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3 years ago

Henry counted 350 lockers in his school.Hayley counted 403 lockers in her school. How does the 3 in 350 compare to the 3 in 403?

Mathematics

2 answers:

andrey2020 [161]3 years ago

7 0

Answer: The value of 3 in 350 is 10 times the value of 3 in 403 .

Step-by-step explanation:

Given : Henry counted 350 lockers in his school.

Hayley counted 403 lockers in her school.

The value of 3 in 350 is 3 hundreds and 300 and the value of 430 is 3 tens and 30.

Since 300 is 10 times 30 because $30\times10=300$

Then, the value of 3 in 350 is 10 times the value of 3 in 403 .

arsen [322]3 years ago

6 0

The 3 in 350 signifies 300 lockers. The 3 in 403 signifies just 3 lockers.

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Answer:

a) $r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

b) $m=\frac{64.5}{2000}=0.03225$

Now we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$

So the line would be given by:

$y=0.3225 x -0.27$

Step-by-step explanation:

Part a

The correlation coeffcient is given by this formula:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=4 $\sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501$

$r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

Part b

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=12000-\frac{200^2}{4}=2000$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=333-\frac{200*5.37}{4}=64.5$

And the slope would be:

$m=\frac{64.5}{2000}=0.03225$

Now we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$

And we can find the intercept using this:

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$

So the line would be given by:

$y=0.3225 x -0.27$

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3 years ago

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Using the geometric mean concept, it is found that the value of a is 18.

----------------------

The geometric mean, of a data-set of n elements, $(n_1, n_2, ..., n_n)$ , is given by:

$G = \sqrt[n]{n_1 \times n_2 \times ... \times n_n}$

That is, the nth root of the multiplication of all elements.

----------------------

In this question:

Two elements(n = 2), a and 34.
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Thus:

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We find the square of each side, so:

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$34a = 36\times17$

Simplifying both sides by 17:

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$a = \frac{36}{2}$

$a = 18$

The value of a is 18.

A similar example is given at brainly.com/question/15010240

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Step-by-step explanation:

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