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Serggg [28]
3 years ago
6

A

Mathematics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

Answer Is A

Hopefully this help

Step-by-step explanation:

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Let
Digiron [165]

What I gather from the question is that X has second moment E(X^2)=81 and variance V(X) = 58, and you're asked to find the expectation and variance of the random variable Y=2X+10.

From the given second moment and variance, we find the expectation of X :

V(X) = E(X^2) - E(X)^2 \implies E(X) = \sqrt{E(X^2) - V(X)} = \sqrt{23}

Expectation is linear, so

E(Y) = E(2X+10) = 2 E(X) + 10 = \boxed{2\sqrt{23} + 10}

Using the same variance identity, we have

V(Y) = V(2X+10) = E((2X+10)^2) - E(2X+10)^2

and

E((2X+10)^2) = E(4X^2 + 40X + 100) = 4E(X^2) + 40E(X) + 100 = 424 + 40\sqrt{23}

so that

V(Y) = V(2X+10) = (424 + 40\sqrt{23}) - (2\sqrt{23} + 10)^2 = \boxed{232}

Alternatively, we can use the identity

V(aX+b) = a^2 V(X) \implies V(2X+10) = 4V(X) = 232

5 0
2 years ago
Kyle is purchasing a refurbished phone. He is looking at the phone models 4, 5, and 6, with a protective case in red (R), blue (
konstantin123 [22]

Answer:

The choices Kyle has are option D.)  {4R, 4B, 4C, 5R, 5B, 5C, 6R, 6B, 6C}

Step-by-step explanation:

The choices available to Kyle either of the models 4, or 5 or 6 and each model can have a protective case which can be either red(R) or blue(B) or camo(C).

So Kyle can have model 4 in red protective case, 4R, or model 4 in blue protective case(4B), or model 4 in camo protective case (4C) or

Kyle can have model 5 in red protective case, 5R, or model 5 in blue protective case(5B), or model 5 in camo protective case (5C) or

Kyle can have model 6 in red protective case, 6R, or model 6 in blue protective case(6B), or model 6 in camo protective case (6C).

So the choices Kyle has are option D.)  {4R, 4B, 4C, 5R, 5B, 5C, 6R, 6B, 6C}

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Answer: I believe that A is the correct answer.

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Hi , two equivalent fractions for 16/48 are 1/3 and 8/24. Equivalent fraction is a fraction that is the same but it has different numbers.
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