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Goryan [66]
3 years ago
14

How many palindromes of length 5 can you form using letters with the following properties: they start with a consonant, and the

consonants and vowels alternate; no letter appears more than twice. (Note: assume letters "a", "e", "i", "o", and "u" are the vowels of the English alphabet).
Mathematics
1 answer:
Zielflug [23.3K]3 years ago
6 0

Answer:

  2100

Step-by-step explanation:

There are 21 consonants that can serve as the first and last letters. There are 5 vowels that can serve as the 2nd and 4th letters. There are 20 remaining consonants that can serve as the 3rd letter. (The same consonant cannot appear in all three places.)

So, the total number of 5-letter palindromes that start with a consonant and alternate with a vowel will be ...

  21×5×20 = 2100 . . . different palindromes

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Answer:

The answer to your question is given below

Step-by-step explanation:

From the question given above, the mean score is 23.

Thus, we can obtain the distance from the mean (absolute deviation) by using the following formula:

Mean deviation = | mean – Score |

Mean = 23

For Score 21:

Mean deviation = | mean – Score |

Mean deviation = | 23 – 21 | = 2

For Score 22:

Mean deviation = | mean – Score |

Mean deviation = | 23 – 22 | = 1

For Score 28:

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For Score 29:

Mean deviation = | mean – Score |

Mean deviation = | 23 – 29 | = 6

SUMMARY:

Score >> Mean >> Absolute deviation

21 >>>>> 23 >>>>> 2

21 >>>>> 23 >>>>> 2

21 >>>>> 23 >>>>> 2

22 >>>>> 23 >>>>> 1

22 >>>>> 23 >>>>> 1

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22 >>>>> 23 >>>>> 1

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29 >>>>> 23 >>>>> 6

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