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mel-nik [20]
3 years ago
9

Which property would be useful in proving that the product of two rational numbers is ALWAYS rational?

Mathematics
2 answers:
aleksandrvk [35]3 years ago
8 0
C. A rational number is a number that can be expressed as the ratio of two whole numbers. The product of the rational numbers a/b and c/d are also equal to a rational number ac/bd.
Brrunno [24]3 years ago
4 0

Answer:

C)\frac{a}{b}\times \frac{c}{d}=\frac{ac}{bd}

Step-by-step explanation:

We have to find the property which would be useful in proving that the product of two rational numbers is always rational.

Rational number: It  is that number which can be expressed as \frac{p}{q} where p and q are both integers, q\neq 0

We know that product of two rational numbers  is always a rational number.

Let \frac{a}{b} and \frac{c}{d} are twp rational numbers.

Product of these two rational numbers is given by

\frac{a}{b}\times \frac{c}{d}

=\frac{ac}{bd}

a, b, c and d are integers and b\neq 0,d\neq 0

Product of two integers is always an integer.

Therefore, ac and bd are integers and bd\neq 0

Hence, we can say that product of two rational numbers is always a rational number.

Option C is true.

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Walking path across the park is represented by the equation Y equals negative 2X -7. A new path will be built perpendicular to t
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using y=mx+b to get value of b

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Carl spent $140 in six days. He spent $20 on the first day. He spent the same amount of money during the next five days Helpppp!
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5 0
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Read 2 more answers
Suppose 44% of the children in a school are girls. If a sample of 727 children is selected, what is the probability that the sam
Harman [31]

Answer:

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Suppose 44% of the children in a school are girls.

This means that p = 0.44

Sample of 727 children

This means that n = 727

Mean and standard deviation:

\mu = p = 0.44

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.44*0.56}{727}} = 0.0184

What is the probability that the sample proportion of girls will be greater than 41%?

This is 1 subtracted by the p-value of Z when X = 0.41. So

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

By the Central Limit Theorem

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

Z = \frac{0.41 - 0.44}{0.0184}

Z = -1.63

Z = -1.63 has a p-value of 0.0516

1 - 0.0516 = 0.9884

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

5 0
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