Answer:
For a = 1.22 there is one solution where y = 1.3
Step-by-step explanation:
Hi there!
Let´s write the system of equations:
a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0
-1.2(x-0.5) = 0
Let´s solve the second equation for x:
-1.2(x-0.5) = 0
x- 0.5 = 0
x = 0.5
Now let´s repalce x = 0.5 and y = 1.3 in the first equation and solve it for a:
a(0.3 - y) + 1.1 +2.4x(y-1.2) = 0
a(0.3 - 1.3) + 1.1 + 2.4(0.5)(1.3 -1.2) = 0
a(-1) + 1.1 + 1.2(0.1) = 0
-a + 1.22 = 0
-a = -1.22
a = 1.22
Let´s check the solution and solve the system of equations with a = 1.22. Let´s solve the first equation for y:
1.22(0.3 - y) + 1.1 +2.4(0.5)(y-1.2) = 0
0.366 - 1.22y + 1.1 + 1.2 y - 1.44 = 0
-0.02y +0.026 = 0
-0.02y = -0.026
y = -0.026 / -0.02
y = 1.3
Then, the answer is correct.
Have a nice day!
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).}
