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Sophie [7]
4 years ago
12

A farm is 9 in wide and 6 in tall. if it is redused to a width of 3 in then how tall will it be

Mathematics
1 answer:
lora16 [44]4 years ago
4 0
It will still be 6 units high. Even if the width is reduced, that does not effect the height of  the farm.
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In a walk fir charity,you. Walk at a rate of 100 meters per minute.how long does it take you to walk 2 kilometers
Ivahew [28]
20 minutes

2 kilometers equal 2000 meters

100/1= 2000/X
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3 years ago
What is the answer to 3x-(2x+4)
sammy [17]

Answer:

= x − 4

Step-by-step explanation:

Let's simplify step-by-step.

3x−(2x+4)

Distribute the Negative Sign:

=3x+−1(2x+4)

=3x+−1(2x)+(−1)(4)

=3x+−2x+−4

Combine Like Terms:

=3x+−2x+−4

=(3x+−2x)+(−4)

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Associations In Data:Question 10
Karo-lina-s [1.5K]

Answer:

\bar X = \frac{62+63+68+72+79+80+83+93+94+95}{10}= 78.9

|62-78.9| = 16.9

|63-78.9| = 15.9

|68-78.9| = 10.9

|72-78.9| = 6.9

|79-78.9| = 0.1

|80-78.9| = 1.1

|83-78.9| = 4.1

|93-78.9| = 14.1

|94-78.9| = 15.1

|95-78.9| = 16.1

MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n}

And replacing we got:

MAD =\frac{16.9+15.9+10.9+6.9+0.1+1.1+4.1+14.1+15.1+16.1}{10}= 10.12

And the best anwer is

10.12

Step-by-step explanation:

We have the following data given:

62 63 68 72 79 80 83 93 94 95

And we need to begin finding the mean with the following formula:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

And replacing we got:

\bar X = \frac{62+63+68+72+79+80+83+93+94+95}{10}= 78.9

Now we can find the mean absolute deviation like this:

|62-78.9| = 16.9

|63-78.9| = 15.9

|68-78.9| = 10.9

|72-78.9| = 6.9

|79-78.9| = 0.1

|80-78.9| = 1.1

|83-78.9| = 4.1

|93-78.9| = 14.1

|94-78.9| = 15.1

|95-78.9| = 16.1

And finally we can find the mean abslute deviation with the following formula:

MAD = \frac{\sum_{i=1}^n |X_i -\bar X|}{n}

And replacing we got:

MAD =\frac{16.9+15.9+10.9+6.9+0.1+1.1+4.1+14.1+15.1+16.1}{10}= 10.12

And the best anwer is

10.12

8 0
4 years ago
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Liula [17]

Answer:

  perpendicular line through a point on a line

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So. points G and H are both equidistant from points B and D. A line between them will intersect point C at right angles to AB.

Segment GH is perpendicular to AB through point C (on AB).

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3 years ago
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