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exis [7]
3 years ago
8

A state of $490,000 is left to three siblings the eldest receives three times as much as the youngest the middle sibling receive

s $15,000 more than the youngest how much did each receive
Mathematics
1 answer:
Elis [28]3 years ago
7 0

Answer:

Step-by-step explanation:

Given

State worth is \$4,90,000

Assume the youngest receives x

So, Eldest receives three times of youngest=3x

Middle receives \$15,000 more than the youngest=x+15000

Sum of their shares= 3x+x+15000+x=4,90,000

\Rightarrow \quad 5x+15000=4,90,000\\\Rightarrow \quad 5x=4,75,000\\\Rightarrow \quad x=\$ 95,000

Eldest receives $2,85,000

Middle receives $1,10,000

Youngest receives $90,000

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Analysing the question one statement at a time.

Before the face with 3 is loaded to be twice likely to come up.

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S = \{1,1,2,2,2,3\}

And the probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{6}

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

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{6}

P(2) = \frac{1}{2}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{6}

P(Odd Number) is then calculated as:

P(Odd\ Number) =  P(1) + P(3)

P(Odd\ Number) = \frac{1}{3} + \frac{1}{6}

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P(Odd\ Number) = \frac{2+1}{6}

P(Odd\ Number) = \frac{3}{6}

P(Odd\ Number) =  \frac{1}{2}

After the face with 3 is loaded to be twice likely to come up.

The sample space becomes:

S = \{1,1,2,2,2,3,3\}

The probability of each is:

P(1) = \frac{n(1)}{n(s)}

P(1) = \frac{2}{7}

P(2) = \frac{n(2)}{n(s)}

P(2) = \frac{3}{7}

P(3) = \frac{n(3)}{n(s)}

P(3) = \frac{1}{7}

P(Odd\ Number) = P(1) + P(3)

P(Odd\ Number) = \frac{2}{7} + \frac{1}{7}

Take LCM

P(Odd\ Number) = \frac{2+1}{7}

P(Odd\ Number) = \frac{3}{7}

Comparing P(Odd Number) before and after

P(Odd\ Number) =  \frac{1}{2} --- Before

P(Odd\ Number) = \frac{3}{7} --- After

<em>We can conclude that the change to the face 3 affects the value of P(Odd Number)</em>

7 0
3 years ago
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