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Volgvan
2 years ago
6

The two toy store owners agree to specialize and to trade 30 puzzles for 10 puppets. The terms of trade are still 3 puzzles for

each puppet. How many puzzles and puppets will each one have after they complete their trade
Mathematics
1 answer:
likoan [24]2 years ago
7 0

Based on the exchange rate, at the end of the trade, Lewis will have 31 puppets and 2 puzzles left over while Geppeto will have 158 puzzles and 4 puppets left over.

<h3>What is the exchange rate of puzzles for puppets?</h3>

The exchange rate of puzzles for puppets is 3 to 1.

Geppeto has 20 puppets to exchange for puzzles.

Lewis has 50 puzzles to exchange for puppets.

Number of times Lewis can exchange puzzles for puppets = 50/3 = 16 times.

Lewis will get 16 puppets in exchange for 48 puzzles.

Therefore;

Lewis will have 16 + 25 puppets = 31 puppets and 2 puzzles left over

Geppeto will have 48 + 100 puzzles = 158 puzzles and 4 puppets left over.

in conclusion, the exchange rate determines the how many puzzles and puppets will each one have after they complete their trade.

Learn more about exchange rate at: brainly.com/question/2202418

#SPJ1

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Baseball's division series is a best of five game series. That is, the first team to win 3 games is the winner. Team A and team
iris [78.8K]

Answer:

1. 27.46% probability that A wins the series in 3 games.

2. There is a 28.84% probability A wins the series in 4 games.

3. There is a 20.18% probability A wins the series in 5 games.

4. There is a 76.48% probability A wins the series.

5. There is a 71.825% probability A wins the series if a "best-of-three" game series is played.

Step-by-step explanation:

For each game, there are these following probabilities:

A 65% probability team A wins.

A 35% probability team B wins.

The combination formula is important to solve this problem:

This is because for example, team A winning games 1,2 and 4 and team B winning game 3 is the same as team A winning games 1,3,4 and team B winning game 2. That is, the order is not important.

C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

1. What the probability that A wins the series in 3 games?

This is team A winning all 3 games. For each game, there is a 65% probability that team A wins. So

P = (0.65)^{3} = 0.2746

There is a 27.46% probability that A wins the series in 3 games.

2. What is the probability A wins the series in 4 games?

This is the team A winning three games and the team B 1. The team B win cannot happen in the fourth game, so the number of possibilities is a combination of 3 by 2. So

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The probability that team A wins the series in 4 games is:

P = 3*(0.65)^{3}*(0.35) = 0.2884

There is a 28.84% probability A wins the series in 4 games.

3.What is the probability A wins the series in 5 games?

This is the team A winning three games and the team B 2. The team B second win cannot happen in the fifth game, so the number of possibilities is a combination of 4 by 2. So

C_{4,2} = \frac{3!}{2!1!} = 6

The probability that team A wins the series in 5 games is:

P = 6*(0.65)^{3}*(0.35)^{2} = 0.2018

There is a 20.18% probability A wins the series in 5 games.

4. What is the probability A wins the series (period)?

They can win the series in 3,4 or 5 games. So this is the sum of the answers for 1,2,3.

So

P = 0.2746 + 0.2884 + 0.2018 = 0.7648

There is a 76.48% probability A wins the series.

5. What is the probability A wins the series if a "best-of-three" game series is played?

These three following outcomes are accepted:

A - A

A - B - A

B - A - A

So

P = (0.65)^{2} + 2*(0.65)^{2}*(0.35) = 0.71825

There is a 71.825% probability A wins the series if a "best-of-three" game series is played.

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Answer:

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