An exam has two probability problems, 1 and 2. if 37% of the students solved problem 1 and 12% of the students solved both problems 1 and 2, what percent of the students who solved problem 1 also solved problem 2?

  about 32%

Step-by-step explanation:

The desired probability is the conditional probability that problem 2 was solved given that problem 1 was solved. That is ...

  P(2|1) = P(1&2)/P(1) = 12%/37% ≈ .324 ≈ 32%

You can also think of it like this. If there were 100 students taking the exam, 37 solved the first problem. Of those 37, 12 solved the second one. 12 is about 32% of 37.

9 x 8 divided by (5-3)=36

step-by-step explanation:

this can be solved first with pythagorean theorem, which tells us to square both legs of the triangle and sum it. this will be equal to the square of the hypotenuse.

the hypotenuse is the side "opposite" the 90 degree angle, which is x.

so we can write using pythagorean theorem,

simplifying:

to further simplify this radical, we need to use the property  

thus