There are 25 students that would like to go on a field trip, but only 18 may go. How many different combinations of students cou
1 answer:
480700. The different combinations of students that could go on the trip with a total of 25 student, but only 18 may go, is 480700.
The key to solve this problem is using the combination formula . This mean the number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed.
The total of students is n and the only that 18 students may go is r:
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No, it's not possible for the sides of a triangle to have those lengths.
According to the triangle inequality theorem, the sum of any two sides of the triangle has to be bigger than the last side. Let's test this.
This inequality satisfies the triangle inequality theorem.
This also satisfies the theorem.
Uh oh. This does not satisfy the triangle inequality theorem. Thus, it is not possible for a triangle to have these side lengths.
According to the triangle inequality theorem, the sum of any two sides of the triangle has to be bigger than the last side. Let's test this.
This inequality satisfies the triangle inequality theorem.
This also satisfies the theorem.
Uh oh. This does not satisfy the triangle inequality theorem. Thus, it is not possible for a triangle to have these side lengths.
Hi there!
Our interval is from 0 to 3, with 6 intervals. Thus:
3 ÷ 6 = 0.5, which is our width for each rectangle.
Since n = 6 and we are doing a right-riemann sum, the points we will be plugging in are:
0.5, 1, 1.5, 2, 2.5, 3
Evaluate:
(0.5 · f(0.5)) + (0.5 · f(1)) + (0.5 · f(1.5)) + (0.5 · f(2)) + (0.5 · f(2.5)) + (0.5 · f(3)) =
Simplify:
0.5( -2.75 + (-3) + (-.75) + 4 + 11.25 + 21) = 14.875