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pantera1 [17]
2 years ago
15

Suppose you know that Mr. and Mrs. Vincent have at most eight children. One day you happen to meet two girls who were the Vincen

t children. When you told the Vincents about this they told you that the chance of this happening is exactly 1/2, i.e., the probability that two of the Vincent children, randomly chosen, are both girls is exactly 1/2. How many children do the Vincent’s have, exactly? Of them what is the exact number of girls?
Mathematics
1 answer:
lianna [129]2 years ago
3 0

Answer:

The number of children are 4 out of which 3 are girls

Step-by-step explanation:

Data provided in the question:

P(Two randomly selected children are girls) = \frac{1}{2}

now,

let the number of children be 'n'

the number of girls be 'x'

thus,

P(Two randomly selected children are girls) = \frac{^xC_2}{^nC_2} = \frac{1}{2}

also,

^nC_r = \frac{n!r!}{(n-r)!}

thus,

\frac{\frac{x!2!}{(x-2)!}}{\frac{n!2!}{(n-2)!}} = \frac{1}{2}

or

\frac{x(x-1)}{n(n-1)}=\frac{1}{2}

or

2x(x-1) = n(n-1)

now

for x = 3 and n = 4

i.e

2(3)(3-1) = 4(4-1)

12 = 12

hence, the relation is justified

therefore,

The number of children are 4 out of which 3 are girls

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A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base
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Answer:

Step-by-step explanation:

Equation for a hyperboloid of one sheet, with center at the origen and axis along z-axis is:

(x/a)²  +  (y/b)²   -  (z/c)²  =  1                         (1)

We have to find a , b, and c

We can express equation (1)

(x/a)²  +  (y/b)²    =  (z/c)² + 1                 (2)

Now if we cut the hyperboloid with planes parallel to xy plane we get for  z = k       ( K = 1 , 2 , 3  and so on ) circles of different radius

(x/a)²  +  (y/b)²    =  (k/c)² + 1

at z = k = 0 at the base of the hyperboloid  d = 300   or r = 150 m

we have

(x/a)²  +  (y/b)²   = 1      

x²  +  y²   =   a²                a² = (150)²       a = b = 150

and    x²  +  y²  = (150)²

Now the other condition is at 200 m above the base d = 500 m   r = 250 m  minimum diameter then in equation (2)  we have:

(x/a)²  +  (y/b)²    =  (z/c)² + 1        

(1/a)² [ x² + y² ]  = (z/c)² + 1  

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2,78  =  40000/c² + 1

2.78c²  =  40000  + c²

1.78c² = 40000

c²  =  40000/1.78

c²  = 22471.91

c = 149,91

Then we finally have the equation:

x²/(150)²   + y² /(150)² - z²/149,91  = 1

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Answer:

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Both solutions are valid.

Explanation:

1. First assumption is that the shape of the fence is <u>rectangular</u>.

2. Second, assum the length parallel to the wall measure y feet, so the other two lengths, y, together with x will add up 100 feet

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3. The, the area of the fence will be:

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4. Now you have two equation with two variables which you can solveL

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