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

For the equity, here as
is percentage of the equity which is to be calculated
is the Risk premium whose value as seen from the attached data for the period 1926-2015 is 8.30%
is the risk aversion factor which is given as 4.
is the standard deviation of the portfolio which from the data for the period 1926-2015 is 20.59
By substituting values.

So the weight of equity is 48.94%.
Now the weight of T bills is given as

So the weight of T-bills is 51.05%.
The portfolio should invest 48.94% in equity while 51.05% in the T-bills.
If the transformer’s primary coil has 20 times as many turns of wire in it as the secondary coil has, then the secondary coil provides a small voltage rise for the large amount of current that flows through it.
Answer: Option B
<u>Explanation:</u>
A transformer has a two types of coils, the first one is primary coils and the second one is secondary coil. A secondary coils with hardly any turns in it provides the charges going through it just limited quantities of energy.
Without a long separation over which to do chip away at the charges streaming in the loop, the transformer delivers just a little ascent in the voltage of those charges. Be that as it may, the coil can give this little voltage to ascend to a huge current without requiring an excess of power supply from the input circuit.
Answer:
Magnitude of the vector is
and the direction is 
Explanation:
Magnitude of first vector = 
Angle = 
Magnitude of second vector = 
Angle = 
x component of first vector

y component of first vector

x component of second vector

y component of first vector

Adding the magnitudes


Magnitude of the sum of the vectors would be

The direction would be

The magnitude of the vector is
and the direction is 
Answer:
1184 kJ/kg
Explanation:
Given:
water pressure P= 28 bar
internal energy U= 988 kJ/kg
specific volume of water v= 0.121×10^-2 m^3/kg
Now from steam table at 28 bar pressure we can write


therefore at saturated liquid we have specific enthalpy at 55 bar pressure.
that the specific enthalpy h = h at 50 bar +(55-50)/(60-50)*( h at 50 bar - h at 60 bar)

h= 1184 kJ/kg