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murzikaleks [220]
3 years ago
12

involving the election of officers on a committee. Assume that the committee consists of 14 members including Tasha. The same th

ree offices are to be filled. (1) In how many different ways can the offices be filled if each person can hold at most one office?
Mathematics
1 answer:
Yanka [14]3 years ago
5 0

Answer:

There are 364 ways of filling the offices.

Step-by-step explanation:

In this case, the order of filling of the offices does not matter, so, we can figure out the different ways of filling the offices by using the combination formula:

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

where n=14 (number of members)

r=3 number of offices

n!=n·(n-1)·(n-2)·...·3·2·1

C^{14} _{3}=\frac{14!}{(14-3)!3!}=\frac{14*13*12*11*10*9*8*7*6*5*4*3*2*1}{(11*10*9*8*7*6*5*4*3*2*1)*(3*2*1)}=\frac{14*13*12}{3*2*1} =364

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ira [324]

find total area  by multiplying  350' x 200' = 70,00 sq ft

area of one block, multiply 25' x 25' = 625 sq ft  = area for one Security guard

divide area for one security guard into total area of facility  70,000 / 625 =

112 security guards

find how many people for each block = 70,000 / 625 = 112 blocks

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3 0
3 years ago
1)How much do you need to subtract from 35/6 to make 5?
nikdorinn [45]

Answer:

1) 5/6

2) 7/10

3) 3 1/8

4) 3 2/7

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Step-by-step explanation:

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3 years ago
When Hiroto is writing, there is 0.920.920, point, 92 probability that there will be no spelling mistakes on a page. One day, Hi
Usimov [2.4K]

Complete question :

When Hiroto is writing, there is 0.92 probability that there will be no spelling mistakes on a page. One day, Hiroto writes an essay that is 11 pages long.

Assuming that Hiroto is equally likely to have a spelling mistake on each of the 11 pages, what is the probability that he will have a spelling mistake on at least one of the pages?

Answer:

0.60

Step-by-step explanation:

The question meets the requirement for a binomial probability distribution :

Recall:

P(x = x) = nCx * p^x * q^(n-x)

Given :

Probability of making a spelling mistake = 1 - p(not making) = 1 - 0.92 = 0.08

Hence,

p = 0.08 ; q = 0.92

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P(x ≥ 1) = 1 - p(x = 0)

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P(x = 0) = 0.399

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P(x ≥ 1) = 1 - 0.399

P(x ≥ 1) = 0.601

P(x ≥ 1) = 0.60

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In the diagram below is a regular pentagon ABCDE, side EB is drawn.
Kaylis [27]

Answer:

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bogdanovich [222]

nearest ten = 63,850

nearest hundred = 63,800

nearest thousand = 64,000

nearest ten thousand = 60,000

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3 years ago
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