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lana66690 [7]
2 years ago
9

Find the MAD of:1,10,7,6,4,8DataMeanDifferenc

Mathematics
1 answer:
pochemuha2 years ago
5 0

Answer:

2.3

Step-by-step explanation:

Mean: 1 + 10 + 7 + 6 + 4 + 8 = 36/6 = 6

6 - 1 = 5

6 - 10 = 4

6 - 7 = 1

6 - 6 = 0

6 - 4 = 2

6 - 8 = 2

mean absolute deviation: 5 + 4 + 1 + 0 + 2 + 2 = 14/6 = 2.3

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Can someone help me I don’t know how to do this it’s hard and confusing me
Svetach [21]

9514 1404 393

Answer:

  • increasing on 0 < x < 2
  • f(2) = 4
  • decreasing on 2 < x < 5

Step-by-step explanation:

You need to know that an even function is symmetrical about the y-axis. The graph to the right of the y-axis is the mirror image of the graph to the left. With that in mind, the questions are easily answered.

It might help to draw the function graph, as we did in the attached. Then it is more straightforward to see what is happening in each of the intervals.

The function is increasing in the interval 0 < x < 2.

The function value at x=2 is 4: f(2) = 4.

The function is decreasing in the interval 2 < x < 5.

4 0
2 years ago
I WILL MARK BRAINLIEST
Ivanshal [37]

Answer:

D

Step-by-step explanation:

<u></u>

A boxed area represents 25% of the data set. This means that 70-85 is 25% and 85-100 is 25%

25%+25%=50%

6 0
2 years ago
3. Seth uses a bowl to fill a container with soil. The
san4es73 [151]
3/4 x 13/1 = 39/4 or 9 3/4 or 9.75
8 0
1 year ago
OMG HELP RN
Brilliant_brown [7]

Answer:

C would be the answer

Hope this helps!

4 0
2 years ago
Let​ T: set of real numbers R Superscript nℝnright arrow→set of real numbers R Superscript mℝm be a linear​ transformation, and
Klio2033 [76]

Answer:

\{T(v_1), T(v_2), T(v_3)\} is linearly dependent set.

Step-by-step explanation:

Given:  \{v_1,v_2,v_3\} is a linearly dependent set in set of real numbers R

To show: the set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

Solution:

If \{v_1,v_2,v_3,...,v_n\} is a set of linearly dependent vectors then there exists atleast one k_i:i=1,2,3,...,n such that k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0

Consider k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

A linear transformation T: U→V satisfies the following properties:

1. T(u_1+u_2)=T(u_1)+T(u_2)

2. T(au)=aT(u)

Here, u,u_1,u_2∈ U

As T is a linear transformation,

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\

As \{v_1,v_2,v_3\} is a linearly dependent set,

k_1v_1+k_2v_2+k_3v_3=0 for some k_i\neq 0:i=1,2,3

So, for some k_i\neq 0:i=1,2,3

k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0

Therefore, set \{T(v_1), T(v_2), T(v_3)\} is linearly dependent.

6 0
3 years ago
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