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NNADVOKAT [17]
3 years ago
8

If a number is odd, then it is an integer and a prime number.

Mathematics
1 answer:
Elza [17]3 years ago
7 0

Answer:

The conclusion of the conditional statement is;

it is an integer and a prime number

Step-by-step explanation:

The statement that the argument claims to be true or the statement that the rest of a conditional statement leads to is the conclusion of a conditional statement

Given a conditional statement, also known as an if-then statement, which can be represented symbolically as p → q, such that the hypothesis is p and the conclusion is q.

In the given statement, the hypothesis, p = a number is odd, while the conclusion, q =  it is an integer and a prime number

Therefore, the conclusion of the conditional statement is it is an integer and a prime number.

However, only the converse of the statement is true, which is given as follows;

If a number is an integer and a prime number, then it is an odd number

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Sheldon wants to buy a laptop that costs $ 450 from an electronics store . He has two coupons to choose from coupon A which give
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Answer:

Coupon B

Step-by-step explanation:

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Required

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Coupon\ A = \$80

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The J.O. Supplies Company buys calculators from a non-US supplier. The probability of a defective calculator is 10 percent. If 3
RSB [31]

Answer:

There is a 24.3% probability that one of the calculators will be defective.

Step-by-step explanation:

For each calculator, there are only two possible outcomes. Either it is defective, or it is not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

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This means that p = 0.1

If 3 calculators are selected at random, what is the probability that one of the calculators will be defective

This is P(X = 1) when n = 3. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

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There is a 24.3% probability that one of the calculators will be defective.

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