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vlabodo [156]
3 years ago
14

You need to arrange 10 of your favorite books on a small shelf. How many different ways can you arrange the books, assyming that

the ordwr of the books make a differe
Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0

Answer:

You can arrange the books 3,628,800 different ways.

Step-by-step explanation:

You can imagine that the shelf has 10 "slots" where you can place a book. Each slot can take 1 of the 10 books. However, after you have placed 1 book, there are only 9 possible books to place. After 2 have been placed, there are only 8 possible.

Going left-to-right, the first slot has 10 available books.

Total combinations: 10

The 2nd slot has 9 available books.

Total combinations: 10 * 9

The 3rd slot has 8 available books.

Total combinations: 10 * 9 * 8

This continues for all 10 slots.

Eventually, for slot 10, there is only 1 book possible.

Total combinations: 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

Multiply all numbers between 10 and 1 with each other is also known as "10 factorial" or "10!"

Thus, the total number of combinations is:

10! = 10*9*8*7*6*5*4*3*2*1 = 3628800

Answer: You can arrange the books 3,628,800 different ways.

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Answer:

84 months

Step-by-step explanation:

Maria is 7 years old, there are 12 months in a year, multiply 12 by 7.

12 * 7 = 84

Maria is 84 months old

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A padlock has a four-digit code that includes digits from 0 to 9, inclusive. What is the probability that the code does not cons
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Since there is no repetition allowed, there are 10 possibilities for the 1st digit, 9 for the 2nd, 8 for the 3rd, and 7 for the 4th. This gives a total of (10)(9)(8)(7) = 5040 four-digit codes.
For all odd digits to be used, there are 5 possibilities for the 1st digit (1,3,5,7,9), 4 for the 2nd, 3 for the 3rd, 2 for the 4th. This gives a total of (5)(4)(3)(2) = 120 codes that only use odd digits.
Therefore there are 5040 - 120 = 4920 codes that do not consist of all odd digits. The probability is 4920/5040 = 41/42.

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3 years ago
Which of the following statements about points are false?
vagabundo [1.1K]

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In an experiment, some diseased mice were randomly divided into two groups. One group was given an experimental drug, and one wa
damaskus [11]

Answer:

D)Yes, because the difference in the means in the actual experiment was more than two standard deviations from 0.

Step-by-step explanation:

We will test the hypothesis on the difference between means.

We have a sample 1 with mean M1=18.2 (drug group) and a sample 2 with mean M2=15.9 (no-drug group).

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This is approximately 2 standards deviation (z=2).

The test statistict=2.09 is bigger than the critical value and lies in the rejection region, so the effect is significant. The null hypothesis would be rejected: the difference between means is significant.

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