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tamaranim1 [39]

2 years ago

13

A computer downloads files at a constant rate. The table shows how many megabytes the computer downloads over specific lengths o

f time. What is the slope of the line that represents the computer downloads? Computer Downloads Minutes 3 5 7 9 Megabytes 11.25 18.75 26.25 33.75

1 answer:

PolarNik [594]2 years ago

5 0

The slope of the line would be 3.75.

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Subtract 2xy-8 from 5x^2+3xy+12

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Answer:

5x^2 + xy + 20

Step-by-step explanation:

Begin with 5x^2+3xy+12, Subtracting 2xy yields 5x^2 + xy + 12. Next, subtracting -8 yields 5x^2 + xy + 12 - (-8), or 5x^2 + xy + 20

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Let x be a random variable representing percentage change in neighborhood population in the past few years, and let y be a rando

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Answer:

$m=\frac{2506.833}{604.833}=4.1447$

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

$\bar x= \frac{\sum x_i}{n}=\frac{77}{6}=12.833$

$\bar y= \frac{\sum y_i}{n}=\frac{587}{6}=97.833$

And we can find the intercept using this:

$b=\bar y -m \bar x=97.833-(4.1447*12.833)=44.644$

So the line would be given by:

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Step-by-step explanation:

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

Σx = 77, Σy = 587, Σx2 = 1593, Σy2 = 70,533, and Σxy = 10040.

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:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=1593-\frac{77^2}{6}=604.833$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=10040-\frac{77*587}{6}=2506.833$

And the slope would be:

$m=\frac{2506.833}{604.833}=4.1447$

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

$\bar x= \frac{\sum x_i}{n}=\frac{77}{6}=12.833$

$\bar y= \frac{\sum y_i}{n}=\frac{587}{6}=97.833$

And we can find the intercept using this:

$b=\bar y -m \bar x=97.833-(4.1447*12.833)=44.644$

So the line would be given by:

$y=4.1447 x +44.644$

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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