Please answer a.s.a.p!

Explain why any of the four operations placed between two terms 5 and -3 sqrt (8) will result in an irrational number

Gnoma [55]

Sum/difference:

Let

$x = 5 + (-3\sqrt{8}) = 5-3\sqrt{8}$

This means that

$3\sqrt{8} = 5-x \iff \sqrt{8} = \dfrac{5-x}{3}$

Now, assume that $x$ is rational. The sum/difference of two rational numbers is still rational (so 5-x is rational), and the division by 3 doesn't change this. So, you have that the square root of 8 equals a rational number, which is false. The mistake must have been supposing that $x$ was rational, which proves that the sum/difference of the two given terms was irrational

Multiplication/division:

The logic is actually the same: if we multiply the two terms we get

$x = -15\sqrt{8}$

if again we assume x to be rational, we have

$\sqrt{8} = -\dfrac{x}{15}$

But if x is rational, so is -x/15, and again we come to a contradiction: we have the square root of 8 on one side, which is irrational, and -x/15 on the other, which is rational. So, again, x must have been irrational. You can prove the same claim for the division in a totally similar fashion.

7 0

2 years ago

What is the radius of a circle given by equation x^2 + y^2 - 2x + 8y - 47 = 0

Lera25 [3.4K]

X^2 + y^2 - 2x + 8y - 47 = 0
x^2 + y^2 - 2x + 8y = 47
(x^2 - 2x) + (y^2 + 8y) = 47
(x^2 - 2(1)x) + (y^2 + 2(4)y) = 47
(x^2 - 2(1)x + 1^2) + (y^2 + 2(4)y + 4^2) = 47 + 1^2 + 4^2
(x - 1)^2 + (y + 4)^2 = 64 = 8^2
r=8

7 0

2 years ago

Read 2 more answers

There is a large number of cubes in a bag.

Alla [95]

Marked(10)/total = 3/20 (from sample)

total/10 = 20/3 (inverting the above proportion)

total = 10•20/3 = 200/3 ≈ 67

7 0

2 years ago

Pls help asapp picture attached

Olin [163]

Answer:

50

Step-by-step explanation:

Angle 1 is 110.

Angle 3 is 2x plus 10.

Angle 1 and 3 are vertically opposite angles.

And vertically opposite angles r equal.

Which means, angle 1=angle 3,and angle 2=angke 4.

Here we only Need angle 2 and 3.

Angke 2= angle 3.

Which is 2x plus 10=110.

Therefore 2x=110-10=100

Therefore x = 100÷2=50

There u have it ma'am.

3 0

3 years ago

Read 2 more answers

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago