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valina [46]
2 years ago
15

A new crew of painters can paint a small apartment in 12 hours.  An experienced crew can paint the small apartment in 6 hours. 

How many hours does it take to paint the apartment when the two crews work together? It takes _____ hours to paint the apartment when the two crews work together
Mathematics
1 answer:
Varvara68 [4.7K]2 years ago
4 0
The new crew works at the rate 1/12 apt/hr.
The old crew works at the rate 1/6 apt/hr.

Let x = hours required to paint the apartment when the two crews work together.

That is, 
(1/12 + 1/6  apt/hr)*(x hr) = (1 apt)
(3/12)x = 1
(1/4)x = 1
x = 4 hours

Answer:  4 hours 

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Scale:
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Answer:

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Step-by-step explanation:

<u><em>Step 1:</em></u>

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<u><em>Step 2:</em></u>

ax^2+bx = -c

<u><em>Step 3:</em></u>

\frac{ax^2+bx}{a} = \frac{-c}{a}

<u><em>Step 4:</em></u>

Adding \frac{b^2}{4a^2} to both sides to complete the square

x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{-c}{a} + \frac{b^2}{4a^2}

<u><em>Step 5:</em></u>

x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{-4ac+b^2}{4a^2}

<u><em>Step 6:</em></u>

Taking square root on both sides

x + \frac{b}{2a} =  \frac{+/ -  \sqrt{b^2-4ac} }{2a}

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Answer:

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First, you'll need to find the marginal distributions of X,Y. By the law of total probability,

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which translates to

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Similarly,

f_Y(y)=\displaystyle\int_0^yf_{X,Y}(x,y)\,\mathrm dx=\begin{cases}2y&\text{for }0

Compute the expectations for both random variables:

E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13

E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\int_0^12y^2\,\mathrm dy=\frac23

Compute the variances and thus standard deviations:

V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2

where

E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16

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Compute the covariance:

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Finally, the correlation:

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