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3 years ago

28,32,47,16,40,35,38,54 what is the MAD of the data

Mathematics

2 answers:

horsena [70]3 years ago

5 0

Answer:

Mean Absolute Deviation (MAD): 8.5

Step-by-step explanation:

Xmean = (28 + 32 + 47 + 16 + 40 + 35 + 38 + 54)/8 = 290/8

Xmean = 36.25

MAD =

(28 - 36.25) + (32 - 36.25) + (47 - 36.25) + (16 - 36.25) + (40 - 36.25) + (35 - 36.25) + (38 - 36.25) + (54 - 36.25) divided by 8

a break down of each ( )

(8.25) + (4.25) + (10.75) + (20.25) + (3.75) + (1.25) + (1.75) + (17.75) divided by 8 = 68

then you take 68/8 = Mean Absolute Deviation 8.5

sergij07 [2.7K]3 years ago

5 0

Answer:

Mean Absolute Deviation (M.A.D): 8.5

Step-by-step explanation:

1. To find the mean absolute deviation of the data, start by finding the mean of the data set.

2. Find the sum of the data values, and divide the sum by the number of data values.

3. Find the absolute value of the difference between each data value and the mean: |data value – mean|.

4. Find the sum of the absolute values of the differences.

5. Divide the sum of the absolute values of the differences by the number of data values.

Hope that helped! :)

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Answer:

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64π

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3 years ago

What is the area of rectangle PQRS with vertices at P(0,2), Q(0,10), R(4,10), and S(4,2)?

nadezda [96]

It’s 4,2 the and the other on PQRS

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2 years ago

The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

and volume,

$V=\frac{1}{3}\pi r^2 h$

To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

$\implies \pi r \sqrt{2r^2+r^2}=8$

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

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finlep [7]

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blagie [28]

Answer:

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Step-by-step explanation:

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