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3 years ago

10

From the fact that virtually every galaxy is moving away from us and more distant galaxies are moving away from us at a faster r

ate than closer ones, we conclude that

1 answer:

dangina [55]3 years ago

6 0

The universe is expanding

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f your risk-aversion coefficient is A = 4 and you believe that the entire 1926–2015 period is representative of future expected

siniylev [52]

Answer:

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

Explanation:

As the complete question is not given here ,the table of data is missing which is as attached herewith.

From the maximized equation of the utility function it is evident that

$Weight=\frac{E_M-r_f}{A\sigma_M^2}$

For the equity, here as

$Weight$ is percentage of the equity which is to be calculated
${E_M-r_f}$ is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
$A$ is the risk aversion factor which is given as 4.
$\sigma_M$ is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59

By substituting values.

$Weight=\frac{E_M-r_f}{A\sigma_M^2}\\Weight=\frac{8.30\%}{4(20.59\%)^2}\\Weight=0.4894 =48.94\%$

So the weight of equity is 48.94%.

Now the weight of T bills is given as

$Weight_{T-Bills}=1-Weight_{equity}\\Weight_{T-Bills}=1-0.4894\\Weight_{T-Bills}=0.5105=51.05\%\\$

So the weight of T-bills is 51.05%.

The portfolio should invest 48.94% in equity while 51.05% in the T-bills.

7 0

2 years ago

What is one property of iodine

mamaluj [8]

Answer: I2

Explanation:

6 0

3 years ago

An atom of arsenic has how many electron-containing orbitals

irina [24]

Answer:

The Arsenic has three electron-containing orbitals. The orbitals s, p and d.

Explanation:

Arsenic is an element with an atomic number equal of 33, it means that it has 33 electrons in its orbitals in the following way:

$1s^{2}$

$2s^{2}$

$2p^{6}$

$3s^{2}$

$3p^{6}$

$3d^{10}$

$4s^{2}$

$4p^{3}$

Therefore, the Arsenic has three electron-containing orbitals (s, p d).

8 0

3 years ago

A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine

AVprozaik [17]

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

a³ / T² = 7.496 × 10⁻⁶ (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

1 a.u = 9.296 × 10⁷ miles

$\frac{4004 * 10^{6}}{9.296 * 10^{7}} = 43.07\ a.u.$

Substituting the values into Kepler's third law equation;

$\frac{(43.07)^{3}}{T^{2}} = 7.496 * 10^{-6}$

$T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}}$ (days)²

$T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}$

T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

7 0

3 years ago

A block of mass 14.9 kg is pulled to the right by an applied force of 39.4 N. If it moves with constant velocity, how much frict

lakkis [162]

The frictional force is 39.4 N

Explanation:

We can solve this problem by applying Newton's 2nd law of motion: in fact, the net force acting on the block is equal to the product between its mass and its acceleration. So we can write

$\sum F = ma$

where

$\sum F$ is the net force

m is the mass

a is the acceleration

Here we know that the box is moving with constant velocity, so its acceleration is zero:

$a=0$

This means that the net force is also zero:

$\sum F=0$

The net force on the block is given by the applied force, forward, and the frictional force, backward:

$\sum F = F_a-F_f=0$

where

$F_a=39.4 N$ is the applied force

$F_f$ is the frictional force

Therefore, solving for $F_f$ ,

$F_f=F_a=39.4 N$

Learn more about friction:

brainly.com/question/6217246

brainly.com/question/5884009

brainly.com/question/3017271

brainly.com/question/2235246

#LearnwithBrainly

8 0

2 years ago