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finlep [7]
3 years ago
8

Factor. 4m6−16p10 (2m3−4p5)(2m3+4p5) (2m3−4p5)2 (2m3−4p5)(2m3−4p5) (2m3+4p5)2

Mathematics
2 answers:
sesenic [268]3 years ago
6 0
I think the answer is
8(3m-80p×(6m-20p)^3×(6m+20p)^2)
Ulleksa [173]3 years ago
6 0

Answer:

\text{The factored form is }(2m^3-4p^5)(2m^3+4p^5)

Step-by-step explanation:

\text{Given the expression }4m^6−16p^{10}

we have to factor the above expression.

4m^6-16p^{10}

(2m^3)^2-(4p^5)^2

Using the identity

a^2-b^2=(a-b)(a+b)

\text{Put }a=2m^3, b=4p^5

(2m^3)^2-(4p^5)^2=(2m^3-4p^5)(2m^3+4p^5)

which is required factored form.

Option 1 is correct.

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Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of er
Vikentia [17]

Answer:

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X =69 represent the sample mean for the sample  

\mu population mean

s=4.0 represent the sample standard deviation

n=64 represent the sample size  

Assuming the X follows a normal distribution

X \sim N(\mu, \sigma)

The sample mean \bar X have the following distribution:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

The margin of error for the sample mean is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Since we have the sample standard deviation we can estimate the margin of error like this:

ME=t_{\alpha/2}\frac{s}{\sqrt{n}}

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025

We can find the degrees of freedom like this:

df=n-1=64-1=63

And we can find the critical value with the following code in excel for example: "=T.INV(0.025,63)" and we got:

t_{\alpha/2}=\pm 1.998

And the margin of error would be :

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

3 0
3 years ago
23 points to whoever can answer this question !!
yKpoI14uk [10]

I do believe that that point is -2.2 because of the fact that the difference between -1.6 and -2.4 is 0.8, and there are 4 marks in between them. So each mark's value is decreased by -0.2.

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Answer:

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Answer:

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Step-by-step explanation:

Using the rule of exponents

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