<img src="https://tex.z-dn.net/?f=%5Chuge%5Cbold%5Cred%7B%7BHELP%7D%7D" id="TexFormula1" title="\huge\bold\red{{HELP}}" alt="\hu

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Oxana [17]

3 years ago

ge\bold\red{{HELP}}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

Travka [436]3 years ago

6 0

$\\ \sf\longmapsto \sqrt{512x^3}=\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 2\times 2x^2x}=16x$
$\\ \sf\longmapsto x-8=-5\implies x=3$
$\\ \sf\longmapsto \sqrt{x-8}=\sqrt{-x+2}\implies x-8=-x+2\implies 2x=10\implies x=5$

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Oxana [17]

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3 years ago

Could anyone solve this equation

Ainat [17]

I)
8 - 3x = 23
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ii)
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5x = 50
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iii)
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Nataly_w [17]

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3 0

3 years ago

Read 2 more answers

A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.

oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

$H_0: \mu=2.4\\\\H_a:\mu> 2.4$

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58$

The degrees of freedom for this sample size are:

$df=n-1=45-1=44$

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>2.5801)=0.0066$

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0

3 years ago

Approximate the value of <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B5%7D%20" id="TexFormula1" title=" \sqrt{5} " alt=" \

k0ka [10]

2.2367 , therefore to the nearest hundredth it would be 2.237

5 0

4 years ago