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Kipish [7]
2 years ago
12

Choose the option that best completes the statement below. In finding the number of permutations for a given number of items, __

___.
A) the number of permutations is limited by the number of items which are alike.
B) it doesn’t matter if some of them are indistinguishable .
C) the number of permutations is determined by the number of items that don’t repeat.
D) the number of permutations isn’t dependent on the items that repeat.
Mathematics
1 answer:
vodomira [7]2 years ago
7 0
Let’s look at the permutations of the letters “ABC.” We can write the letters in any of the following ways:
ABC
ACB
BAC
BCA
CBA
CAB
Since there are 3 choices for the first spot, two for the next and 1 for the last we end up with (3)(2)(1) = 6 permutations. Using the symbolism of permutations we have: 3 P_{3}=(3)(2)(1)=6. Note that the first 3 should also be small and low like the second one but I couldn’t get that to look right.

Now let’s see how this changes if the letters are AAB. Since the two As are identical, we end up with fewer permutations.
AAB
ABA
BAA
To make the point a bit better let’s think of one A are regular and one as bold A.
A
BA and ABA look different now because we used bold for one of the As but if we don’t do this we see that these are actual the same. If they represented a word they would be the same exact word.

So in this case the formula would be \frac{3 P_{3} }{2!}= \frac{(3)(2)(1)}{(2)(1)}= \frac{6}{2}=3. We use 2! In the denominator because there are 2 repeating letters. If there were three we would use 3!


Hopefully, this is enough to let you see that the answer is A. The number of permutations is limited by the number of items that are identical.



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the time taken by a student to the university has been shown to be normally distributed with mean of 16 minutes and standard dev
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Answer:

a) 2.84% probability that he is late for his first lecture.

b) 5.112 days

Step-by-step explanation:

When the distribution is normal, we use 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.

In this question, we have that:

\mu = 16, \sigma = 2.1

a. Find the probability that he is late for his first lecture.

This is the probability that he takes more than 20 minutes to walk, which is 1 subtracted by the pvalue of Z when X = 20. So

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

Z = \frac{20 - 16}{2.1}

Z = 1.905

Z = 1.905 has a pvalue of 0.9716

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2.84% probability that he is late for his first lecture.

b. Find the number of days per year he is likely to be late for his first lecture.

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0.0284*180 = 5.112 days

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