Suppose that 10 sophomores, 3 juniors, and 10 seniors are candidates for a prestigious mathematics award of which three will be given out. in how many different ways can the recipients of these three awards be chosen?

1771 possible ways

Step-by-step explanation:

In this case, we need to know first how many candidates are in total:

10 + 3 + 10 = 23 candidates in total.

Now, we need to choose 3 of them to receive an award. In this case, we have several scenarios, but as it's an award we can also assume that the order in which the candidates are chosen do not matter, so, the formula to use is the following:

C = m! / n! (m - n)!

Where m is the total candidates and n, is the number of candidates to be chosen. Replacing this data we have:

C = 23! / 3! (23 - 3)!

C = 2.59x10^22 / 6(2.43x10^18)

C = 1771

So we have 1771 ways of choose the candidates.

in simplified version   it would be x ≤-4

step-by-step explanation:

multiply each term in x ≤-4 by -1