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3 years ago

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A fish tank at an aquarium has a volume of 1,568 cubic feet and a depth of 8 feet. If the base of the tank is square, what is th

e length of each site of the tank?

1 answer:

Mice21 [21]3 years ago

7 0

Answer:

The base is a 14 by 14 square

Step-by-step explanation:

Here is why, if you take 1,568 and divide by 8 you get 196. Square rooting that would give you 14. You need to square root since from the question the equation is clearly x^2*8 = 1568

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The senior classes at High School A and High School B planned separate trips to Lake Compounce. The senior class at High School

storchak [24]

Answer:

Number of students in vans =15 and bus =46

Step-by-step explanation:

use variables for the vans and buses

let x= vans and y=buses

Now solve the 2 equations simultaneously

<em><u>remember</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>your</u></em><em><u> </u></em><em><u>equations</u></em>

6x+7y=412..........1

6x+9y=504.........2

make x the subject of one equation

and label it equation 3

6x=412—7y

$x = \frac{206}{3} - \frac{7y}{6} .............3$

subs 3 into 2

$6( \frac{206}{3} - \frac{7y}{6} ) + 9y = 504$

$412 - 7y + 9y = 504$

$- 7y + 9y = 504 - 412$

$2y = 92$

$y = 46...................4$

subs 4 into 1

$6x + 7(46) = 412$

$6x + 322 = 412$

$6x = 412 - 322$

$6x = 90$

$x = 15$

Therefore there are 15 students in each van and 46 students in each bus.

3 0

2 years ago

The accompanying data on x = current density (mA/cm2) and y = rate of deposition (m/min)μ appeared in a recent study.

gtnhenbr [62]

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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Answer:

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Step-by-step explanation:

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