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Travka [436]
2 years ago
15

A 56-foot piece of siding is cut into three pieces so that the second piece is twice as long as the first piece and the third pi

ece is four times as long as the first plece. If x
represents the length of the first plece, find the lengths of all three pieces.
The length of the first piece of siding is
long, the length of the second piece is
long, and the length of the third piece is
long.
Mathematics
1 answer:
ELEN [110]2 years ago
5 0

The lengths of the first, second and third piece are respectively 8 ft, 16 ft and 32 ft

<h3>How to solve Algebra word Problems?</h3>

Let the length of the first piece be x.

Let the length of the second piece be 2x

Let the length of the thrid piece be 4x

Since the total length of the piece is 56 ft, then we can say that;

x + 2x + 4x = 56

7x = 56

x = 56/7

x = 8 ft

Thus, length of first piece = 8 ft

Length of second piece = 2 * 8 = 16 ft

Length of third piece = 4 * 8 = 32 ft

Thus, the lengths of the first, second and third piece are respectively 8 ft, 16 ft and 32 ft

Read more about Algebra Word Problems at; brainly.com/question/13818690

#SPJ1

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DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

<h2>(a)</h2>

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At the beginning we have b+w balls. We want to pick a white one, so we have a probability of \frac{w}{b+w} of picking a white one.

If this happens, we're left with w-1 white balls and still b black balls, for a total of b+w-1 balls. So, now, the probability of picking a white ball is

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The exact same logic leads to a probability of

\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}

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This leads to an overall probability of

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