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serg [7]
2 years ago
8

How many distinct products can be formed using two different integers from the given set: {–6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4

, 5}?
Mathematics
1 answer:
zhannawk [14.2K]2 years ago
6 0

Number of distinct products that can be formed is 144

<h3>Permutation</h3>

Since we need to multiply two different integers to be selected from the set which contains a total of 12 integers. This is a permutation problem since we require distinct integers.

Now, for the first integer to be selected for the product, since we have 12 integers, it is to be arranged in 1 way. So, the permutation is ¹²P₁ = 12

For the second integer, we also have 12 integers to choose from to be arranged in 1 way. So, the permutation is ¹²P₁  = 12.

<h3>Number of distinct products</h3>

So, the number of distinct products that can be formed from these two integers are ¹²P₁ × ¹²P₁ = 12 × 12 = 144

So, the number of distinct products that can be formed is 144

Learn more about permutation here:

brainly.com/question/25925367

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The editor of a textbook publishing company is deciding whether to publish a proposed textbook. Information on previous textbook
kogti [31]

Answer:

34.86% probability that it will be huge​ success

Step-by-step explanation:

Bayes Theorem:

Two events, A and B.

P(B|A) = \frac{P(B)*P(A|B)}{P(A)}

In which P(B|A) is the probability of B happening when A has happened and P(A|B) is the probability of A happening when B has happened.

In this question:

Event A: Receiving a favorable review.

Event B: Being a huge success.

Information on previous textbooks published show that 20 % are huge​ successes

This means that P(B) = 0.2

99 % of the huge successes received favorable​ reviews

This means that P(A|B) = 0.99

Probability of receiving a favorable review:

20% are huge​ successes. Of those, 99% receive favorable reviews.

30% are modest​ successes. Of those, 70% receive favorable reviews.

30% break​ even. Of those, 40% receive favorable reviews.

20% are losers. Of those, 20% receive favorable reviews.

Then

P(A) = 0.2*0.99 + 0.3*0.7 + 0.3*0.4 + 0.2*0.2 = 0.568

Finally

P(B|A) = \frac{P(B)*P(A|B)}{P(A)} = \frac{0.2*0.99}{0.568} = 0.3486

34.86% probability that it will be huge​ success

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3 years ago
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24y/6 = 4y

3. Write as a factored expression
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