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Brrunno [24]
4 years ago
11

Compute the permutation. A chemist has eight test tubes to examine. In how many orders can he do this examination? 16,777,216 20

,160 40,320
Mathematics
1 answer:
Talja [164]4 years ago
6 0

Answer:

40,320

Step-by-step explanation:

He can pick any one of the eight for the first examination.

He can pick any one of the remaining 7 for the 2nd examination.

He can pick any one of the remaining 6 for the 3rd examination.

...

He can pick any one of the remaining 2 for the 7th examination.

There's one left for the 8th examination.

Thus, there are 8×7×6×5×4×3×2×1 = 8! = 40,320 different possible orders in which the tubes can be examined.

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If HJ=2x+5, JK=3x-7, and KH=18 what is the value of x
weqwewe [10]

Answer:

  • x = 4

Step-by-step explanation:

<u>Given</u>

  • HJ=2x+5
  • JK=3x-7
  • KH=18

<u>We see J is between H and K, so </u>

  • KH = HJ + JK
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3 years ago
Tim tiene 39 pares de audífonos y 13 reproductores de música. Tim quiere vender todos los audífonos y los reproductores de músic
Lena [83]

Answer: 13 packages.

Step-by-step explanation:

This can be translated to:

Tim has 39 pairs of headphones and 13 music players. Tim wants to sell all the headphones and music players in identical packages. What is the largest number of packages that Tim can do?

To solve this we must find the largest common factor between 39 and 13.

to do this, we divide the smaller number by different integers, and see if the result also divides the larger number into an integer.

13/1 =13 and 39/13 = 3

So 13 is the largest common factor.

This means that we can make 13 packages, each with:

39/13= 3 headphones

13/13 = 1 music players

5 0
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Step-by-step explanation:

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