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True [87]
3 years ago
9

How many different committees can be formed from 10 teachers and 37 students of the committee consist of three teachers and two

students? how many ways can the committee of five members be selected?
Mathematics
1 answer:
pashok25 [27]3 years ago
3 0

Answer:

<em>79,920 different ways</em>

Step-by-step explanation:

Combination has to do with selection:

If we are to select 3 teachers from a pool of 10 teachers to form a committee, this can be done in 10C3 number of ways.

10C3 = 10!/(10-3)!3!

10C3 = 10!/7!3!

10C3 = 10*9*8*7!/7!3!

10C3 = 10*9*8/3*2

10C3 = 720/6

10C3 = 120 ways

Similarly, selecting five 2 students from a pool of 37 students to form a committee, this can be done in 37C2 difference ways;

37C2 = 37!/(37-2)!2!

37C2 = 37!/35!2!

37C2 = 37*36*35!/35!*2

37C2 = 37*36/2

37C2 = 37 * 18

37C2 = 666 ways

<em>Hence the total number of ways that the 5 committees can be selected is expressed as 10C3 * 37C3 = 120 * 666</em>

<em> 120 * 666 = 79,920 ways</em>

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In the figure, if the measure of 28 = 72°, what's the measure of 214?
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Answer:

c = 13

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m∡B = 30°

Step-by-step explanation:

This is a 5-12-13 triangle. However, to make sure, I will put the steps.

Allow for each sides to be denoted as a-b-c, in which c is the hypotenuse (longest side). Set the equation:

a² + b² = c²

Plug in the corresponding numbers to the corresponding variables:

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Simplify. First, solve the exponents, and then add:

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Note the equal sign, what you do to one side, you do to the other. Isolate the variable, c, by rooting both sides:

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13 is your answer for c.

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2 years ago
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