Constant proportanalities in the equation y=rx 15,5 25,8 1/3 33,11

Use a proof by contradiction to show that the square root of 3 is national You may use the following fact: For any integer kirke

Ierofanga [76]

Answer:

1. Let us proof that √3 is an irrational number, using <em>reductio ad absurdum</em>. Assume that $\sqrt{3}=\frac{m}{n}$ where $m$ and $n$ are non negative integers, and the fraction $\frac{m}{n}$ is irreducible, i.e., the numbers $m$ and $n$ have no common factors.

Now, squaring the equality at the beginning we get that

$3=\frac{m^2}{n^2}$ (1)

which is equivalent to $3n^2=m^2$ . From this we can deduce that 3 divides the number $m^2$ , and necessarily 3 must divide $m$ . Thus, $m=3p$ , where $p$ is a non negative integer.

Substituting $m=3p$ into (1), we get

$3= \frac{9p^2}{n^2}$

which is equivalent to

$n^2=3p^2$ .

Thus, 3 divides $n^2$ and necessarily 3 must divide $n$ . Hence, $n=3q$ where $q$ is a non negative integer.

Notice that

$\frac{m}{n} = \frac{3p}{3q} = \frac{p}{q}$ .

The above equality means that the fraction $\frac{m}{n}$ is reducible, what contradicts our initial assumption. So, $\sqrt{3}$ is irrational.

2. Let us prove now that the multiplication of an integer and a rational number is a rational number. So, $r\in\mathbb{Q}$ , which is equivalent to say that $r=\frac{m}{n}$ where $m$ and $n$ are non negative integers. Also, assume that $k\in\mathbb{Z}$ . So, we want to prove that $k\cdot r\in\mathbb{Z}$ . Recall that an integer $k$ can be written as

$k=\frac{k}{1}$ .

Then,

$k\cdot r = \frac{k}{1}\frac{m}{n} = \frac{mk}{n}$ .

Notice that the product $mk$ is an integer. Thus, the fraction $\frac{mk}{n}$ is a rational number. Therefore, $k\cdot r\in\mathbb{Q}$ .

3. Let us prove by <em>reductio ad absurdum</em> that the sum of a rational number and an irrational number is an irrational number. So, we have $x$ is irrational and $p\in\mathbb{Q}$ .

Write $q=x+p$ and let us suppose that $q$ is a rational number. So, we get that

$x=q-p$ .

But the subtraction or addition of two rational numbers is rational too. Then, the number $x$ must be rational too, which is a clear contradiction with our hypothesis. Therefore, $x+p$ is irrational.

7 0

4 years ago

. What is the sum of the first 4 multiples of 3?

KIM [24]

Answer is 30 .I think it’s will help you

6 0

3 years ago

Read 2 more answers

A sample of 200 observations from the first population indicated that X1 is 170. A sam- ple of 150 observations from the second

nikitadnepr [17]

Answer:

a. If the P-value is smaller than the significance level, the null hypothesis is rejected.

b. Pooled proportion = 0.8

c. z = 2.7

d. As the P-value (0.0072) is smaller than the significance level (0.05), the null hypothesis is rejected.

There is enough evidence to support the claim that the proportions differ significantly.

Step-by-step explanation:

This is a hypothesis test for the difference between proportions.

We will use the P-value approach, so the decision rule is that if the P-value is lower than the significance level, the null hypothesis is rejected.

The claim is that the proportions differ significantly.

Then, the null and alternative hypothesis are:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=200 has a proportion of p1=0.85.

$p_1=X_1/n_1=170/200=0.85$

The sample 2, of size n2=150 has a proportion of p2=0.7333.

$p_2=X_2/n_2=110/150=0.7333$

The difference between proportions is (p1-p2)=0.1167.

$p_d=p_1-p_2=0.85-0.7333=0.1167$

The pooled proportion, needed to calculate the standard error, is:

$p=\dfrac{X_1+X_2}{n_1+n_2}=\dfrac{170+110}{200+150}=\dfrac{280}{350}=0.8$

The estimated standard error of the difference between means is computed using the formula:

$s_{p1-p2}=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.8*0.2}{200}+\dfrac{0.8*0.2}{150}}\\\\\\s_{p1-p2}=\sqrt{0.0008+0.00107}=\sqrt{0.00187}=0.0432$

Then, we can calculate the z-statistic as:

$z=\dfrac{p_d-(\pi_1-\pi_2)}{s_{p1-p2}}=\dfrac{0.1167-0}{0.0432}=\dfrac{0.1167}{0.0432}=2.7$

This test is a two-tailed test, so the P-value for this test is calculated as (using a z-table):

$P-value=2\cdot P(z>2.7)=0.0072$

As the P-value (0.0072) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the proportions differ significantly.

6 0

3 years ago

12 over 68 is equal to 43 over d? what is d?

pentagon [3]

12/68= 43/d

Divide 43 by 12 which is 3.583

then multiply 68 by 3.583

So d is 243.6

5 0

3 years ago

Read 2 more answers

X^2 + bx + 49 is a perfect squad trinomial what is one possible value of b? & I need help with the others also due soon!

Svetlanka [38]

20. (2) 14

A perfect square trinomial will factor into two expressions that are the same, for example: x^2 + 6x + 9 = (x + 3)(x + 3). Since this problem has a C value of 49, it will factor into (x + 7)(x + 7). 7 doubled is 14, therefore one possible value of B is 7.

21. (4) 2, -12

x^2 + 10x + 25 = 24 + 25

(x + 5)^2 = 49

x + 5 = +/- 7

x = 2, -12

22. (3) 3 + sqrt(17)

x^2 - 6x = 8

Complete the Square

x^2 - 6x + 9 = 8 + 9

(x - 3)^2 = 17

x - 3 = +/- sqrt(17)

x = 3 + sqrt(17), 3 - sqrt(17)

23. (1) 1, -5

x^2 + 4x - 5 = 0

x^2 + 4x = 5

x^2 + 4x + 4 = 5 + 4

(x + 2)^2 = 9

x + 2 = +/- 3

x = 1, -5

Hope this helps!

6 0

3 years ago