Answer:
2000 nickels
Explanation:
One way to solve proportionality problems, direct and inverse: the simple 3 rule.
If the relationship between the magnitudes is direct (when one magnitude increases so does the other), the simple direct rule of three must be applied.
On the contrary, if the relationship between the magnitudes is inverse (when one magnitude increases the other decreases) the rule of three simple inverse applies.
The simple 3 rule is an operation that helps us quickly solve proportionality problems, both direct and inverse.
To make a simple rule of three we need 3 data: two magnitudes proportional to each other, and a third magnitude. From these, we will find out the fourth term of proportionality.
In the simple three rule, therefore, the proportionality relationship between two known values A and B is established, and knowing a third value C, a fourth value D is calculated.
A -> B
C -> D
Calculation
1 nickel --> 5 g
X? nickel --> 10000g
X = (10000 g * 1 nickel) / 5 g
X = 2000 nickels
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.
Answer:
38.4 m/s
Explanation:
a) at t = 3.2s. 
b) at t = 3.2 + Δt. 
c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt
As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s
Known variables
d=4.6m
initial velocity=0m/s
downward acceleration=-9.8m/s2
d=1/2gt2
4.6=1/2 -9.8 t2
t=0.93s
Answer:
110.9 m/s²
Explanation:
Given:
Distance of the tack from the rotational axis (r) = 37.7 cm
Constant rate of rotation (N) = 2.73 revolutions per second
Now, we know that,
1 revolution = 
radians
So, 2.73 revolutions = 
Therefore, the angular velocity of the tack is, 
Now, radial acceleration of the tack is given as:
Plug in the given values and solve for 
. This gives,
![a_r=(17.153\ rad/s)^2\times 37.7\ cm\\a_r=294.225\times 37.7\ cm/s^2\\a_r=11092.28\ cm/s^2\\a_r=110.9\ m/s^2\ \ \ \ \ \ \ [1\ cm = 0.01\ m]](https://tex.z-dn.net/?f=a_r%3D%2817.153%5C%20rad%2Fs%29%5E2%5Ctimes%2037.7%5C%20cm%5C%5Ca_r%3D294.225%5Ctimes%2037.7%5C%20cm%2Fs%5E2%5C%5Ca_r%3D11092.28%5C%20cm%2Fs%5E2%5C%5Ca_r%3D110.9%5C%20m%2Fs%5E2%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%5B1%5C%20cm%20%3D%200.01%5C%20m%5D)
Therefore, the radial acceleration of the tack is 110.9 m/s².
