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2 years ago

Permutations of the word "Congratulations"

Mathematics

1 answer:

Mice21 [21]2 years ago

5 0

Answer:

There are 15 letters, but if the two A's must always be together, that's the same as if they're just one letter, so our "base count" is 14! ; note that this way of counting means that we also don't need to worry about compensating for "double counting" identical permutations due to transposition of those A's, because we don't "count" both transpositions. However, that counting does "double count" equivalent permutations due to having two O's, two N's, and two T's, so we do need to compensate for that. Therefore the final answer is 14!/(23)=10,897,286,400

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Solve 2x + 5y = 20; 3x - 10y = 37 by elimination. How do you do this? Thank-you!

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2 years ago

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2 years ago

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The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.9 mg and standard deviation 0.1

Ad libitum [116K]

Answer:

The answer to the question is;

The probability that the resulting sample mean of nicotine content will be less than 0.89 is 0.1587 or 15.87 %.

Step-by-step explanation:

The mean of the distribution = 0.9 mg

The standard deviation of the sample = 0.1 mg

The size of the sample = 100

The mean of he sample = 0.89

The z score for sample mean is given by

$Z =\frac{X-\mu}{\sigma/ \sqrt{n} }$ where

X = Mean of the sample

μ = Mean of the population

σ = Standard deviation of the population

Therefore Z = $\frac{0.89-0.90}{0.1/\sqrt{100} }$ = -1

From the standard probabilities table we have the probability for a z value of -1.0 = 0.1587

Therefore the probability that the resulting sample mean will be less than 0.89 = 0.1587 That is the probability that the mean is will be less than 0.89 is 15.87 % probability.

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3 years ago

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We know that 1 serving = 4 bananas
3 apples
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That means we also needs to half the other ones
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1/2 of 3 = 1.5 3+ 1.5=4.5
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2 years ago

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Nikolay [14]

Answer:292820839382

Step-by-step explanation:

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2 years ago

Permutations of the word "Congratulations"​

Permutations of the word "Congratulations"