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Lunna [17]
1 year ago
14

Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen

numbers are the side lengths of a triangle with positive area
Mathematics
1 answer:
k0ka [10]1 year ago
5 0

Let a,b,c be the randomly selected lengths. Without loss of generality, suppose a[tex]P(A + B \ge C) = P(A + B - C \ge 0)

where A,B,C are independent random variables with the same uniform distribution on [0, 1].

By their mutual independence, we have

P(A=a,B=b,C=c) = P(A=a) \times P(B=b) \times P(C=c)

so that the joint density function is

P(A=a,B=b,C=c) = \begin{cases}1 & \text{if }(a,b,c)\in[0,1]^3 \\ 0 & \text{otherwise}\end{cases}

where [0,1]^3=[0,1]\times[0,1]\times[0,1] is the cube with vertices at (0, 0, 0) and (1, 1, 1).

Consider the plane

a + b - c = 0

with (a,b,c)\in\Bbb R^3. This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)

The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region a+b-c\ge0, since 0+0-1=-1. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.

Then we have

\displaystyle P(A+B-C\ge0) = 1 - P(A+B-C < 0) \\\\ ~~~~~~~~ = 1 - \int_0^1\int_0^{1-a}\int_{a+b}^1 P(A=a,B=b,C=c) \, dc\,db\,da \\\\ ~~~~~~~~ = 1 - \int_0^1 \int_0^{1-a} (1 - a - b) \, db \, da \\\\ ~~~~~~~~ = 1 - \int_0^1 \frac{(1-a)^2}2\,da \\\\ ~~~~~~~~ = 1 - \frac16 = \boxed{\frac56}

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Answer:

This is true

Step-by-step explanation:

Take for example 10=5x, you would divide both sides by 5 without changing anything to the equation.

In a case like 5=0x, you can't divide both sides by 0, so the solution is undefined.

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2 years ago
On the first part of her trip Natalie rode her bike 16 miles and on the second part of the trip she rode her bike 42 miles. Her
-BARSIC- [3]

Answer:

14 mph   ( average speed during the second part of the trip )

Step-by-step explanation:

Let´s call  "x"  the average speed during the first part then

t = 5 hours

t  =  t₁  +  t₂        t₁   and  t₂   times during part 1 and 2 respectively

l = t*v           (  distance is speed by time )      t =  l/v

First part

t₁  = 16/x        and      t₂  = 42 / ( x + 6)

Then

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5 *x * ( x + 6 )  =  16*x  + 96 +  42 x

5*x² + 30*x  - 58*x - 96  =  0

5*x²  -  28*x  -  96  =  0

We obtained a second degree equation, we will solve for x and dismiss any negative root since negative time has not sense

x₁,₂  =   [28 ± √ (28)² + 1920  ] / 10

x₁,₂  = ( 28  ± √2704 )/ 10

x₁  = 28  -  52 /10        we dismiss that root

x₂  = 80/10

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3 years ago
Shana wants to use all 62 feet of the fencing she has to make a rectangular run for her dog. She decides to make the length of t
Andreyy89

Answer:

Width of the rectangular Park = 11 feet

Step-by-step explanation:

Given that the total length of the fencing is 62 feet. As has to be fenced around a rectangular park , it would be the perimeter of the rectangular park.

Also Shana wants the length of the run to be 20 feet. Hence the length of the park is 20 feet.

Here we will use the formula for perimeter to find the width of the run

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l+w = \frac{62}{2}

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Hence the width of the run for her dog in park would be 11 feet.

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Answer:

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The overall percent off compared to the original price was 6.5%.

Answer: 6.5%

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