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nadezda [96]
4 years ago
10

In how many ways can a

Mathematics
1 answer:
Tomtit [17]4 years ago
6 0

Answer:

The answer is 364. There are 364 ways of choosing a recorder, a facilitator and a questioner froma club containing 14 members.

This is a Combination problem.  

Combination is a branch of mathematics that deals with the problem relating to the number of iterations which allows one to select a sample of elements which we can term "<em>r</em>" from a collection or a group of distinct objects which we can name "<em>n</em>". The rules here are that replacements are not allowed and sample elements may be chosen in any order.

Step-by-step explanation:

Step I

The formula is given as

C (n,r) = \frac{n}{r} = \frac{n!}{(r!(n-r)!)}

n (objects) = 14

r (sample) = 3

Step 2 - Insert Figures

C (14, 3) = (\frac{14}{3}) = \frac{14!}{(3!(14-3)!)}

= \frac{87178291200}{(6 X 39916800)}

= \frac{87178291200}{239500800}

= 364

Step 3

The total number of ways a recorder, a facilitator and a questioner can be chosen in a club containing 14 members therefore is 364.

Cheers!

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4 years ago
Which point is not on the graph of the equation y=10+x
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Answer:

C. (8, 2)

Step-by-step explanation:

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When plugged into the equation, the point (8, 2) makes it incorrect:

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The equation is incorrect because 2 is not equal to 18.

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4 years ago
Can you write a rule for finding the output (absolute guessing error) given the input (a guess)?
igor_vitrenko [27]

Answer:

The rule for the plotted line is y = \left | x - 50 \right |

Step-by-step explanation:

The coordinates when y = 0, x = 50

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The equation in slope and intercept form is y - 10 = 1 × (x - 60)

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Also, we have

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