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bixtya [17]
3 years ago
5

If X is a normal random variable with parameters µ = 10 and σ 2 = 36, compute

Mathematics
1 answer:
alex41 [277]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that X is a normal random variable with parameters µ = 10 and σ 2 = 36,

X is N(10, 6)

Or z = \frac{x-10}{6}

is N(0,1)

a)  P(X > 5),

=P(Z>-0.8333)\\=0.7977

(b) P(4 < X < 16),

=P(|z|

(c) P(X < 8),

=P(Z

(d) P(X < 20),

=P(Z

(e) P(X > 16).

=P(Z>-0.6667)

= 0.2524

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Vsevolod [243]
You know that you have the Opposite side and the Adjacent side. therefore you would use Tangent (T= O/A)
because you are finding the angle you would go Tan^-1(12÷20) =31°
it is 12÷20 because O/A and the 12 is the opposite side therefore It would be O and 20 is the adjacent side therefore equaling 20 so O/A = 12/20
Your answer is 31°
8 0
2 years ago
(951² - 159²)÷(7539²-357²) x (258²-852²)
Marysya12 [62]

Answer:

(879,120)÷(56709072) ×(-659340)=-10221.274

Step-by-step explanation:

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2 years ago
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nadezda [96]
<h2>\large\mathfrak{{\pmb{\underline{\red{Given}}{\red{:}}}}}</h2>
  • ∠RSU and ∠USV are complementary angles
  • ∠RSU = (4x – 15)°
  • ∠USV = 20°

<h2>\large\mathfrak{{\pmb{\underline{\color{red}{To~find}}{\color{red}{:}}}}}</h2>
  • The value of x

<h2>\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}</h2>
  • ∠RSU and ∠USV are complementary angles

\sf\implies{ \angle RSU + \angle USV = 90 \degree }

\sf\implies{ (4x-15) + 20 = 90 }

\sf\implies{ 4x-15+20=90}

\sf\implies{4x+5=90}

\sf\implies{4x=90-5}

\sf\implies{4x=85}

\sf\implies{x={\dfrac{85}{4}}}

\sf\implies{ {\orange{\underline{\boxed{\sf{\pmb{x=21.25}}}}}}}

<h2>\large\mathfrak{\therefore{\pmb{\underline{\color{gold}{Required~answer}}{\color{gold}{:}}}}}</h2>
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5 0
3 years ago
Poisson Distribution LEARNING OBJECTIVE: Calculate the probability of a poisson distribution. 3.00 The average number of bridge
ira [324]

Answer:

the required probability is 0.09

Option  a) 0.09 is the correct Answer.

Step-by-step explanation:

Given that;

mean μ  = 7  

x = 4

the probability of exactly 4 bridge construction projects taking place at one time in this state = ?

Using the Poisson probability formula;

P( X=x ) = ( e^-μ × u^x) / x!

we substitute

P(X = 4) = (e⁻⁷ × 7⁴) / 4!

= 2.1894 / 24

= 0.0875 ≈ 0.09

Therefore the required probability is 0.09

Option  a) 0.09 is the correct Answer.

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2 years ago
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PIT_PIT [208]

Answer:

No. The amounts of change are the same, but the original amounts are different.

6 0
3 years ago
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