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3 years ago

5. If the covariance between the variables x and y is 18 and the variance

Mathematics

1 answer:

klasskru [66]3 years ago

5 0

Answer:

The coefficient of correlation=0.5

Step-by-step explanation:

We are given that

Covariance between the variable x and y=18

Variance of x=16

Variance of y=81

We have to find the coefficient of correlation

We know that

Coefficient of correlation

$r=\frac{covariance(x,y)}{\sqrt{variance(x)}\times \sqrt{variance(y)}}$

Using the formula

$r=\frac{18}{\sqrt{16}\times \sqrt{81}}$

$r=\frac{18}{4\times 9}$

$r=0.5$

Hence, the coefficient of correlation=0.5

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There are 12 marbles in a box. 8 are purple, and 4 are green. What is the probability of picking a purple marble and then a gree

Bad White [126]

Answer: 8/33

Step-by-step explanation:

So the probability of picking a purple marble is 8/12, after that there are 11 marbles left in the box

Now the probability of picking a green marble without placing the first marble back is 4/11

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1 year ago

twice the sum of a number and 2 is equal to three times the difference of the number and 8. Find the number.

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Step-by-step explanation:

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3 years ago

What values for q (0 ≤q≤2π)<br> satisfy the equation?<br><br> 22√sin q + 2 = 0

Vesna [10]

Answer:
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sin(q) = $\frac{- \sqrt{2} }{2}$

Now, we know that:
sin (45) = $\frac{ \sqrt{2} }{2}$

From the ASTC rule, we know that the sine function is negative in the third and fourth quadrant.
This means that:
either q = 90 + 45 = 135° which is equivalent to $\frac{3 \pi }{4}$
or q = 270 + 45 = 315° which is equivalent to $\frac{7 \pi }{4}$

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3 years ago

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2 years ago

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can someone show me how to find the general solution of the differential equations? really need to know how to do it for the upc

mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

$\mu(x)P(x)=\dfrac{\mathrm d\mu(x)}{\mathrm dx}\iff\dfrac{\mathrm d\mu(x)}{\mu(x)}=P(x)\,\mathrm dx$
$\implies\ln|\mu(x)|=\displaystyle\int P(x)\,\mathrm dx$
$\implies\mu(x)=\exp\left(\displaystyle\int P(x)\,\mathrm dx\right)$

and so on.

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3 years ago