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3 years ago

Restless Leg Syndrome and Fibromyalgia

Mathematics

1 answer:

Anna11 [10]3 years ago

3 0

Answer:

The probability that a person with restless leg syndrome has fibromyalgia is 0.183.

Step-by-step explanation:

Denote the events as follows:

F = a person with fibromyalgia

R = a person having restless leg syndrome

The information provided is as follows:

P (R | F) = 0.33

P (R | F') = 0.03

P (F) = 0.02

Consider the tree diagram attached below.

Compute the probability that a person with restless leg syndrome has fibromyalgia as follows:

$P(F|R)=\frac{P(R|F)P(F)}{P(R|F)P(F)+P(R|F')P(F')}$

$=\frac{(0.33\times 0.02)}{(0.33\times 0.02)+(0.03\times 0.98)}\\\\=\frac{0.0066}{0.0066+0.0294}\\\\=0.183333\\\\\approx 0.183$

Thus, the probability that a person with restless leg syndrome has fibromyalgia is 0.183.

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Step-by-step explanation:

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3 years ago

(x^2-4x)^2+7x^2-28x+12=0

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Answer:

$x^4-9x^2-28x=-12$

Step-by-step explanation:

$(x^2-4x)^2+7x^2-28x+12=0$

$(x^4-16x^2)+7x^2-28x=-12$

$x^4-9x^2-28x=-12$

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3 years ago

Find the unit tangent vector T and the principal unit normal vector N for the following parameterized curve.

const2013 [10]

Answer:

$T(t) = ( -sin(t^2), cos(t^2) )\\\\N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

Step-by-step explanation:

Remember that for any curve $r(t)$

The tangent vector is given by

$T(t) = \frac{r'(t) }{| r'(t)| }$

And the normal vector is given by

$N(t) = \frac{T'(t)}{|T'(t)|}$

For this case, using the chain rule

$r'(t) = ( -10*2tsin(t^2) , 102t cos(t^2) )\\$

And also remember that

$|r'(t)| = \sqrt{(-10*2tsin(t^2))^2 + ( 10*2t cos(t^2) )^2} \\\\ = \sqrt{400 t^2*( sin(t^2)^2 + cos(t^2) ^2 })\\=\sqrt{400t^2} = 20t$

Therefore

$T(t) = r'(t) / |r'(t) | = ( -10*2tsin(t^2) , 10*2t cos(t^2) )/ 20t\\\\ = ( -10*2tsin(t^2)/ 20t , 10*2t cos(t^2) / 20t )\\= ( -sin(t^2), cos(t^2) )$

Similarly, using the quotient rule and the chain rule

$T'(t) = ( -2t cos(t^2) , -2t sin(t^2))$

And also

$|T'(t)| = \sqrt{ ( -2t cos(t^2))^2 + (-2t sin(t^2))^2} = \sqrt{ 4t^2 ( ( cos(t^2))^2 + ( sin(t^2))^2)} = \sqrt{4t^2} \\ = 2t$

Therefore

$N(t) = T'(t) / |T'(t) | = (-cos(t^2) , -sin(t^2))$

Notice that

1. $|N(t)| = |T(t) | = \sqrt{ cos(t^2)^2 + sin(t^2)^2 } = \sqrt{1} = 1$

2. $N(t)*T(T) = cos(t^2) sin(t^2 ) - cos(t^2) sin(t^2 ) = 0$

Simlarly

$r'(t) = (2t,-6,0) \\$

and

$|r'(t)| = \sqrt{(2t)^2 + 6^2} = \sqrt{4t^2 + 36}$

Therefore

$T(t) =r'(t)/|r'(t)| = (2t/ \sqrt{4t^2 + 36} , -6/\sqrt{4t^2 + 36},0)$

Then

$T'(t) = (9/(9 + t^2)^{3/2} , (3 t)/(9 + t^2)^{3/2},0)$

and also

$|T'(t)| = \sqrt{ ( (9/(9 + t^2)^{3/2} )^2 + ( (3 t)/(9 + t^2)^{3/2})^2 + 0^2 }\\= 3/(t^2 + 9 )$

And since

$N(t) =T(t)/|T'(t)| = (3/(9 + t^2)^{1/2} , t/(9 + t^2)^{1/2},0)$

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3 years ago

What is the y-intercept of the quadratic function f(x) = (x – 8)(x + 3)? (8,0) (0,3) (0,–24) (–5,0)

DiKsa [7]

The correct answer is: [C]: " (0, 24) " .
___________________________________________________________

Explanation:
___________________________________________________________

Given the quadratic function:
___________________________________________________________

→ " y = (x − 8) (x + 3) " ; ← Note: Replace the "f(x)" with: "y" ;

→ Find the "y-intercept".
___________________________________________________________

→ Note: The "y-intercept" is the coordinate of the point(s) of the graph of the equation at which the graph crosses the "x-axis" when "x = 0" .

→ So; we set plug in "0" for "x" into our equation; and solve for "y" ;

 → " y = (x − 8) (x + 3) " ;

 → y = (0 − 8) (0 + 3) ;

 → y = (-8) * (3) ;

 → y = - 24 ;
___________________________________________________________

So, the "y -intercept" of the given quadratic function is:

the point at which: "x = 0 ; y = -24 " ;

→ that is; the point the coordinates: " (0, - 24) " ;
___________________________________________________________
→ which is: Answer choice: [C]: " (0, - 24) " .
___________________________________________________________

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3 years ago

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