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IgorC [24]
4 years ago
5

A system of linear equations with fewer equations than unknowns is sometimes called an underdetermined system. Can such a system

have a unique​ solution? Explain. Choose the correct answer below. A. ​No, it cannot have a unique solution. Because there are more variables than​ equations, there must be at least one free variable. If the linear system is consistent and there is at least one free​ variable, the solution set contains infinitely many solutions. If the linear system is​ inconsistent, there is no solution.
Mathematics
1 answer:
elena55 [62]4 years ago
3 0

Answer:

No, it cannot have a unique solution. Because there are more variables than​ equations, there must be at least one free variable. If the linear system is consistent and there is at least one free​ variable, the solution set contains infinitely many solutions. If the linear system is​ inconsistent, there is no solution.

Step-by-step explanation:

the questionnaire options are incomplete, however the given option is correct

We mark this option as correct because in a linear system of equations there can be more than one solution, since the components of the equations, that is, the variables are multiple, leaving free variables which generates more alternative solutions, however when there is no consistency there will be no solution

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Read 2 more answers
RANDOMLY CHOOSING TWO POTENTIAL CLIENTS
gregori [183]
Let <span>Jacob, Carol, Geraldo, Meg, Earvin, Dora, Adam, and Sally be represented by the letters J, C, G, M, E, D, A, and S respectively. </span>

<span>In part IV we are asked:

</span><span>What is the sample space of the pairs of potential clients that could be chosen?
</span><span>
Since the Sample Space is the set of all possible outcomes, we need to make a set (a list) of all the possible pairs, which are as follows:

{(J, C), (J, G), (J, M), (J, E), (J, D), (J, A), (J, S)

          , </span>(C, G), (C, M), (C, E), (C, D), (C, A), (C, S)
<span>                   
</span>                      , (G, M), (G, E), (G, D), (G, A), (G, S)
<span>             
                                    ,</span>(M, E), (M, D), (M, A), (M, S)    
<span> 
                                               , </span>(E, D),  (E, A),  (E, S) <span>   
                                                          
                                                           , </span>(D, A), (D, S)
              
                                                                       , (A, S).}

We can check that the number of the elements of the sample space, n(S) is 

1+2+3+4+5+6+7=28.


This gives us the answer to the first question: <span>How many pairs of potential clients can be randomly chosen from the pool of eight candidates? 

(Answer: 28.)


II) </span><span>What is the probability of any particular pair being chosen?
</span>
The probability of a particular pair to be picked is 1/28, as there is only one way of choosing a particular pair, out of 28 possible pairs.

III) <span>What is the probability that the pair chosen is Jacob and Meg or Geraldo and Sally? 

The probability of choosing (J, M) or (G, S) is 2 out of 28, that is 1/14.


Answers:

I) 28
II) 1/28</span>≈0.0357
III) 1/14≈0.0714
IV)


{(J, C),  (J, G), (J, M),  (J, E),   (J, D),  (J, A),   (J, S)
          , (C, G), (C, M), (C, E),   (C, D), (C, A),  (C, S)
                      , (G, M), (G, E),  (G, D), (G, A),  (G, S)
                                   ,(M, E),  (M, D), (M, A),  (M, S)    
                                               , (E, D),  (E, A),  (E, S)    
                                                            , (D, A), (D, S)
                                                                        , (A, S).}
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