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blondinia [14]
3 years ago
7

S=2*3.14r^2 2*3.14rh for h

Mathematics
2 answers:
melisa1 [442]3 years ago
4 0
\sf\\S=2\pi r^2+2\pi rh\\\\Subtract\ 2\pi r^2\ on\ both\ sides.\\\\S-2\pi r^2=2\pi rh\\\\Divide\ by\ 2\pi r\ on\ both\ sides.\\\\h= \frac{S}{2\pi r}-\frac{2\pi r^2}{2\pi r}\\\\\\{\boxed{h=\frac{S}{2\pi r}-r}
Artyom0805 [142]3 years ago
3 0
If you would like to solve <span>S = 2 * 3.14 * r^2 + 2 * 3.14 * r * h for h, you can do this using the following steps:

</span>S = 2 * 3.14 * r^2 + 2 * 3.14 * r * h
S - 2 * 3.14 * r^2 = 2 * 3.14 * r * h       /2
S/2 - 3.14 * r^2 = 3.14 * r * h              /3.14
S/6.28 - r^2 = r * h                           /r
h = S/(6.28*r) - r

The correct result would be h = S/(6.28*r) - r.
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photoshop1234 [79]

Answer:

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4 0
3 years ago
What is the value of 9a + 6b?​
Svetlanka [38]

i) 3a+2b=9     /*3

9a+6b=27

ii) 8x+6y=60   /: 2

4x+3y=30

6 0
3 years ago
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Veronika [31]

Answer:

48

Step-by-step explanation:

4 0
2 years ago
Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf
kobusy [5.1K]

Answer:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

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".  

We have the following data given:

\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by \alpha=1-0.93=0.07 and \alpha/2 =0.035. And the critical value would be given by:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

4 0
3 years ago
Please help, 50 points! :) <br>PDF attached below
Sveta_85 [38]

Answer:

See below

Step-by-step explanation:

Attachment 1 : (a) Remember that it mentions x is the years since 1900. That would mean that the table is a bit different. To create this " new table " simply subtract 1900 from the years provided, and substitute.

To create this equation we will need a regression calculator. The equation will be as follows.

y = 0.125873x - 7.11916 ( note that you can double check this equation be substituting points from the table in the attachment )

(b) 2025 - 1900 = 126 years,

y = 0.125873(125) - 7.11916 = $ 8.614965

Minimum Wage : $ 8.614965

Attachment 2 : The rest of the problems can be solved similarly...

(a) Quadratic Regression Equation : - 0.49311x² + 23.2798x + 996.029

(b) - 0.49311(20)² + 23.2798(20) + 996.029 = 1264.381 mg/cm³

Attachment 3 : (a) Exponential Regression : 9.08292(1.09965)ˣ

(b) 9.08292(1.09965)⁶⁰ = 2713.27743\dots ( About 2713 recommendations )

8 0
3 years ago
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