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If m is the product of all the integers from 2 to 11, inclus
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04 Oct 2018, 16:52
04 Oct 2018, 16:52
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94% (00:46) correct
5% (01:29) wrong based on 76 sessions
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If m is the product of all the integers from 2 to 11, inclusive, and n is the product of all the integers from 4 to 11, inclusive, what is the value of \(\frac{n}{m}\)?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{1}{6}\)
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Re: If m is the product of all the integers from 2 to 11, inclus
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13 Oct 2018, 07:25
13 Oct 2018, 07:25
2
sandy wrote:
If m is the product of all the integers from 2 to 11, inclusive, and n is the product of all the integers from 4 to 11, inclusive, what is the value of \(\frac{n}{m}\)?
Give your answer as a fraction.
Give your answer as a fraction.
Show: :: OA
\(\frac{1}{6}\)
m is the product of all the integers from 2 to 11, inclusive
So, m = (2)(3)(4)(5).....(10)(11)
n is the product of all the integers from 4 to 11, inclusive
So, n = (4)(5).....(10)(11)
What is the value of n/m?
n/m = (4)(5).....(10)(11)/(2)(3)(4)(5).....(10)(11)
= 1/(2)(3)
= 1/6
Answer: 1/6
If m is the product of all the integers from 2 to 11, inclus
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25 Feb 2024, 08:20
25 Feb 2024, 08:20
1
\(\dfrac{n}{m} = \dfrac{4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11}{2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10 \times 11}\)
cancelling all the common factors in the numerator and denominator,
\(\dfrac{n}{m} = \dfrac{1}{2 \times 3}\)
\(\dfrac{n}{m} = \dfrac{1}{6}\)
cancelling all the common factors in the numerator and denominator,
\(\dfrac{n}{m} = \dfrac{1}{2 \times 3}\)
\(\dfrac{n}{m} = \dfrac{1}{6}\)
