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Elza [17]
3 years ago
14

How many anagrams are there to the word FRIDAY? In how many of these do the letters 'DAY' appear consecutively? (Disregard wheth

er the word actually exists in the dictionary or can even be pronounced)
Mathematics
1 answer:
QveST [7]3 years ago
7 0

There are 720 total different combinations, and 24 of those will have the letters DAY consecutively within all those combinations.

<u>Explanation:</u>

FRIDAY is a six letter word.

To form all combinations of letters, we would have six choices of letters to fill the first spot. Whatever letter you choose for the first spot, you then have five choices of letters to fill the second spot. Four choices of letters to fill the third spot. And so on.

So, the total number of different combinations of the six letters would be 6!

6! = 6 X 5 X 4 X 3 X 2 X 1

   = 720

DAY could appear consecutively starting in 4 different spots: the 1st, 2nd, 3rd, and 4th spots of all combinations. DAY will appear consecutively 6 times in each of those 4 starting spots.

6 X 4 = 24.

So, there are 720 total different combinations, and 24 of those will have the letters DAY consecutively within all those combinations.

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Given functions f and g, find f-g.
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5 0
2 years ago
Suppose the horses in a large stable have a mean weight of 975lbs, and a standard deviation of 52lbs. What is the probability th
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Answer:

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 975, \sigma = 52, n = 31, s = \frac{52}{\sqrt{31}} = 9.34

What is the probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable?

pvalue of Z when X = 975 + 15 = 990 subtracted by the pvalue of Z when X = 975 - 15 = 960. So

X = 990

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{990 - 975}{9.34}

Z = 1.61

Z = 1.61 has a pvalue of 0.9463

X = 960

Z = \frac{X - \mu}{s}

Z = \frac{960 - 975}{9.34}

Z = -1.61

Z = -1.61 has a pvalue of 0.0537

0.9463 - 0.0537 = 0.8926

0.8926 = 89.26% probability that the mean weight of the sample of horses would differ from the population mean by less than 15lbs if 31 horses are sampled at random from the stable

7 0
3 years ago
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