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LenaWriter [7]
2 years ago
12

Jason wanted to place all 540 marbles in the 3 newly bought boxes. If he wants to

Mathematics
1 answer:
kotegsom [21]2 years ago
7 0

Answer:

The first box will have 90 marbles, the second 180 and the third 270.

Step-by-step explanation:

Division in a ratio of 1:2:3

1 + 2 + 3 = 6

So

The first box will have 1/6 of the marbles.

The second box will have 2/6 = 1/3 of the marbles

The third box will have 3/6 = 1/2 of the marbles.

First box:

One sixth, so:

(1/6)*540 = 540/6 = 90

Second box:

One third, so:

(1/3)*540 = 540/3 = 180

Third box:

One half, so:

(1/2)*540 = 540/2 = 270

The first box will have 90 marbles, the second 180 and the third 270.

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An urn contains two blue balls (denoted B1 and B2) and three white balls (denoted W1, W2, and W3). One ball is drawn, its color
alekssr [168]

Answer:

(a) Shown below.

(b) The probability that the first ball drawn is blue is 0.40.

(c) The probability that only white balls are drawn is 0.36.

Step-by-step explanation:

The balls in the urn are as follows:

Blue balls: B₁ and B₂

White balls: W₁, W₂ and W₃

It is provided that two balls are drawn from the urn, with replacement, and their color is recorded.

(a)

The possible outcomes of selecting two balls are as follows:

B₁B₁          B₂B₁          W₁B₁          W₂B₁          W₃B₁

B₁B₂         B₂B₂          W₁B₂         W₂B₂          W₃B₂

B₁W₁         B₂W₁         W₁W₁         W₂W₁         W₃W₁

B₁W₂        B₂W₂         W₁W₂        W₂W₂         W₃W₂

B₁W₃        B₂W₃         W₁W₃        W₂W₃         W₃W₃

There are a total of N = 25 possible outcomes.

(b)

The sample space for selecting a blue ball first is:

S = {B₁B₁, B₁B₂, B₁W₁, B₁W₂, B₁W₃, B₂B₁, B₂B₂, B₂W₁, B₂W₂, B₂W₃}

n (S) = 10

Compute the probability that the first ball drawn is blue as follows:

P(\text{First ball is Blue})=\frac{n(S)}{N}=\frac{10}{25}=0.40

Thus, the probability that the first ball drawn is blue is 0.40.

(c)

The sample space for selecting only white balls is:

X = {W₁W₁, W₂W₁, W₃W₁, W₁W₂, W₂W₂, W₃W₂, W₁W₃, W₂W₃, W₃W₃}

n (X) = 9

Compute the probability that only white balls are drawn as follows:

P(\text{Only White balls})=\frac{n(X)}{N}=\frac{9}{25}=0.36

Thus, the probability that only white balls are drawn is 0.36.

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<u>Explanation:</u>

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