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dlinn [17]
2 years ago
10

The three angles of a quadrilaterals are 540, 970,860.Find the fourth angle?

Mathematics
1 answer:
m_a_m_a [10]2 years ago
4 0

Answer:123 degrees

Step-by-step explanation:

360-54-97-86=123

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8x-3 equals -24. Your correct answer is -24
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What is the solution to the inequality?<br> <br> 10x + 18 &lt; -2
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Let's solve your inequality step-by-step.

<span><span><span>10x</span>+18</span><<span>−2

</span></span>Step 1: Subtract 18 from both sides.

<span><span><span><span>10x</span>+18</span>−18</span><<span><span>−2</span>−18

</span></span><span><span>10x</span><<span>−20

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3 years ago
9.014 which digit is in the hundredths place
jekas [21]

Answer:

Step-by-step explanation:

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WILL GIVE BRAINLIEST! SUPER CONFUSED!!!
kondor19780726 [428]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

               ⬆                ⬆                 ⬆

         Low IQR      Median       High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR  is average of 59.3 and 59.6 =  59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

                    ⬆                  ⬆                      ⬆

             Low IQR         Median          High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

5 0
4 years ago
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