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Dennis_Churaev [7]

3 years ago

8

Which expression is equivalent to StartRoot StartFraction 2 x Superscript 5 Baseline Over 18 EndFraction EndRoot? Assume x great

er-than-or-equal-to 0. StartFraction x squared StartRoot x EndRoot Over 3 EndFraction StartFraction 3 StartRoot x EndRoot Over x squared EndFraction StartFraction StartRoot x EndRoot Over 3 x squared EndFraction StartFraction 2 x StartRoot x EndRoot Over 3 EndFraction

2 answers:

german3 years ago

6 0

Answer:

A. $\frac{x^2\sqrt{x}}{3}$

Step-by-step explanation:

We have been given an expression $\sqrt{\frac{2x^5}{18}}$ . We are asked to find which expression is equivalent to our given expression.

Let me simplify our given expression. First of all we will divide 18 by 2 as:

$\sqrt{\frac{2x^5}{2\cdot 9}}$

$\sqrt{\frac{x^5}{9}}$

Now we will rewrite $x^5 \text{ as } x^4\cdot x$ .

$\sqrt{\frac{x^4\cdot x}{9}}$

Now we will write $x^4\text{ and } 9$ as perfect squares.

$\sqrt{\frac{(x^2)^2\cdot x}{3^2}}$

Let us factor our perfect square from square-root.

$\frac{x^2\sqrt{x}}{3}$

Therefore, option A is the correct choice.

Elodia [21]3 years ago

5 0

Answer:

it's A on edge

Step-by-step explanation:

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A field researcher is gathering data on the trunk diameters of mature pine and spruce trees in a certain area. The following are

lidiya [134]

Answer:

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

Step-by-step explanation:

Using a .10 significance level, can he conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree

The null hypothesis is that they are equal(that is, subtraction between the means is 0), while the alternate hypothesis is that they are greater(that is, subtraction between the means is larger than 0). So

$H_0: \mu_1 - \mu_2 = 0$

$H_a: \mu_1 - \mu_2 > 0$

In which $\mu_1$ is the mean pine trees height while $\mu_2$ is the mean spruce trees height.

The test statistic is:

$z = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

Mean trunk diameter (cm) 35 30:

Pine trees - 35

Spruce trees - 30

The sample mean is given by the subtraction of the means. So

$X = 35 - 30 = 5$

Sample Size 40 80

Population variance 160 160

This means that the standard error for each sample is given by:

$s_1 = \frac{\sqrt{160}}{\sqrt{40}} = 2$

$s_1 = \frac{\sqrt{160}}{\sqrt{80}} = \sqrt{2}$

The standard error of the difference is the square root of the sum squared of the standard error of each sample.

$s = \sqrt{2^2 + (\sqrt{2})^2} = \sqrt{6} = 2.4495$

Test statistic:

$z = \frac{X - \mu}{s}$

$z = \frac{5 - 0}{2.4495}$

$z = 2.04$

Pvalue of the test:

Probability of a sample mean larger than 5, which is 1 subtracted by the pvalue of Z when X = 5.

Looking at the z-table, z = 2.04 has a pvalue of 0.9793.

1 - 0.9793 = 0.0207

As the pvalue of the test is 0.02097 < 0.1, using a .10 significance level, he can conclude that the average trunk diameter of a pine tree is greater than the average diameter of a spruce tree.

8 0

3 years ago

Can yall help, whoever answers correctly will get brainllest!

melamori03 [73]

Answer:

A,C,D

Step-by-step explanation:

4 0

3 years ago

How do you solve this question?

dezoksy [38]

What is the x interval??

8 0

3 years ago

What is the greatest common factor of 40 and 27?

Olenka [21]

To find the greatest common factor of 40 and 27, we can first start by making a list of factors for each number.

The factors of 27 are: 1, 3, 9, 27

The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40

When looking at these numbers, we can see that 1 is the greatest common factor of both 40 and 27.

7 0

3 years ago

Solve for xxx. Enter the solutions from least to greatest. x^2 +7=43x 2 +7=43x, squared, plus, 7, equals, 43 \text{lesser }x =le

vaieri [72.5K]

Answer:

Step-by-step explanation:

x²+7=43x

x²-43x+7=0

$x=\frac{43 \pm \sqrt{(43)^2-4(1)(7)} }{2(1)} \\=\frac{43 \pm \sqrt{1849-28} }{2} \\=\frac{43 \pm \sqrt{1821} }{2} \\\approx \frac{43 \pm 42.673}{2}\\\approx 0.1635,42.8365$

4 0

3 years ago