Umbilical
point.
An
umbilic point, likewise called just an umbilic, is a point on a surface at
which the arch is the same toward any path.
In
the differential geometry of surfaces in three measurements, umbilics or
umbilical focuses are focuses on a surface that are locally round. At such
focuses the ordinary ebbs and flows every which way are equivalent,
consequently, both primary ebbs and flows are equivalent, and each digression
vector is a chief heading. The name "umbilic" originates from the
Latin umbilicus - navel.
<span>Umbilic
focuses for the most part happen as confined focuses in the circular area of
the surface; that is, the place the Gaussian ebb and flow is sure. For surfaces
with family 0, e.g. an ellipsoid, there must be no less than four umbilics, an
outcome of the Poincaré–Hopf hypothesis. An ellipsoid of unrest has just two
umbilics.</span>
Answer:
-2/3 ; - 3/5 ; - 1/2 ; 5/8
Step-by-step explanation:
Given that:
We could convert the numbers to decimal to obtain a better view of the values :
-2/3 = - 0.66666
5/8 = 0.625
-3/5 = - 0.6
-1 /2 = - 0.5
Negative values are always arranged first on left of a number line.
-0.66666 < - 0.6 < - 0.5 <0.625
Hence,
-2/3 ; - 3/5 ; - 1/2 ; 5/8
Answer: -1
The negative value indicates a loss
============================================================
Explanation:
Define the three events
A = rolling a 7
B = rolling an 11
C = roll any other total (don't roll 7, don't roll 11)
There are 6 ways to roll a 7. They are
1+6 = 7
2+5 = 7
3+4 = 7
4+3 = 7
5+2 = 7
6+1 = 7
Use this to compute the probability of rolling a 7
P(A) = (number of ways to roll 7)/(number total rolls) = 6/36 = 1/6
Note: the 36 comes from 6*6 = 36 since there are 6 sides per die
There are only 2 ways to roll an 11. Those 2 ways are:
5+6 = 11
6+5 = 11
The probability for event B is P(B) = 2/36 = 1/18
Since there are 6 ways to roll a "7" and 2 ways to roll "11", there are 6+2 = 8 ways to roll either event.
This leaves 36-8 = 28 ways to roll anything else
P(C) = 28/36 = 7/9
-----------------------------
In summary so far,
P(A) = 1/6
P(B) = 1/18
P(C) = 7/9
The winnings for each event, let's call it W(X), represents the prize amounts.
Any losses are negative values
W(A) = amount of winnings if event A happens
W(B) = amount of winnings if event B happens
W(C) = amount of winnings if event C happens
W(A) = 18
W(B) = 54
W(C) = -9
Multiply the probability P(X) values with the corresponding W(X) values
P(A)*W(A) = (1/6)*(18) = 3
P(B)*W(B) = (1/18)*(54) = 3
P(C)*W(C) = (7/9)*(-9) = -7
Add up those results
3+3+(-7) = -1
The expected value for this game is -1.
The player is expected to lose on average 1 dollar per game played.
Note: because the expected value is not 0, this is not a fair game.
