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enot [183]
4 years ago
15

I need help on this question

Mathematics
1 answer:
Irina-Kira [14]4 years ago
3 0
18,15,11,14,19,15,6

Have The Median Of 15 Because When You Put Them In Order It Will Become Like This

6,11,14,15,15,18,19
⬆
This Is
The Median

If You Want To Know The Definition Of Median Well The Definition Of Median

Median-To Find The Median Place The Number You Are Given in value order And Find the middle number

The Middle number is 15 so the Median is 15

(If there are two of the same numbers average them)



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-x + 2y = 3<br> 2x – 3y = -6
s2008m [1.1K]

Answer:

x = -3

y = 0

Step-by-step explanation:

<u>Given</u><u> </u><u>equations</u><u> </u><u>:</u><u>-</u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>i</u><u> </u><u>)</u><u> </u><u> </u>

<u>-x</u><u> </u><u>+</u><u> </u><u>2</u><u>y</u><u> </u><u>=</u><u> </u><u>3</u><u> </u>

<u>-x</u><u> </u><u>=</u><u> </u><u>3</u><u> </u><u>-</u><u> </u><u>2</u><u>y</u><u> </u>

<u>x</u><u> </u><u>=</u><u> </u><u>2</u><u>y</u><u> </u><u>-</u><u> </u><u>3</u><u> </u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u>

<u>From</u><u> </u><u>(</u><u> </u><u>ii</u><u> </u><u>)</u><u> </u>

<u>2</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>y</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u>

<u>2</u><u>x</u><u> </u><u>=</u><u> </u><u>-</u><u>6</u><u> </u><u>+</u><u> </u><u>3</u><u>y</u><u> </u>

<u>x =  \frac{ - 6 + 3y}{2}</u>

<u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>Equating</u><u> </u><u>(</u><u> </u><u>iii</u><u> </u><u>)</u><u> </u><u>and</u><u> </u><u>(</u><u> </u><u>iv</u><u> </u><u>)</u>

<u>x</u><u> </u><u>=</u><u> </u><u>x</u><u> </u>

<u>2y - 3 =  \frac{ - 6 + 3y}{2}</u>

4y - 6 = -6 + 3y

4y - 3y = -6 + 6

y = 0

Putting value of y in ( iii )

x = 2y - 3

x = 2 ( 0 ) - 3

x = -3

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