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r-ruslan [8.4K]
3 years ago
10

Sasha recorded the height and weight of all of the girls in her physical education course.

Mathematics
2 answers:
docker41 [41]3 years ago
7 0

Answer:

0.9156

Step-by-step explanation:

In the picture attached, the data is plotted (I made it using MS Excel, you can use similar programmes or a calculator).

The line that best correlates the data has the following equation:

y = 0.1037x + 50.922

and its correlation coefficient is:

R² = 0.9156

 

baherus [9]3 years ago
6 0

Answer:

Correlation coefficient, r = 0.957

Step-by-step explanation:

Let y represent the weight

Let x represent the height

y                        x                       x²                 y²                     xy

99                     61                      3721            9801               6039

104                    61                      3721            10816              6344

110                     62                     3844           12100              6820

133                     64                     4096           17689             8512

130                     65                     4225           16900            8550

142                     68                     4629           20169            9656

146                     66                     4356           21316              9636

153                     67                     4489           23409            10251

184                     69                     4761            33856            12696

185                     70                     4900           34225            12950

\sum y = 1386, \sum x = 653, \sum x^2 = 42737, \sum y^2 = 200276, \sum xy = 91454

Correlation coefficient formula,

r = \frac{n \sum xy - \sum x \sum y}{\sqrt{(n \sum x^2 - (\sum x)^2) (n \sum y^2 - (\sum y)^2)} } \\r = \frac{10* 91454 - 653*1386}{\sqrt{(10* 42737 - (653)^2) (10* 200276 - (1386)^2)} }        

Correlation coefficient, r = 0.957

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Solve the equation below for x: <br><br> logx+log(x-2)=log(3x)
Ugo [173]

Answer:

x=5

Step-by-step explanation:

Given equation is

\log\left(x\right)+\log(x-2)=\log(3x)

Apply formula: \log A+\log B=\log\left(AB\right)

\log\left(x(x-2)\right)=\log(3x)

\log\left(x^2-2x)\right)=\log(3x)

Since both sides have equal bases so we can drop the log

x^2-2x=3x

x^2-2x-3x=0

x^2-5x=0

x(x-5)=0

Which gives x=0 and x-5=0

or x=0 and x=5

Original problem \log\left(x\right)+\log(x-2)=\log(3x) is not defined at x=0 henc only x=5 is the final answer.

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4 years ago
Consider a set of one-dimensional points: 6, 12, 18, 24, 30, 42, 48.
Vinvika [58]

Answer:

1) With initial centroids 10-40, final clusters are

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second cluster (42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 378

2) With initial centroids 10-20, final clusters are

first cluster (6,12,18,24)  

second cluster (30,42,48)

And the Sum of Squared Errors (SSE) for the clustering result is 348

Step-by-step explanation:

K-means works as follows

  • the points will be clustered according to their distance to the centroids.
  • Then centroids are updated as the cluster means.
  • This process continues until clusters doesn't change anymore

1)  <u><em>initial centroids</em></u> 10-40

first cluster (6,12,18,24)  mean:15

second cluster (30,42,48) mean:40

<u><em>new centroids</em></u> 15-40

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

<u><em>final centroids</em></u> 18-45

first cluster (6,12,18,24,30) mean:18

second cluster (42,48) mean:45

Sum of Squared errors = (18-6)^{2}+(18-12)^{2}+(18-18)^{2}+(18-24)^{2}+(18-30)^{2}+(45-42)^{2}+(45-48)^{2}=378

2)<u><em>initial centroids</em></u> 10-20

first cluster (6,12)  mean:9

second cluster (18,24,30,42,48)  mean:32.4

<u><em>new centroids</em></u> 9-32.4

first cluster (6,12,18) mean:12

second cluster (24,30,42,48) mean:36

<u><em>new centroids</em></u> 12-36

first cluster (6,12,18,24) mean:15

second cluster (30,42,48) mean:40

<u><em>final centroids</em></u> 15-40

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second cluster (30,42,48) mean:40

Sum of Squared errors = (15-6)^{2}+(15-12)^{2}+(15-18)^{2}+(15-24)^{2}+(40-30)^{2}+(40-42)^{2}+(40-48)^{2}=348

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AysviL [449]

Answer:

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Step-by-step explanation:

<u><em>Explanation</em></u><u>:-</u>

Given data  chi-square test for independence is being used to evaluate the relationship between two variables

<em>Given "A" is classified into 3 categories</em>

<em>Second 'B' is classified into 4 categories</em>

In this chi-square test, we test if two attributes A and B under consideration are independent or not

We will assume that

<em>Null Hypothesis : H₀</em>: The two variables are independent

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<em>ν = ( r-1) (s-1)</em>

<em>Given data 'r' = 3  and  's' = 4</em>

<em>Degrees of freedom for independence </em>

<em>ν = ( r-1) (s-1) = ( 3-1) ( 4-1) = 2×3 =6</em>

<em>Test statistic</em>

<em>                        χ ²  =  ∑  </em>\frac{(O-E)^{2} }{E}<em></em>

<em></em>

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