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2 years ago

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Mathematics

1 answer:

vekshin12 years ago

3 0

The answer is 6*square rooted*2

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I have 6 eggs, I broke 2 eggs, fried 2 eggs, and ate 2 eggs. How many eggs do I have left.

ohaa [14]

Answer:

4 eggs are left.

Step-by-step explanation:

It is actually a tricky question and the trick is that the person, who initially have 6 eggs, has broken two eggs out of them.

Those two broken eggs were then fried and later he ate those.

It is a single incident but in question it appears that 3 different incidents took place in which the person have used all 6.

So actually, he only ate 2 eggs out of 6.

4 0

2 years ago

Read 2 more answers

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:

$y=0.3225 x -0.27$

4 0

2 years ago

Please help. I will give brainliest immediately.

DerKrebs [107]

Answer:

I'm pretty sure it is d, 5/22

7 0

2 years ago

Write the slope intercept 4x-y=1

Sergio [31]

Answer:

y = 4x - 1

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

A builder wants to build the bridge whose cross section is shown in the diagram. Two companies offer simple bids on building the

leva [86]

The length of the bridge is the distance from the beginning to the end.

The distance b between each beam is 9ft.

Let:

$I \to$ I-Beam

$d_I \to$ distance between I beam and the bridge

$b \to$ distance between each I beam

Given that:

$I = \frac 34 ft\\$

$d_I = 3 ft$

$Length = 55ft\ 6in$ --- length of the bridge

From the diagram (see attachment), there are: 6 I-beams.

So, the length of the 6 I-beams is:

$L_1 = 6 \times I$

$L_1 = 6 \times \frac 34$

$L_1 = \frac {18}4$

$L_1 = 4.5ft$

There are 2 I-beams beside the bridge

So, the distance between the 2 I-beams and the bridge is:

$d_1 =2 \times d_I$

$d_1 =2 \times 3ft$

$d_1 =6ft$

There are 5 spaces between the I-beams

So, the length of the total spaces is:

$L_2 = 5 \times b$

$L_2 = 5b$

The total length is:

$Length = L_1 + d_1 + L_2$

So, we have:

$4.5ft + 6ft + 5b = 55ft\ 6in$

Collect like terms

$5b = 55ft\ 6in - 4.5ft - 6ft$

$5b = 44.5ft\ 6in$

Convert inches to feet

$5b = 44.5ft\ + \frac{6}{12}ft$

$5b = 44.5ft\ + 0.5ft$

$5b = 45ft$

Divide both sides by 5

$b = 9ft$

Hence, the distance (b) between each beam is 9ft.

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