All categories

3 years ago

How does the mean absolute deviation (MAD) of the data in set 1 compare to the mean absolute deviation of the data in set 2?

Mathematics

1 answer:

frutty [35]3 years ago

4 0

Answer:

Mean Absolute deviation for (set 1): 4

Mean Absolute deviation for (set 2): 9

Set 2 has a lower (MAD) mean absolute deviation that Set 1.

Step-by-step explanation:

(set 1)

Mean: 80 + 82 + 90 = 252/3 = 84

84 - 80 = 4

84 - 82 = 2

84 - 90 = 6

Mean Absolute Deviation: 2 + 4 + 6 = 12/3 = 4

(set 2)

Mean: 60 + 80 + 82 + 90 = 312/4 = 78

78 - 60 = 18

78 - 80 = 2

78 - 82 = 4

78 - 90 = 12

Mean Absolute Deviation: 2 + 4 + 12 + 18 = 36/4 = 9

You might be interested in

Please help me!!!!!!!!!

serious [3.7K]

Download the app photomath it will help you

6 0

2 years ago

How can I do this exercise? <img src="https://tex.z-dn.net/?f=%5Cfrac%7B-2%7D%7B12%7D%20%3D%20%5Cfrac%7Bx%20%2B%201%7D%7Bx%20%2B

matrenka [14]

Answer:

-8/7 =x

Step-by-step explanation:

-2 x+1

----- = -----------

12 x+2

Using cross products

-2 (x+2) = 12 (x+1)

Distribute

-2x -4 = 12x+12

Add 2x to each side

-2x+2x -4 = 12x+2x+12

-4 = 14x +12

Subtract 12 from each side

-4-12 = 14x

-16 = 14x

Divide each side by 14

-16/14 =14x/14

-8/7 =x

3 0

3 years ago

45% of 20 is what I need help and I need u to explain it please thank u

lora16 [44]

45% of 20
20 times 0.45
= 9
The answer is 9

5 0

2 years ago

Pia needs 2/3 yard of fabric to cover a bench. Which amount of fabric is greater than

Fudgin [204]

Answer:

5/6 and 7/9

Step-by-step explanation:

A and D are less than 2/3 so B and C is correct.

Hope this helps plz mark brainliest :D

6 0

3 years ago

Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o

Ivahew [28]

Answer:

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

Step-by-step explanation:

Optimizing With Lagrange Multipliers

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

$\displaystyle C_i=3x_i+\frac{i}{40}x_i^2$

It means the cost for each generator is expanded as

$\displaystyle C_1=3x_1+\frac{1}{40}x_1^2$

$\displaystyle C_2=3x_2+\frac{2}{40}x_2^2$

$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

$g: x_1+x_2+x_3=1000$

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

$\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)$

We find all the partial derivatives of f and equate them to 0

$\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0$

$\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0$

$\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0$

$f_\lambda=x_1+x_2+x_3-1000=0$

Solving for \lambda in the three first equations, we have

$\displaystyle \lambda=3+\frac{2}{40}x_1$

$\displaystyle \lambda=3+\frac{4}{40}x_2$

$\displaystyle \lambda=3+\frac{6}{40}x_3$

Equating them, we find:

$x_1=3x_3$

$\displaystyle x_2=\frac{3}{2}x_3$

Replacing into the restriction (or the fourth derivative)

$x_1+x_2+x_3-1000=0$

$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

$\displaystyle \frac{11}{2}x_3=1000$

$x_3=181.8\ MW$

And also

$x_1=545.5\ MW$

$x_2=272.7\ MW$

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

5 0

3 years ago