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jarptica [38.1K]
3 years ago
5

PLEASE HELP

Mathematics
1 answer:
almond37 [142]3 years ago
8 0

Answer:

x + 6y = -3

Step-by-step explanation:

Explained in the paper.

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Step-by-step explanation:

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Read 2 more answers
Let X1, X2 and X3 be three independent random variables that are uniformly distributed between 50 and 100.
sergejj [24]

Answer:

a) the probability that the minimum of the three is between 75 and 90 is 0.00072

b) the probability that the second smallest of the three is between 75 and 90 is 0.396

Step-by-step explanation:

Given that;

fx(x) = { 1/5 ; 50 < x < 100

              0, otherwise}

Fx(x) = { x-50 / 50 ; 50 < x < 100

                          1 ;   x > 100

a)

n = 3

F(1) (x) = nf(x) ( 1-F(x)^n-1

= 3 × 1/50 ( 1 - ((x-50)/50)²

= 3/50 (( 100 - x)/50)²

=3/50³ ( 100 - x)²

Therefore P ( 75 < (x) < 90) =  ⁹⁰∫₇₅ 3/50³ ( 100 - x)² dx

= 3/50³ [ -2 (100 - x ]₇₅⁹⁰

= (3 ( -20 + 50)) / 50₃

= 9 / 12500 = 0.00072

b)

f(k) (x) = nf(x)  ( ⁿ⁻¹_k₋ ₁) ( F(x) )^k-1 ; ( 1 - F(x) )^n-k

Now for n = 3, k = 2

f(2) (x) = 3f(x) × 2 × (x-50 / 50) ( 1 - (x-50 / 50))

= 6 × 1/50 × ( x-50 / 50) ( 100-x / 50)

= 6/50³ ( 150x - x² - 5000 )

therefore

P( 75 < x2 < 90 ) = 6/50³ ⁹⁰∫₇₅ ( 150x - x² - 5000 ) dx

= 99 / 250 = 0.396

3 0
3 years ago
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