At a track meet 2/5 of the first place finishers went to Ohio State. One half of Ohio State's first place finishers were women. If 2/9 of the first place finishers who did NOT attend Ohio State were women, what fractional part of the total number of first place finishers were men?

2/3

Step-by-step explanation:

The first place finishers can be divided into two groups: those who went to Ohio State, and those who did not attend Ohio State.

2/5 of the first place finishers went to Ohio State.  1/2 of these were women, which means 1/2 were men.  The fraction of first place finishers who went to Ohio State AND were men is:

1/2 (2/5) = 1/5

3/5 of the first place finishers did not attend Ohio State.  2/9 of these were women, which means 7/9 were men.  The fraction of first place finishers who did not attend Ohio State AND were men is:

7/9 (3/5) = 7/15

So the fraction of first place finishers who were men is:

1/5 + 7/15

= 3/15 + 7/15

= 10/15

= 2/3

your answer is the last one

step-by-step explanation:

divide the length of a'c' by the length of ac to get a scale factor of two

& then reflect vertically (over the y-axis)

so your answer is dilation by a scale factor of 2 followed by reflection over the y-axis

answer is:

1.75 pounds