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Sati [7]
3 years ago
12

Please help me, you have to simply it

Mathematics
1 answer:
allsm [11]3 years ago
7 0

\bf \sqrt{720x^2y^3}~~ \begin{cases} 720=2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3\cdot 5\\ \qquad 2^2\cdot 2^2\cdot 3^2\cdot 5\\ \qquad (2^2)^2\cdot 3^2\cdot 5\\ \qquad 4^2\cdot 3^2\cdot 5\\ \qquad (4\cdot 3)^2\cdot 5\\ \qquad 12^2\cdot 5\\ y^3=y^{2+1}\\ \qquad y^2y \end{cases}\implies  \begin{array}{llll} \sqrt{12^2\cdot 5x^2y^2y} \\\\\\ 12xy\sqrt{5y} \end{array}

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What is 100 divided by 2
LenaWriter [7]
50 Is the answer.........
4 0
3 years ago
Read 2 more answers
A box of pencils is 5 1/4 inches wide. Seven pencils, laid side by side, take up 2 5/8 inches of the width. How many inches of t
Andre45 [30]

Answer:   Width of box is not taken up by pencils =  2\dfrac58\text{ inches}

Width of each pencil =\dfrac38\text{ inches}

Step-by-step explanation:

Given: Width of pencil box = 5\dfrac14\text{ inches}=\dfrac{21}{4}\text{ inches}

Width of seven pencils = 2\dfrac58\text{ inches}=\dfrac{16+5}{8}\text{ inches}=\dfrac{21}{8}\text{ inches}

Width of box is not taken up by pencils =  Width of pencil box  - Width of seven pencils

\left \{ {{y=2} \atop {x=2}} \right. \dfrac{21}{4}-\dfrac{21}{8}\\\\=\dfrac{21\times2-21}{8}\\\\=\dfrac{21}{8}\text{ inches}=2\dfrac58\text{ inches}

Width of box is not taken up by pencils =  2\dfrac58\text{ inches}

Width of each pencil = (Width of seven pencils ) ÷ 7

=\dfrac{21}{8}\div7\\\\=\dfrac{21}{8}\times\dfrac17\\\\=\dfrac38\text{ inches}

Width of each pencil =\dfrac38\text{ inches}

3 0
3 years ago
Read 2 more answers
Angle A and Angle B are supplementary. Angle A has a measure of 3x - 6 and Angle B has a measure of 54. What is the value of x?
Anni [7]

Answer:

44 degrees

Step-by-step explanation:

Since they are supplementary, they add up to 180. This means that

3x - 6 + 54 = 180

Solving,

3x + 48 = 180

subtract 48 from both sides

3x = 132

divide both sides by 3

x = 44 degrees

8 0
3 years ago
A recent study at UGA (March 5, 2008) suggests that students with cell phones may take more risks than students that do not have
Ira Lisetskai [31]

Answer:

(a) -2.8

b) (1) Reject H0: p = 0.50 in favor of HA: p < 0.50; there is sufficient evidence to conclude that less than half of all UGA female students who, if they had a cell phone, would be willing to walk somewhere after dark that they would normally not go.

Step-by-step explanation:

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

H0: p = 0.50

This means that:

\mu = 0.5, \sigma = \sqrt{0.5*0.5} = 0.5

In a random sample of 305 UGA female students, 128 responded that, if they had a cell phone, they would be willing to walk somewhere after dark that they would normally not go.

This means that n = 305, X = \frac{128}{305} = 0.4197

a) Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.4197 - 0.5}{\frac{0.5}{\sqrt{305}}}

z = -2.8

Pvalue:

We are testing the hypothesis that the proportion is less than 0.5, which means that the pvalue of the test is the pvalue of z = -2.81.

Looking at the z-table, z = -2.8 has a pvalue of 0.0026

(b) What is the correct conclusion for this hypothesis test at the 0.05 level of significance?

0.0026 < 0.05, which means that we reject the null hypothesis, that the proportion is 0.5, and accept the alternate hypothesis, that the proportion is less than 0.5, option (1).

6 0
3 years ago
I WILL GIVE YOU BRAINLIEST, AND 30 POINTS SO PLEASE HELP ME, IT'S URGENT!
Keith_Richards [23]
A is the right answer.
5 0
4 years ago
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