The numbers -3, -4, and 6 are used once and substitued at random for a, b, and c in the equation ax+b=c. What is the probability that x turns out to be negative.
I'm not sure where to begin with this. Can you help?
First solve for x.
x = (c-b)/a
Then make a table of all possible a,b, c permutations and calculate the corresponding x.
a b c x
_____________
-3.-4. 6..-10/3
-3. 6. -4.. 10/3
-4.-3...6.. -9/4
-4..6..-3.. 9/4
6..-3..-4.. -1/6
6..-4..-3.. 1/6
Half of the possibilties lead to a positive value of x.
ahhhhhh thanks!
Of course! Let's break down the problem step by step.
First, we need to determine the possible values for a, b, and c. Since the numbers -3, -4, and 6 are used once each, we have three possible values for each variable.
Now, let's consider the equation ax + b = c. We want to find the probability that x turns out to be negative. To do that, we need to find the values of a, b, and c that satisfy this condition.
To determine when x is negative, we can rewrite the equation in terms of x as follows: x = (c - b) / a. Notice that if (c - b) is positive and a is negative, then x will be negative.
Now, let's consider the different possibilities:
1) When a = -3, b = -4, and c = 6:
Here, (c - b) = 6 - (-4) = 10, which is positive. Since a = -3 is also negative, x will be negative.
2) When a = -3, b = 6, and c = -4:
Here, (c - b) = (-4) - 6 = -10, which is negative. Therefore, x will not be negative in this case.
3) When a = -4, b = -3, and c = 6:
Here, (c - b) = 6 - (-3) = 9, which is positive. Similarly, a = -4 is negative, so x will be negative.
4) When a = -4, b = 6, and c = -3:
Here, (c - b) = (-3) - 6 = -9, which is negative. So, x will not be negative.
As we have checked all the possible combinations, we can see that in two out of the four cases, x turns out to be negative.
Therefore, the probability that x is negative is 2 out of 4, or 2/4, which simplifies to 1/2.
So, the probability that x turns out to be negative is 1/2.