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aleksandrvk [35]

3 years ago

15

There are two twins. One twin wants to explore the planet near Proxima Centauri and leaves on a spaceship traveling at 90% of th

e speed of light, while the other twin stays home on Earth. How much does the twin on Earth age while the other twin travels to Proxima Centauri?

1 answer:

miskamm [114]3 years ago

7 0

Answer:

<em>4.7 years</em>

Step-by-step explanation:

Distance of planet from Earth = 1.3 parsec

We know that 1 parsec = 3.26 light years

And a light year is the distance traveled by light in time period of 1 year.

So, Distance of planet from Earth = 1.3 $\times$ 3.26 = 4.24 light years

Given that, the person travels at a speed equal to 90% of the light speed.

To travel to the planet from Earth, Light takes 4.24 years.

To travel the distance $D$ , at a speed $s$ light takes time of 4.24 years.

To travel the distance $D$ , at a speed $0.90s$ , the person takes time:

$\dfrac{4.24}{0.90} = \bold{4.7\ years}$

So, the twin on Earth ages by <em>4.7 years.</em>

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Answer:

Do I really need it?

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Step-by-step explanation:

When you are trying to build up your savings you need to to spent money only on the necessary things to be able to save more money every time you get paid. So, according to this, the three questions that Aria should ask herself before making a purchase are:

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3 years ago

A player of a video game is confronted with a series of opponents and has an 80% probability of defeating each one. Success with

Vikentia [17]

Answer:

(a) The PMF of <em>X</em> is: $P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b) The probability that a player defeats at least two opponents in a game is 0.64.

(c) The expected number of opponents contested in a game is 5.

(d) The probability that a player contests four or more opponents in a game is 0.512.

(e) The expected number of game plays until a player contests four or more opponents is 2.

Step-by-step explanation:

Let <em>X</em> = number of games played.

It is provided that the player continues to contest opponents until defeated.

(a)

The random variable <em>X</em> follows a Geometric distribution.

The probability mass function of <em>X</em> is:

$P(X=k)=(1-p)^{k-1}p;\ p>0, k=0, 1, 2, 3....$

It is provided that the player has a probability of 0.80 to defeat each opponent. This implies that there is 0.20 probability that the player will be defeated by each opponent.

Then the PMF of <em>X</em> is:

$P(X=k)=(1-0.20)^{k-1}0.20;\ k=0, 1, 2, 3....$

(b)

Compute the probability that a player defeats at least two opponents in a game as follows:

P (X ≥ 2) = 1 - P (X ≤ 2)

= 1 - P (X = 1) - P (X = 2)

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20\\=1-0.20-0.16\\=0.64$

Thus, the probability that a player defeats at least two opponents in a game is 0.64.

(c)

The expected value of a Geometric distribution is given by,

$E(X)=\frac{1}{p}$

Compute the expected number of opponents contested in a game as follows:

$E(X)=\frac{1}{p}=\frac{1}{0.20}=5$

Thus, the expected number of opponents contested in a game is 5.

(d)

Compute the probability that a player contests four or more opponents in a game as follows:

P (X ≥ 4) = 1 - P (X ≤ 3)

= 1 - P (X = 1) - P (X = 2) - P (X = 3)

$=1-(1-0.20)^{1-1}0.20-(1-0.20)^{2-1}0.20-(1-0.20)^{3-1}0.20\\=1-0.20-0.16-0.128\\=0.512$

Thus, the probability that a player contests four or more opponents in a game is 0.512.

(e)

Compute the expected number of game plays until a player contests four or more opponents as follows:

$E(X\geq 4)=\frac{1}{P(X\geq 4)}=\frac{1}{0.512}=1.953125\approx 2$

Thus, the expected number of game plays until a player contests four or more opponents is 2.

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3 years ago

Suppose that E(θˆ1) = E(θˆ2) = θ, V(θˆ 1) = σ2 1 , and V(θˆ2) = σ2 2 . Consider the estimator θˆ 3 = aθˆ 1 + (1 − a)θˆ 2. a Show

katen-ka-za [31]

Answer:

Step-by-step explanation:

Given that:

$E( \hat \theta _1) = \theta \ \ \ \ E( \hat \theta _2) = \theta \ \ \ \ V( \hat \theta _1) = \sigma_1^2 \ \ \ \ V(\hat \theta_2) = \sigma_2^2$

If we are to consider the estimator $\hat \theta _3 = a \hat \theta_1 + (1-a) \hat \theta_2$

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$E ( \hat \theta_3) = E ( a \hat \theta_1+ (1-a) \hat \theta_2)$

$E ( \hat \theta_3) = aE ( \theta_1) + (1-a) E ( \hat \theta_2)$

$E ( \hat \theta_3) = a \theta + (1-a) \theta = \theta$

b) If $\hat \theta _1 \ \ and \ \ \hat \theta_2$ are independent

$V(\hat \theta _3) = V (a \hat \theta_1+ (1-a) \hat \theta_2)$

$V(\hat \theta _3) = a ^2 V ( \hat \theta_1) + (1-a)^2 V ( \hat \theta_2)$

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$a (\sigma_1^2 + \sigma_2^2) = \sigma^2_2$

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As such; $a = \dfrac{1}{2}$ if $\sigma_1^2 \ \ = \ \ \sigma_2^2$

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