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zhuklara [117]
3 years ago
13

uppose the correlation between two variables, math attitude (x) and math achievement (y) was found to be .78. Based on this stat

istic, we know that the proportion of the variability seen in math achievement that can be predicted by math attitude is:
Mathematics
2 answers:
dalvyx [7]3 years ago
5 0

Answer:

r=SSY'/SSY

Step-by-step explanation:

Nana76 [90]3 years ago
3 0

Answer:

The proportion of the variability seen in math achievement that can be predicted by math attitude is 0.78, the same value as the correlation coefficient.

Step-by-step explanation:

The correlation coefficient r between this two variables is found to be 0.78.

This coefficient can be calculated as:

r=\dfrac{SSY'}{SSY}

where SSY' is the sum of the squares deviation from the mean for the predicted value and SSY is the sum of the squares deviation from the mean for the criterion variable.

Then, the value of the coefficient r is giving the proportion of the variability seen in the criterion value Y that can be explained by the predictor variable X.

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Vinil7 [7]
From the beginning we know we are dealing with two different rates. Distance is equal to (Speed X Time). We have two different speeds though. We will call the min speed = x and the max speed = y.  With those two ideas in place we can make two equations for both Tony and Rae.
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2x+3.5y=355
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2x+3y=320
Now you can solve the equations by subtracting Rae's equation from Tony's

.5y=35 solve for y and you get 70
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The product of a number w and 737
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Wewaii [24]
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~kaikers
4 0
4 years ago
Please solve this, will rate 5 stars and mark as STAR!​
Nina [5.8K]

Answer:

\boxed{5 \cdot \sqrt{2}  \cdot \sqrt[6]{5} }

Step-by-step explanation:

\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}

\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}

\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }

250=2 \cdot 5^3

\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5  \sqrt[3]{2}

Once

\sqrt[6]{2}  \cdot \sqrt[6]{5} = \sqrt[6]{10}

We have

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5}

We can proceed considering the common base of exponentials

\sqrt[3]{2}  \cdot \sqrt[6]{2}  =  2^{\frac{1}{3}} \cdot  2^{\frac{1}{6} }  = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}

Therefore,

5  \sqrt[3]{2} \cdot \sqrt[6]{2}  \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2}  \cdot \sqrt[6]{5}

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