Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussels sprouts. of these students, \small \frac{2}{3} dislike lima beans; and of those who dislike lima beans, \small \frac{3}{5} also dislike brussels sprouts. how many of the students like brussels sprouts but dislike lima beans?

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How I got that

Create a two-way table showing the breakdown of students who like or dislike either vegetable. This is shown in the attached image below. Anywhere there is a letter indicates we aren't able to determine that value. Luckily we dont need to know those values to answer the question.

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Let's say we had 3000 students. I picked some large number that is a multiple of 3. That way when we multiply by 2/3, we get a whole number.

Take 2/3 of 3000 and you should get 2000. So there are 2000 students who dislike lima beans. Write "2000" at the end of the "dislike lima beans" row which is where the total is. Above that we'll have 1000 people who like lima beans, but we dont need to use this value.

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Of those 2000 students, 3/5 of them also dont like sprouts either. So (3/5)*2000 = 0.6*2000 = 1200 students do not like both veggies. Furthermore, 2000 - 1200 = 800 like sprouts but dont like lima beans.

Alternatively, 2/5 of the 2000 students like sprouts but dont like lima beans, so (2/5)*2000 = 0.4*2000 = 800.

the answer will be

the expression (1.26)2t reveals the approximate rate of increases in the number of plants semiannually.

hope this : )