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Murljashka [212]
3 years ago
7

In one year, three awards (research, teaching, and service) are given to a class of 25 graduate students in a statistics departm

ent. If each student can receive at most one award, how many possible selections are there?
Mathematics
1 answer:
allochka39001 [22]3 years ago
3 0

Answer:

13,800

Step-by-step explanation:

This is a permutation, not a combination, because order matters.

Here's why:

Arbitrarily, let's say that the three students chosen for the award are Abby, Bob, and Charlie. The order of which each award is given to the three of them matters, because otherwise they would be receiving different awards. For example, if Abby gets the research award, Bob gets the teaching award, and Charlie gets the service award, this would be notably different than if Abby got the teaching award, Bob got the service award, and Charlie got the research award.

Therefore, for the first award, we have 25 people to choose from. After we select that person, we have 24 people to choose from, since the problem stipulates that each student can receive at most one award. Then 23, and so on.

Since we're choosing three people to give awards to, there are:

25\cdot 24\cdot 23=\boxed{13,800} permutations. Because order matters (refer to explanation above), this is our final answer.

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Please help!! What is the solution to the quadratic inequality? 6x2≥10+11x
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Answer:

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

Step-by-step explanation:

First of all, let simplify and factorize the resulting polynomial:

6\cdot x^{2} \geq 10 + 11\cdot x

6\cdot x^{2}-11\cdot x -10 \geq 0

6\cdot \left(x^{2}-\frac{11}{6}\cdot x -\frac{10}{6} \right)\geq 0

Roots are found by Quadratic Formula:

r_{1,2} = \frac{\left[-\left(-\frac{11}{6}\right)\pm \sqrt{\left(-\frac{11}{6} \right)^{2}-4\cdot (1)\cdot \left(-\frac{10}{6} \right)} \right]}{2\cdot (1)}

r_{1} = \frac{5}{2} and r_{2} = -\frac{2}{3}

Then, the factorized form of the inequation is:

6\cdot \left(x-\frac{5}{2}\right)\cdot \left(x+\frac{2}{3} \right)\geq 0

By Real Algebra, there are two condition that fulfill the inequation:

a) x-\frac{5}{2} \geq 0 \,\wedge\,x+\frac{2}{3}\geq 0

x \geq \frac{5}{2}\,\wedge\,x \geq-\frac{2}{3}

x \geq \frac{5}{2}

b) x-\frac{5}{2} \leq 0 \,\wedge\,x+\frac{2}{3}\leq 0

x \leq \frac{5}{2}\,\wedge\,x\leq-\frac{2}{3}

x\leq -\frac{2}{3}

The solution of the inequation 6\cdot x^{2} \geq 10 + 11\cdot x is \left(-\infty,-\frac{2}{3}\right]\cup\left[\frac{5}{2},+\infty\right).

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3 years ago
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