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3 years ago

How to simplify 24yz^5/32y^2z

Mathematics

1 answer:

Maurinko [17]3 years ago

8 0

24y*z^5/32y^2*z

24/32 = 3/4
y/y^2 = 1/y
z^5/z = z^4/1

Soo now that they are all simplified the answer is 3z^4/4y

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From the sum of 3a^2−ab−2b^2 and 2a^2+5ab−3b^2 subtract a^2−3ab−4b^2

Lubov Fominskaja [6]

$(3a^2-ab-2b^2) + (2a^2+5ab-3b^2) - (a^2-3ab-4b^2)$

First we distribute the negative throughout the last polynomial:

$(3a^2-ab-2b^2) + (2a^2+5ab-3b^2) + (-a^2+3ab+4b^2)$

Combine like terms:

$4a^2 + 7ab - b^2$

3 0

2 years ago

Explain type i error and give an example. explain type ii error and give an example. what is the best way to reduce both kinds o

Leviafan [203]

Type I error says that we suppose that the null hypothesis exists rejected when in reality the null hypothesis was actually true.

Type II error says that we suppose that the null hypothesis exists taken when in fact the null hypothesis stood actually false.

<h3>What is Type I error and Type II error?</h3>

In statistics, a Type I error exists as a false positive conclusion, while a Type II error exists as a false negative conclusion.

Making a statistical conclusion still applies uncertainties, so the risks of creating these errors exist unavoidable in hypothesis testing.

The probability of creating a Type I error exists at the significance level, or alpha (α), while the probability of making a Type II error exists at beta (β). These risks can be minimized through careful planning in your analysis design.

Examples of Type I and Type II error

Type I error (false positive): the testing effect says you have coronavirus, but you actually don’t.
Type II error (false negative): the test outcome says you don’t have coronavirus, but you actually do.

To learn more about Type I and Type II error refer to:

brainly.com/question/17111420

#SPJ4

7 0

1 year ago

Graph the solution x

mixas84 [53]

Answer:

I think you need to put more info

Step-by-step explanation:

8 0

2 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Mkey [24]

For this case, we have to:

By definition, we know:

The domain of $f (x) = \sqrt [3] {x}$ is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root. Thus, it will always be defined.

So, we have:

$y = \sqrt [3] {x-2}$ with $x = 0$ : $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x+2}$ with $x = 0:\ y = \sqrt [3] {2}$ is also defined.

$f (x) = \sqrt {x}$ has a domain from 0 to ∞.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .