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Drupady [299]
4 years ago
13

Layla used 0.482 gram of salt in her experiment. Maurice use 0.51 gram of salt. Who used the greater amount of salt?

Mathematics
2 answers:
dem82 [27]4 years ago
8 0
0.51>0.482
The number (5) in the tenth column (the number on the right of the decimal)
is greater than the other number in the tenth column (4)

0.51 is greater than 0.482

Hope this helps! A thanks/brainiest answer would be appreciated :)
BigorU [14]4 years ago
4 0

Answer:  Maurice used greater amount of salt.

Step-by-step explanation:

Since we have given that

Quantity of salt in her experiment that Layla used = 0.482 grams

Quantity of salt in her experiment that Maurice used = 0.51 grams

As we know that

0.51=\dfrac{51}{100}\\\\and\\\\0.482=\dfrac{482}{1000}

So, for comparison we will write 0.51 as \dfrac{510}{1000}

Now, we can compare easily:

\dfrac{510}{1000}>\dfrac{482}{1000}\\\\=0.51>0.482

Hence, Maurice used greater amount of salt.

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M = 15.625 km

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σ ≈ 15.92 km

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Step-by-step explanation:

The given data are;

Depth {}                                 Magnitude

0.76 {}                                    0.84

4.93 {}                                    0.47

8.16 {}                                     0.35

33.58 {}                                  1.32

21.2 {}                                     1.61

35.03 {}                                  4.57

10.05 {}                                   5.52

47.91 {}                                    1.99

For the Depth, we have;

The mean, μ = (0.76+4.93+8.16+33.58+21.2+35.03+10.05+47.91)/8 =20.2025 km

The median, M = The (n + 1)/2th term after arranging the term in increasing order as follows;

0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 , the median is therefore;

(8 + 1)/2th term or the 4.5th term which is 10.05 + (21.2 - 10.05)/2 = 15.625 km

The Range = The highest - The lowest value = 47.91 - 0.76 = 47.15 km

The Standard deviation of, σ, is given as follows;

\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu  \right )^{2} }{N}}

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x_i = The individual data point = (0.76, 4.93, 8.16, 10.05, 21.2, 33.58, 35.03, 47.91 )

N = The total number of data point = 8

Substituting, (using Microsoft Excel) we get;

\sigma =\sqrt{\dfrac{\sum \left (x_i-20.2025  \right )^{2} }{8}} \approx 15.92 \ km

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