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garik1379 [7]
3 years ago
7

A bag of marbles contains 6 blue marbles, 2 yellow marbles, 4 red marbles, and 1 green marble. what is the probability of reachi

ng into the bag and selecting a yellow marble?
Mathematics
2 answers:
Anastasy [175]3 years ago
8 0
Its probably 2 to 13
andre [41]3 years ago
7 0

Answer: \frac{2}{13}

Step-by-step explanation:

Given: The number of yellow marbles in the bag = 2

The total number of marbles in the bag = 6+2+4+1=13

Now, the probability of reaching into the bag and selecting a yellow marble is given by:-

P(\text{yellow})=\frac{\text{yellow marble}}{\text{total marbles}}\times100\\\\\Rightarrow\ P(\text{yellow})=\frac{2}{13}

Hence, the probability of reaching into the bag and selecting a yellow marble=\frac{2}{13}

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If m angle ECD is six less than five times m angle BCE, and mangle BCD = 162º. find each
Usimov [2.4K]

Answer:

see explanation

Step-by-step explanation:

let ∠BCE = x then ∠ECD = 5x - 6 and

∠BCE + ∠ECD = ∠BCD, substitute values

x + 5x - 6 = 162, that is

6x - 6 = 162 ( add 6 to both sides )

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Hence

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5 0
3 years ago
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What is the probability that the number of “HEADs” on these four coins is equal to 3? (4 points)
slavikrds [6]

Answer:

a. \frac{1}{4}  

Step-by-step explanation:  

We are asked to find the probability of getting 3 heads on 4 flips.

\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

Since we know that flipping a fair coin has 2 equally likely possible outcomes, so flipping four coins will have 2*2*2*2=16 possible outcomes.

Sample space of possible outcomes.

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,

THHH, THHT, THTH,THTT, TTHH, TTHT, TTTH, TTTT.

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Therefore, the probability of getting 3 heads on 4 coins will be \frac{1}{4} and option a is the correct choice.

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