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klasskru [66]
2 years ago
10

Lauren is shopping and finds 6 bracelets, 2 pairs of earrings, and 3 necklaces that she likes. How many ways can Lauren select 4

pieces of jewelry to buy?
Mathematics
2 answers:
Grace [21]2 years ago
5 0
Because it does not specify that she has to choose one or two from each category all pieces of jewelry can be treated as choices in a combination set.

the formula for a combination is

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

where n is a total number of choices ( 11 ) 
and r is number chosen out of those choices (4)
 so

C(11,4)= \frac{11!}{4!(11-4)!} 

She has 330 possible choices  to come home with 4 pieces of jewelry



Rama09 [41]2 years ago
3 0

Answer:

the correct answer is b)

Step-by-step explanation:

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the length of the highway is 1500 miles. If 1 inch represents 150 miles, what is the length of the highway on the map?
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10 inches.

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A box is in the shape of a rectangular prism. Nicole is trying to ship boxes of her homemade peanut butter cookies. The box she
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A company has 125 personal computers. The probability that any one of them will require repair on a given day is 0.15.
mariarad [96]

Answer:

a. the probability that any one of the computers will require repair on a given day is constant

Step-by-step explanation:

The following properties must be true in order for a distribution to be binomial:

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- Each trial is independent of the others (one computer requiring repair does not interfere with the likelihood of another requiring repair).

There are only two outcomes (requires repair or do not require repair)

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8 0
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Problem 5 (4+4+4=12) We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely. 1) Fin
tekilochka [14]

Answer:

1

p(b) =  \frac{1}{6}

2

p(k) =  \frac{1}{3}

3

P(a) =  \frac{1}{3}

Step-by-step explanation:

Generally when two fair 6-sided dice is rolled the doubles are

(1 1) , ( 2 2) , (3 3) , (4 4) , ( 5 5 ), (6 6)

The total outcome of doubles is N = 6

The total outcome of the rolling the two fair 6-sided dice is

n = 36

Generally the probability that doubles (i.e., having an equal number on the two dice) were rolled is mathematically evaluated as

p(b) =  \frac{N}{n}

p(b) =  \frac{6}{36}

p(b) =  \frac{1}{6}

Generally when two fair 6-sided dice is rolled the outcome whose sum is 4 or less is

(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)

Looking at this outcome we see that there are two doubles present

So

The conditional probability that doubles were rolled is mathematically represented as

p(k) =  \frac{2}{6}

p(k) =  \frac{1}{3}

Generally when two fair 6-sided dice is rolled the number of outcomes that would land on different numbers is L = 30

And the number of outcomes that at least one die is a 1 is W = 10

So

The conditional probability that at least one die is a 1 is mathematically represented as

P(a) =  \frac{W}{L}

=> P(a) =  \frac{10}{30}

=> P(a) =  \frac{1}{3}

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2 years ago
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