70% of kids who visit a doctor have a fever, and 40% of kids who visit a doctor have a sore throat. if the probability that a kid has a sore throat given that he has a fever is 30%, what is the probability that he has a fever given that he has a sore throat?

52.5%

Step-by-step explanation:

Let A be kids who have a fever, and B be kids who have a sore throat, so

Based on the given information we have

P(A) = 70%, P(B) = 40%, and P(B|A) = 30%   (B|A is read as "B given A", which means probability of B happening given that A has already happened)

We are asked to find P(A|B), the probability of a kid having a fever given that we already know he has a sore throat.

Step 1:  First find P(A and B), this is the probability of a kid having both a fever and a sore throat.  We use the formula

        P(B|A) = P(A and B)/P(A)    

We have 2 of these values listed above, so we plug them in...

       30% = P(A and B)/70%

 This give us a value for P(A and B) which equals 21%  [multiply both sides by 70% to isolate P(A and B), (30%)(70%) = 21%]

Now we flip the equation to P(A|B), which is

     P(A|B) = P(A and B)/P(B)

We have 2 of the values, so we plug them in...

     P(A|B) =  21%/40%

  This gives us a value of 52.5%   (divide 21% by 40%)  

2(2x + 3) = -4 + 2(2x + 5)

4x + 6 = -4 + 4x + 10

4x + 6 = 4x + 6

4x - 6 + 6 = 4x - 6 + 6

4x = 4x

x = x

unlimited solutions.