The graph shows motion of a train. Enter a value to complete the statement
2 answers:
You might be interested in
<u>Answer:</u>
X and Y are stochastically dependent RVs .
<u>Step-by-step explanation:</u>
Let ,
X = sum of the values that come up after throwing n (≥ 1) fare dice.
Y = number of times an odd number come up.
Let, n = 3
then, P(X =6) = p (say) clearly 0 < p < 1
and P (Y = 3) =
And,
P( X = 6, Y = 3) = 0 ≠
Hence, X and Y are stochastically dependent RVs
Answer:
(a, b, c) = (-2, -3, 5)
Step-by-step explanation:
The idea is to find values of a, b, and c that satisfy all three equations. Any of the methods you learned for systems of 2 equations in 2 variables will work with 3 equations in 3 variables. Often, folks find the "elimination" method to be about the easiest to do by hand.
Here, we notice that the coefficients in general are not nice multiples of each other. However, the coefficient of "a" in the second equation is 1, so we can use that to eliminate "a" from the other equations.
__
Add 3 times the second equation to the first.
3(a +3b -4c) +(-3a -4b +2c) = 3(-31) +(28)
3a +9b -12c -3a -4b +2c = -93 +28 . . . . eliminate parentheses
5b -10c = -65 . . . . . . . . . . . . . . . . . . . . . . .collect terms
b - 2c = -13 . . . . . . . divide by 5 (because we can) "4th equation"
Subtract 2 times the second equation from the third.
(2a +3c) -2(a +3b -4c) = (11) -2(-31)
2a +3c -2a -6b +8c = 11 +62
-6b +11c = 73 . . . . . . "5th equation"
Now, we have two equations in b and c that we can solve in any of the ways we know for 2-variable equations. Once again, it looks convenient to use the first of these (4th equation) to eliminate the b variable. (That is why we made its coefficient be 1 instead of leaving it as 5.)
Add 6 times the 4th equation to the 5th equation:
6(b -2c) + (-6b +11c) = 6(-13) +(73)
6b -12c -6b +11c = -78 +73 . . . . . . . eliminate parentheses
-c = -5 . . . . . . . . . . . . . . . . . . . . . . . . collect terms
c = 5 . . . . . multiply by -1
We can now work backwards to find the other variable values. Substituting into the 4th equation, we have ...
b -2(5) = -13 . . . .substitute for c
b = -3 . . . . . . . . . add 10
And the values for b and c can be substituted into the 2nd equation.
a + 3(-3) -4(5) = -31
a -9 -20 = -31 . . . . . eliminate parentheses
a = -2 . . . . . . . . . . . . add 29
__
The solution to this set of equations is (a, b, c) = (-2, -3, 5).
_____
<em>Comment on steps</em>
At each stage, we made choices calculated to simplify the process. By using equations that had a variable coefficient of 1, we avoided messy fractions or multiplying by more numbers than necessary when we used the elimination process. That is, the procedure is guided by the idea of <em>elimination</em>, but the specific steps are <em>ad hoc</em>. Using <em>these same specific steps</em> on different equations will likely be useless.
_____
<em>Alternate solution methods</em>
The coefficients of these equations can be put into the form called an "augmented matrix" as follows:
Many graphing and/or scientific calculators are able to solve equations written in this form. The function used is the one that puts this matrix into "reduced row-echelon form". The result will look like ...
where the rightmost column is the solution for the variables in the same order they appear in the equations. The vertical line in the body of the matrix may or may not be present in a calculator view. The square matrix to the left of the vertical bar is an identity matrix (1 in each diagonal element) when there is exactly one solution.