(c) P (97 < X < 106) = 0.5328.

(d) P (X < 85.6 or X > 111.4) = 0.0369.

(e) P (X > 103) = 0.3085.

(f) P (X < 98.2) = 0.3821.

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Step-by-step explanation:

It is provided that <em>X</em> follows a Normal distribution with mean, <em>μ</em> = 100 and standard deviation, <em>σ</em> = 6.

(a)

Compute the value of P (X > 89.2) as follows:

P (X>89.2)=P(\frac{X-\mu}{\sigma}>\frac{89.2-100}{6})\\=P(Z>-1.80)\\=P(Z

Thus, the value of P (X > 89.2) is 0.9641.

(b)

Compute the value of P (X < 109.78) as follows:

P (X

Thus, the value of P (X < 109.78) is 0.9484.

(c)

Compute the value of P (97 < X < 106) as follows;

P (97 < X < 106) = P (X < 106) - P (X < 97)

                          =P(\frac{X-\mu}{\sigma}

Thus, the value of P (97 < X < 106) is 0.5328.

(d)

Compute the value of P (X < 85.6 or X > 111.4) as follows;

P (X < 85.6 or X > 111.4) = P (X < 85.6) + P (X > 111.4)

                                       =P(\frac{X-\mu}{\sigma}\frac{111.4-100}{6})\\=P(Z1.9)\\=0.0082+0.0287\\=0.0369

Thus, the value of P (X < 85.6 or X > 111.4) is 0.0369.

(e)

Compute the value of P (X > 103) as follows:

P (X>103)=P(\frac{X-\mu}{\sigma}>\frac{103-100}{6})\\=P(Z>0.50)\\=14-P(Z

Thus, the value of P (X > 103) is 0.3085.

(f)

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Thus, the value of P (X < 98.2) is 0.3821.

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Compute the value of P (100 < X < 124) as follows;

P (100< X < 124) = P (X < 124) - P (X < 100)

                          =P(\frac{X-\mu}{\sigma}

Thus, the value of P (100 < X < 124) is 0.5000.

(h)

Compute the value of <em>x</em>₁ and <em>x</em>₂ as follows if P (<em>x</em>₁ < X < <em>x</em>₂) = 0.80 as follows:

P(X_{1}

The value of <em>z</em> is ± 1.282.

The value of <em>x</em>₁ and <em>x</em>₂ are:

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Thus, the middle 80% of all heights of 5 year old children fall between 92.31 and 107.70.

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