How do you divide 12 by 5/6

Mathematics

2 answers:

sergeinik [125]3 years ago

7 0

72/5 will be your answer

Svetlanka [38]3 years ago

6 0

Put a one under the 12 to make a fraction so it will look like this 12/1 5/6

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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

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7 0

1 year ago

Write a word sentence as an inequality. A number k minus 8.4 is greater than 21.

Travka [436]

Answer: k-8.4>21

Step-by-step explanation: K subtracted by 8.4 is greater than 21, so we start the equality off with our unknown variable K. Because K-8.4 is greater than 21, our symbol faces the direction of K-8.4 (symbol always faces the direction of the greater values).

8 0

3 years ago

Complete the table of values

Agata [3.3K]

Both problems give you a function in the second column and the x-values. To find out the values of a through f, you need to plug in those x-values into the function and simplify!

You need to know three exponent rules to simplify these expressions:
1) The negative exponent rule says that when a base has a negative exponent, flip the base onto the other side of the fraction to make it into a positive exponent. For example, $3^{-2} =
\frac{1}{3^{2} }$ .
2) Raising a fraction to a power is the same as separately raising the numerator and denominator to that power. For example, $(\frac{3}{4}) ^{3} = \frac{ 3^{3} }{4^{3} }$ .
3) The zero exponent rule says that any number raised to zero is 1. For example, $3^{0} = 1$ .


Back to the Problem:
Problem 1
The x-values are in the left column. The title of the right column tells you that the function is $y = 4^{-x}$ . The x-values are:
1) x = 0
Plug this into $y = 4^{-x}$ to find letter a:
$y = 4^{-x}\\
y = 4^{-0}\\
y = 4^{0}\\
y = 1$

2) x = 2
Plug this into $y = 4^{-x}$ to find letter b:
$y = 4^{-x}\\ 
y = 4^{-2}\\ 
y = \frac{1}{4^{2}} \\ 
y= \frac{1}{16}$

3) x = 4
Plug this into $y = 4^{-x}$ to find letter c:
$y = 4^{-x}\\ 
y = 4^{-4}\\ 
y = \frac{1}{4^{4}} \\ 
y= \frac{1}{256}$


Problem 2
The x-values are in the left column. The title of the right column tells you that the function is $y = (\frac{2}{3})^x$ . The x-values are:
1) x = 0
Plug this into $y = (\frac{2}{3})^x$ to find letter d:
$y = (\frac{2}{3})^x\\
y = (\frac{2}{3})^0\\
y = 1$

2) x = 2
Plug this into $y = (\frac{2}{3})^x$ to find letter e:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^2\\ y = \frac{2^2}{3^2}\\
y = \frac{4}{9}$

3) x = 4
Plug this into $y = (\frac{2}{3})^x$ to find letter f:
$y = (\frac{2}{3})^x\\ y = (\frac{2}{3})^4\\ y = \frac{2^4}{3^4}\\ y = \frac{16}{81}$

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Answers:
a = 1
b = $\frac{1}{16}$ 
c = $\frac{1}{256}$
d = 1
e = $\frac{4}{9}$
f = $\frac{16}{81}$