Among a large group of patients recovering from shoulder injuries, it is found that 22% visit both a physical therapist and a chiropractor, whereas 12% visit neither of these. the probability that a patient visits a chiropractor exceeds by 0.14 the probability that d le to 010.0 validedo a patient visits a physical therapist. oblodiog bemion-l dans h an n itidadong determine the probability that a randomly chosen member of this group visits a physical therapist.

0.48

Step-by-step explanation:

First of all. let's recall probability formula needed for solving the problem.

We know that, If the events A and B are not mutually exclusive, the probability is:  

probability of event (A or B) = probability of event A + probability of event B – p(A and B).

This formula will be used in calculations of this problem.

Let's consider that event A is patient visiting physical therapist; event B is patient visiting chiropractor.

P (A∩B) = 0.22 (visiting both physical therapist and chiropractor) ⇔ p(A and B)

P(B) = P(A) + 0.14 [ Probability patient visiting chiropractor is 0.14 more than probability visiting physical therapist]

Patients visiting none of these is 12%

Those who visit either therapist or chiropractor are amount to 1 - 0.12 = 0.88

Now we should use the formula mentioned in the beginning of the text:

0.88 = P(A) + P(B) - 0.22 ⇒ P(A) + P(B) = 1.1

If we replace P(B) by P(A) + 0.14 ⇒ 2P(A) + 0.14 = 1.1 ⇒ 2P(A) = 1.1 - 0.14 = 0.96

So, P(A) = 0.96/2 = 0.48 [Probability of patients visiting therapist]

the answer is 5/14.

hope this .

one peck is 9.309 quarts

9.309x 2