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MArishka [77]
3 years ago
8

Can someone help this is hard

Mathematics
1 answer:
Svetlanka [38]3 years ago
3 0
They are linear pairs
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Can you help with Issue A?
defon

Answer:

$16,532.98

Step-by-step explanation:

Her last deposit = 500, second last = 500 x 1.05, third last = 500 x 1.05^2................ first deposit = 500 x 1.05^19

=500 +500 x 1.05 + 500 x 1.05^2 + 500 x 1.05^19

= 500(1.05^20-1) divided by (1.05-1)

= 16532.98

7 0
3 years ago
Two variables are correlated with r = 0.44.
butalik [34]
It's moderate positive because when the number is closer to zero the weaker it is and the closer to one its stronger so since 0.44 is in more in the middle it moderate and since there is no negative sign which is - then it moderate positive. Btw I took the test and it was right. 
3 0
3 years ago
AC = 3x -8, BC = -7 + 2x, and AB = 8.<br> Find BC.
Rainbow [258]

Answer:

AB+BC+AC=180°

8 - 7 + 2x +3x -8= 180°

-7+5x=180°

x=180°+7/5

x=37.4°

again

BC=-7+2*37.4

=67.8°

4 0
1 year ago
What is the height of a cylinder with a volume of 1,899.7 cubic inches and a radius of 11 inches?
FrozenT [24]


Not 100% sure. 

Height = 37.81inches


8 0
3 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
3 years ago
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