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3 years ago

Find the value of x x that makes the expressions equivalent. 2(x+9); 30+18

Mathematics

2 answers:

Vlada [557]3 years ago

7 0

Answer:

$x=15$

Step-by-step explanation:

We are given the equations $2(x+9), 30+18$ to search for a value of $x$ that yields an equivalent statement. $2(x+9)=30+18\\2(x+9)=48\\x+9=24\\x=15$ .

Marina CMI [18]3 years ago

6 0

Hope this helps with the answer to your question

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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

Graph the image of triangle PQR after a reflection across the line y = 2

ololo11 [35]

Answer: Refer to the diagram below

P ' is at (-7, 9)
Q ' is at (1, 9)
R ' is at (-6, 4)

================================================

Explanation:

The diagram shows how point R moves to R'. We move up 2 units going from R(-6,0) to (-6,2). This lands us on the line of reflection. Then we move another 2 units up to land on (-6,4) which is the location of point R'.

The other points P and Q follow the same idea. Though the distances will be different from R. For P and Q, we'll move 7 units up to arrive at the line of reflection, then another 7 units to arrive at the proper locations of P' and Q', which are (-7,9) and (1,9) respectively.