Answer:
8
Step-by-step Explanation:
Given:
- Initially, there are 5 red counters and [tex]\sf x [/tex] total counters.
- After turning one red and one green counter, we have:
- Red counters = [tex]\sf 5 - 1 = 4 [/tex]
- Green counters = [tex]\sf x - 5 + 1 = x - 4 [/tex]
We are given the probability condition:
Let's use the corrected equation and solve for [tex]\sf x[/tex]:
[tex]\sf \dfrac{5}{x} \times \dfrac{x - 4}{x } + \dfrac{x - 6}{x} \times \dfrac{5}{x } = \dfrac{19}{32} [/tex]
Expand the Equation:
[tex]\sf \dfrac{5(x - 4)}{x^2} + \dfrac{6(x - 5)}{x^2} = \dfrac{19}{32} [/tex]
Combine the Fractions:
[tex]\sf \dfrac{5(x - 4) + 6(x - 5)}{x^2} = \dfrac{19}{32} [/tex]
[tex]\sf \dfrac{5x - 20 + 6x - 30}{x^2} = \dfrac{19}{32} [/tex]
[tex]\sf \dfrac{11x - 50}{x^2} = \dfrac{19}{32} [/tex]
Cross Multiply:
[tex]\sf 32(11x - 50) = 19x^2 [/tex]
[tex]\sf 352x - 1600 = 19x^2 [/tex]
Rearrange the Equation:
[tex]\sf 19x^2 - 352x + 1600 = 0 [/tex]
Solve the Quadratic Equation:
Use the quadratic formula:
[tex]\sf x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/tex]
where [tex]\sf a = 19 [/tex], [tex]\sf b = -352 [/tex], and [tex]\sf c = 1600 [/tex].
[tex]\sf x = \dfrac{-(-352) \pm \sqrt{(-352)^2 - 4 \times 19 \times 1600}}{2 \times 19} [/tex]
[tex]\sf x = \dfrac{352 \pm \sqrt{123904 - 121600}}{38} [/tex]
[tex]\sf x = \dfrac{352 \pm \sqrt{2304}}{38} [/tex]
[tex]\sf x = \dfrac{352 \pm 48}{38} [/tex]
Simplify:
[tex]\sf x_1 = \dfrac{352 + 48}{38} = \dfrac{400}{38} [/tex]
[tex]\sf x_2 = \dfrac{352 - 48}{38} = \dfrac{304}{38} [/tex]
[tex]\sf x_1 = \dfrac{200}{19} [/tex]
[tex]\sf x_2 = \dfrac{152}{19} [/tex]
Since [tex]\sf x[/tex] represents the number of counters, it must be a positive integer.
Thus, [tex]\sf x = \dfrac{200}{19}[/tex] is not a valid solution.
Therefore, [tex]\sf x = \dfrac{152}{19} = 8[/tex]
Since [tex]\sf x [/tex] represents the total number of counters, it must be an integer.
Therefore, the appropriate value for [tex]\sf x [/tex] in this context would be [tex]\sf \boxed{8} [/tex].
