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3 years ago

How to rewright 16^3/4

Mathematics

2 answers:

ser-zykov [4K]3 years ago

4 0

8 is the answer to the question.

ivolga24 [154]3 years ago

3 0

It is 1024 i believe. 16^3/4 = 1024

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Guyys!! Im helping my lil sister here but I completely forgot it?

iren [92.7K]

Answer:B:72

Step-by-step explanation:

Because you are doing multiplication and division in this problem

5 0

3 years ago

to make an apple pie, mimi needs 2 cups of flour and 3 cups of apples. how many cups of ingredients does mimi need to make 5 pie

Lady_Fox [76]

For one apple pie, Mimi needs 2 cups of flour and 3 cups of apples, which is 5 cups in all, in order to produce how many cups she would need to make five pies you simply times the number given, number of cups, by 5. So you times 2 by 5, which is 10 and 3 by 5, which is 15 and together that makes 25 cups, 10 cups of flour and 15 cups of apples. If you had three cups of flour and needed to make 20 pies, all you would do is times three by 20, which is 60.

Hope I explained this correctly, if you need any more help, please ask.

7 0

3 years ago

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3 and 5 please with work

Crank

No need thaNK you for asking but to be honest idk

5 0

3 years ago

Use 6.84 days as a planning value for the population standard deviation. a. Assuming 95% confidence, what sample size would be r

vlabodo [156]

Answer:

The sample size required is $n = (\frac{1.96*6.84}{M})^2$ , in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.95}{2} = 0.025$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.025 = 0.975$ , so Z = 1.96.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Use 6.84 days as a planning value for the population standard deviation.

This means that $\sigma = 6.84$ .

What sample size would be required to obtain a margin of error of M days?

This is n for the given value of M. So

$M = z\frac{\sigma}{\sqrt{n}}$

$M = 1.96\frac{6.84}{\sqrt{n}}$

$M\sqrt{n} = 1.96*6.84$

$\sqrt{n} = \frac{1.96*6.84}{M}$

$(\sqrt{n})^2 = (\frac{1.96*6.84}{M})^2$

$n = (\frac{1.96*6.84}{M})^2$

The sample size required is $n = (\frac{1.96*6.84}{M})^2$ , in which M is the desired margin of error. If n is a decimal number, it is rounded up to the next integer.

8 0

3 years ago

What is 25% divided by 800

mihalych1998 [28]

It would be 0.0003125

4 0

3 years ago

Read 2 more answers