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tiny-mole [99]
3 years ago
6

Solve the given inequality 11x-20<_13​

Mathematics
2 answers:
Lunna [17]3 years ago
7 0

\huge{ \mathfrak{ Answer }}

  • 11x - 20 \leqslant 13

  • 11x  \leqslant  20 + 13

  • x  \leqslant  33 \div 11

  • x \leqslant 3

\#TeeNForeveR

RUDIKE [14]3 years ago
3 0

Answer:

x ≤ 3

Step-by-step explanation:

11

x

−

20

≤

13

Step 1: Add 20 to both sides.

11

x

−

20

+

20

≤

13

+

20

11

x

≤

33

Step 2: Divide both sides by 11.

11

x

11

≤

33

11

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Step-by-step explanation:

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2 years ago
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

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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

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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))

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= 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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