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Sedbober [7]
3 years ago
8

A bag has 5 yellow marbles, 3 red marbles and 2 blue marbles. Quinn randomly picks a marble from the bag and returns it before a

nother is picked. How many times would Quinn expect to pick a blue marble if he picks a marble 200 times?
Mathematics
2 answers:
agasfer [191]3 years ago
6 0

Answer:

Quinn can expect to pick blue 40 times

Step-by-step explanation:

the probability of picking blue is stated as 2/10 which can be reduced to 1/5

multiply the number of times they pick a marble ( shown as 200) by the probability of picking a blue one ( 1/5)

200 x 1/5 = 200 / 5 = 40

Quinn can expect to pick blue 40 times

Ksju [112]3 years ago
3 0

Answer:40 blue marbles

Step-by-step explanation:

First find the probability of a blue marble

P (blue) = number of blue marbles / total marbles

            = 2 blue/ ( 5+3+2) total

             = 2/10

              =1/5

We can use this since we return the marble to the bag each time

We multiply this by 200 for the number of times it is picked

200 * 1/5 = 40

We should expect to pick 40 blue marbles

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The product is 104 x^{4}+16 \sqrt{30} x^{4}

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First, we shall simplify the given expression.

Thus, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2

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Expanding the expression, we have,

\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)

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\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}

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Multiplying, we get,

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