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ValentinkaMS [17]
3 years ago
6

i have two questions. Is y= (4 2/3) x proportonal? and is y= 3(x - 1) proportional? If it’s proportional, what’s the constant va

riable?
Mathematics
1 answer:
Dmitrij [34]3 years ago
7 0

\bf \qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ y = 4\frac{2}{3}x\qquad \qquad yes\qquad \checkmark\qquad \qquad k = 4\frac{2}{3} \\\\[-0.35em] ~\dotfill\\\\ y=3(x-1)\implies \stackrel{\textit{distributing}}{y=3x-3}\qquad \qquad yes\qquad \checkmark \qquad \qquad k=3

bear in mind that, direct proportional equations have a y-intercept.

for y = kx, is pretty much y = kx + 0, where 0 = y-intercept.

and the "k" constant of proportionality, is pretty much just its slope.

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a. f_{X,Y} is a joint density function if its integral over the given support is 1:

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so the answer is yes.

b. We should first find the density of the marginal distribution, f_Y(y):

f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy

f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}

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or about 0.2019.

For the other probability, we can use the joint PDF directly:

P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}

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E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5

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3 years ago
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Step-by-step explanation:

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