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Ugo [173]
2 years ago
7

Each of the numbers 1 through 7 is written on a small piece of paper and placed in a box. What is the probability of picking a p

aper with an odd number?
A. 12


B. 47


C. 37


D. 14



I NEED ASAP PLS 20 pnts!
Mathematics
2 answers:
gregori [183]2 years ago
5 0

Answer:

Either C or B I’m not quite sure which one. I’m sorry

Step-by-step explanation:

Sidana [21]2 years ago
3 0
B because there are 4 odd number and only 3 even numbers
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Question 8 Maurice won 40% of the races he ran in this season. He won 14 races this season How many races did Maurice run in tot
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Answer:

D)35

Step-by-step explanation:

14%4=3.5

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6 0
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3/6 + 1/2 what is the answer.​
Simora [160]

Answer:

reduce 3/6 to lowest terms which would be, 1/2 because 3 goes into 3 once and 3 goes into 6 twice.  

Making your equation simplified to

1/2 + 1/2

which is equal to 1

1/2 + 1/2 = 1

Step-by-step explanation:

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2 years ago
A bag contains different colored candies. There are 50 candies in the bag, 28 are red, 10 are blue, 8 are green and 4 are yellow
Mrrafil [7]

Answer:

\displaystyle \frac{54}{5405}.

Step-by-step explanation:

How many unique combinations are possible in total?

This question takes 5 objects randomly out of a bag of 50 objects. The order in which these objects come out doesn't matter. Therefore, the number of unique choices possible will the sames as the combination

\displaystyle \left(50\atop 5\right) = 2,118,760.

How many out of that 2,118,760 combinations will satisfy the request?

Number of ways to choose 2 red candies out a batch of 28:

\displaystyle \left( 28\atop 2\right) = 378.

Number of ways to choose 3 green candies out of a batch of 8:

\displaystyle \left(8\atop 3\right)=56.

However, choosing two red candies out of a batch of 28 red candies does not influence the number of ways of choosing three green candies out of a batch of 8 green candies. The number of ways of choosing 2 red candies and 3 green candies will be the product of the two numbers of ways of choosing

\displaystyle \left( 28\atop 2\right) \cdot \left(8\atop 3\right) = 378\times 56 = 21,168.

The probability that the 5 candies chosen out of the 50 contain 2 red and 3 green will be:

\displaystyle \frac{21,168}{2,118,760} = \frac{54}{5405}.

3 0
3 years ago
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