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Pani-rosa [81]
3 years ago
14

Which axiom is used to prove that the product of two rational numbers is rational

Mathematics
1 answer:
aliina [53]3 years ago
5 0

Answer:

First, a rational number is defined as the quotient between two integer numbers, such that:

N = a/b

where a and b are integers.

Now, the axiom that we need to use is:

"The integers are closed under the multiplication".

this says that if we have two integers, x and y, their product is also an integer:

if x, y ∈ Z ⇒ x*y ∈ Z

So, if now we have two rational numbers:

a/b and c/d

where a, b, c, and d ∈ Z

then the product of those two can be written as:

(a/b)*(c/d) = (a*c)/(b*d)

And by the previous axiom, we know that a*c is an integer and b*d is also an integer, then:

(a*c)/(b*d)

is the quotient between two integers, then this is a rational number.

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The triangle in the right side of the first row, the triangle in the left side of the second row and the triangle in the left side of the third row are labeled correctly.

<h3>What triangles are represented correctly?</h3>

Triangles can be defined in terms of their base and their height. Both are linear measures. The base is one side of the triangle and the height is a line perpendicular to the base that meets the vertex opposite to the base.

Based on this explanation we find that the triangle on the right of the first row, the triangle on the left of the second row and the triangle on the left of the third row are labelled correctly.

To learn more on representation of triangles: brainly.com/question/18884053

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A random sample of 200 shoppers at a local grocery store found that 135 of the 200 sampled
VMariaS [17]

Using the z-distribution, as we are working with a proportion, it is found that samples of 937 should be taken.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which:

  • \pi is the sample proportion.
  • z is the critical value.
  • n is the sample size.

The margin of error is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

In this problem, we have that:

  • The estimate is of \pi = 0.675.
  • The margin of error is of M = 0.03.
  • 95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

Then, we solve for n to find the minimum sample size.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = 1.96\sqrt{\frac{0.675(0.325)}{n}}

0.03\sqrt{n} = 1.96\sqrt{0.675(0.325)}

\sqrt{n} = \frac{1.96\sqrt{0.675(0.325)}}{0.03}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.675(0.325)}}{0.03}\right)^2

n = 936.4

Rounding up, it is found that samples of 937 should be taken.

More can be learned about the z-distribution at brainly.com/question/25890103

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