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Dmitry [639]
3 years ago
12

Really easy seventh grade math can I get this problem answered? Thanks!!

Mathematics
2 answers:
gizmo_the_mogwai [7]3 years ago
5 0
20/8=x/6
x=15
they are similar so the proportion is the same
Karo-lina-s [1.5K]3 years ago
5 0

x = 15

since the figures are similar then the ratios of corresponding sides are equal

\frac{x}{6} = \frac{20}{8} ( cross- multiply )

8x = 6 × 20 = 120 ( divide both sides by 8 )

x = 15


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grin007 [14]
Answer would be a (-1, ten-thirds) :)
8 0
2 years ago
-0.312 as a fraction in simplest form
Black_prince [1.1K]

Answer:

It's -39/125

Step-by-step explanation:

Rewrite the decimal number as a fraction with 1 in the denominator

Multiply to remove 3 decimal places. Here, you multiply top and bottom by 103 = 1000

Find the Greatest Common Factor (GCF) of 312 and 1000, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 8

Therefore

X=39/125

In conclusion,

−0.312=−39/125

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3 years ago
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RUDIKE [14]
C it’s what I chose for it
6 0
2 years ago
Consider the two data sets below:
alina1380 [7]

Answer:

<u><em>Option c) The data sets will have the same values of their interquartile range.</em></u>

<u><em></em></u>

Explanation:

<u>1. The values are in order: </u>they are in increasing oder, from lowest to highest value.

<u>2. Calculate the interquartile range.</u>

<em />

<em>Interquartile range</em>, IQR, is the third quartile, Q3, less the first quartile Q1:

  • IQR = Q3 - Q1

To find the first and the third quartile, first find the median:

<u>Data Set 1</u>: 19, 25, 35, 38, 41, 49, 50, 52, 59

             [19, 25, 35, 38],  41,  [49, 50, 52, 59]

                                         ↑

                                     median = 41

   

<u>Data Set 2</u>: 19, 25, 35, 38, 41, 49, 50, 52, 99

             [19, 25, 35, 38] , 41,  [49, 50, 52, 99]

                                         ↑

                                      median = 41

Now find the median of each subset: the values below the median and the values above the median.

Data set 1: <u>First quartile</u>

                [19, 25, 35, 38],

                            ↑

                           Q1 = [25 + 35] / 2 = 30

                   <u>Third quartile</u>

                   [49, 50, 52, 59]

                                ↑

                                Q3 = [50 + 52] / 2 = 51

                     IQR = Q3 - Q1 = 51 - 30 = 21

Data set 2: <u> First quartile</u>

                   [19, 25, 35, 38]

                               ↑

                               Q1 = [25 + 35] / 2= 30

                  <u>Third quartile</u>

                   [49, 50, 52, 99]

                                ↑

                                Q3 = [52 + 50]/2 = 51

                   IQR = 51 - 30 = 21

Thus, it is shown that the data sets have will have the same values for the interquartile range: IQR = 21. (option c)

This happens because replacing one extreme value (in this case the maximum value) by other extreme value does not affect the median.

<em>An outlier will change the range</em> because the range is the maximum value less the minimum value.

5 0
2 years ago
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