<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%20%2B%5Csqrt%7B5%7D%20%3D%5Csqrt%7B12%7D" id="TexFormula1" title="\sqrt{x} +\sq

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nataly862011 [7]

3 years ago

rt{5} =\sqrt{12}" alt="\sqrt{x} +\sqrt{5} =\sqrt{12}" align="absmiddle" class="latex-formula">

Mathematics

2 answers:

creativ13 [48]3 years ago

4 0

The value of x is 17 - 4sqrt(15)

Pie3 years ago

3 0

Answer:

x=17-4*\sqrt{15}

Step-by-step explanation:

$\sqrt{x}+\sqrt{5}=\sqrt{12}\\\sqrt{x}=\sqrt{12}-\sqrt{5}\\x=(\sqrt{12}-\sqrt{5})^{2}\\x=12+5-4*\sqrt{15}\\x=17-4*\sqrt{15}.$

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Assume, for the sake of this question, that the data were collected through a well-designed, well-implemented random sampling me

amid [387]

Answer:

Since the pvalue of the test is 0.2743 > 0.1, the threshold probably was met.

Step-by-step explanation:

The widget manufacturing company had established a threshold of 60% preferring the proposed new widget to move forward with producing the new widgets.

This means that at the null hypothesis we test if the proportion is at least 60%, that is:

$H_{0}: p \geq 0.6$

And the alternate hypothesis is:

$H_{a}: p < 0.6$

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.

0.6 is tested at the null hypothesis:

This means that:

$\mu = 0.6$

$\sigma = \sqrt{0.6*0.4}$

Three hundred thirty-eight of 575 respondents reported preferring the proposed new widget.

This means that $n = 575, X = \frac{338}{575} = 0.5878$

Value of the test-statistic:

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

$z = \frac{0.5878 - 0.6}{\frac{\sqrt{0.6*0.4}}{\sqrt{575}}}$

$z = -0.6$

Pvalue of the test and decision:

We want to find the probability of a proportion of 0.5878 or lower, which is the pvalue of z = -0.6.

Looking at the z-table, z = -0.6 has a pvalue of 0.2743.

Since 0.2743 > 0.1, the threshold probably was met.

3 0

3 years ago

Evaluate 5xyz + 3y for x=2.5, y=12, and z=4

Reika [66]

Answer:

636

Step-by-step explanation:

5xyz+3y

5(2.5)(12)(4)+3(12)

(12.5)(48)+36

600+36

636

8 0

4 years ago

A bank sampled its customers to determine the proportion of customers who use their debit card at least once each month. A sampl

Zinaida [17]

Answer:

(0.084,0.396)

Step-by-step explanation:

The 99% confidence interval for the proportion of customers who use debit card monthly can be constructed as

$p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }$

$p=\frac{x}{n}$

$p=\frac{12}{50}$

$p=0.24$

$q=1-p=1-0.24=0.76$

$\frac{\alpha }{2} =\frac{\0.01 }{2}=0.005$

$p-z_{\frac{\alpha }{2}} \sqrt{\frac{pq}{n} }$

$0.24-z_{0.005} \sqrt{\frac{0.24*0.76}{50} }$

$0.24-2.58(0.0604)$

$0.24-0.155832$

By rounding to three decimal places we get,

$0.084$

The 99% confidence interval for the proportion of customers who use debit card monthly is (0.084,0.396).

3 0

3 years ago

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gavmur [86]

Answer:

68/35 hope this helps

Step-by-step explanation:

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4 years ago

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LiRa [457]

8:15 is the answer assuming the clock is at PM.

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3 years ago

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