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7nadin3 [17]
3 years ago
11

Solve for a.

Mathematics
2 answers:
FinnZ [79.3K]3 years ago
8 0

Answer:

iii) a=3b/2

Step-by-step explanation:

7a-2b= 5a+b

7a-5a=2b+b

2a=3b

a=3b/2

pychu [463]3 years ago
8 0

Answer:

\large\boxed{a=\dfrac{3b}{2}=\dfrac{3}{2}b=1.5b}

Step-by-step explanation:

7a-2b=5a+b\qquad\text{add}\ 2b\ \text{to both sides}\\\\7a-2b+2b=5a+b+2b\\\\7a=5a+3b\qquad\text{subtract}\ 5a\ \text{from both sides}\\\\7a-5a=5a-5a+3b\\\\2a=3b\qquad\text{divide both sides by 2}\\\\\dfrac{2a}{2}=\dfrac{3b}{2}\\\\a=\dfrac{3b}{2}

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Each day, X arrives at point A between 8:00 and 9:00 a.m., his times of arrival being uniformly distributed. Y arrives independe
astraxan [27]

Answer:

Y will arrive earlier than X one fourth of times.

Step-by-step explanation:

To solve this, we might notice that given that both events are independent of each other, the joint probability density function is the product of X and Y's probability density functions. For an uniformly distributed density function, we have that:

f_X(x) = \frac{1}{L}

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Now, as  X is distributed over a 1 hour interval, and Y is distributed over a 0.5 hour interval, we have:

f_X(x) = 1\\\\f_Y(y)=2.

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It's useful to draw a diagram here, I have attached one in which you can see the integration region.

You can see there a box, that represents all possible outcomes for Y and X. There's a diagonal coming from the box's upper right corner, that diagonal represents the cases in which both X and Y arrive at the same time, under that line we have that Y arrives before X, that is our integration region.

Let's set up the integration:

\iint_A f_{X,Y} (x,y) dx\, dy\\\\\iint_A f_{X} (x) \, f_{Y} (y) dx\, dy\\\\2 \iint_A  dx\, dy

We have used here both the independence of the events and the uniformity of distributions, we take the 2 out because it's just a constant and now we just need to integrate. But the function we are integrating is just a 1! So we can take the integral as just the area of the integration region. From the diagram we can see that the region is a triangle of height 0.5 and base 0.5. thus the integral becomes:

2 \iint_A  dx\, dy= 2 \times \frac{0.5 \times 0.5 }{2} \\\\2 \iint_A  dx\, dy= \frac{1}{4}

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3 years ago
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Answer:

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Step-by-step explanation:

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ella [17]

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Answer:

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Answer:

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