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Vsevolod [243]
3 years ago
11

For each correlation coefficient below, calculate what proportion of variance is shared by the two correlated variables:

Mathematics
1 answer:
Viefleur [7K]3 years ago
4 0

Answer:

A. 0.0625

B. 0.1089

C. 0.81

D. 0.0196

Step-by-step explanation:

The proportion of variance is shared by the two correlated variables is given as the r².

Therefore,

For A. r = 0.25

Proportion of variance is shared by the two correlated variables = 0.25²

= 0.0625

For B. r = 0.33

Proportion of variance is shared by the two correlated variables = 0.33²

= 0.1089

For C. r = 0.90

Proportion of variance is shared by the two correlated variables = 0.90²

= 0.81

For D. r = 0.14

Proportion of variance is shared by the two correlated variables = 0.14²

= 0.0196

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Answer:

14 Wood

Step-by-step explanation:

50/5 = 10 Wood = 4 days

70/5 = 14 Wood = 6 Days

I hope this is correct!

3 0
2 years ago
Calculus Problem
Roman55 [17]

The two parabolas intersect for

8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2

and so the base of each solid is the set

B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, |x^2-(8-x^2)| = 2|x^2-4|. But since -2 ≤ x ≤ 2, this reduces to 2(x^2-4).

a. Square cross sections will contribute a volume of

\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x

where ∆x is the thickness of the section. Then the volume would be

\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x

We end up with the same integral as before except for the leading constant:

\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx

Using the result of part (a), the volume is

\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x

and using the result of part (a) again, the volume is

\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}

7 0
2 years ago
An alloy consists of nickel, zinc, and copper in the ratio 2:7:9.
almond37 [142]
Zinc = 7/18
Divide 4.9 by 7 = 0.7
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6 0
3 years ago
Find the equation (in terms of <br> x<br> ) of the line through the points (-2,6) and (1,0)
AysviL [449]

Answer:

2by5=,78

Step-by-step explanation:

if you 9thru7. into8 you get 2by 7

7 0
2 years ago
Read 2 more answers
Plz help but not if you don't know
balandron [24]
I hope this helps you



Euclid



h^2=4.6



h=2 square root of 6



h^2+6^2=base^2


24+36=base^2


base=2square root of 15


h^2+4^2=height^2


24+16=height^2


height=2 square root of 10


5 0
3 years ago
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