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3 years ago

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A baseball team has won 8 out of the first 20 games and has 40 games remaining. The team needs a winning percentage of 62.5% mak

e the playoffs. How many more wins does the team need in order to get into the playoffs? If the team wins the number of games to get into the playoffs, what is the final win percentage if the team wins 2 and loses 3 games in the playoff?

1 answer:

Korvikt [17]3 years ago

3 0

They’ve already played 20 games and there are 40 remaining, so 60 total games.
80 * .625 = 37.5, so they need to win at least 38 games to make the playoffs. So they must win 30 more games. If they do so they’ll have won 38 out of 60. Winning 2 games out of 5 in the playoffs means their total is 40 out of 65. 40/65 = .6153 = 61.5%.

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If n is a positive integer, how many 5-tuples of integers from 1 through n can be formed in which the elements of the 5-tuple ar

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Answer:

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

Step-by-step explanation:

Lets divide it in cases, then sum everything

Case (1): All 5 numbers are different

In this case, the problem is reduced to count the number of subsets of cardinality 5 from a set of cardinality n. The order doesnt matter because once we have two different sets, we can order them descendently, and we obtain two different 5-tuples in decreasing order.

The total cardinality of this case therefore is the Combinatorial number of n with 5, in other words, the total amount of possibilities to pick 5 elements from a set of n.

${n \choose 5 } = \frac{n!}{5!(n-5)!}$

Case (2): 4 numbers are different

We start this case similarly to the previous one, we count how many subsets of 4 elements we can form from a set of n elements. The answer is the combinatorial number of n with 4 ${n \choose 4} .$

We still have to localize the other element, that forcibly, is one of the four chosen. Therefore, the total amount of possibilities for this case is multiplied by those 4 options.

The total cardinality of this case is $4 * {n \choose 4} .$

Case (3): 3 numbers are different

As we did before, we pick 3 elements from a set of n. The amount of possibilities is ${n \choose 3} .$

Then, we need to define the other 2 numbers. They can be the same number, in which case we have 3 possibilities, or they can be 2 different ones, in which case we have ${3 \choose 2 } = 3$ possibilities. Therefore, we have a total of 6 possibilities to define the other 2 numbers. That multiplies by 6 the total of cases for this part, giving a total of $6 * {n \choose 3}$

Case (4): 2 numbers are different

We pick 2 numbers from a set of n, with a total of ${n \choose 2}$ possibilities. We have 4 options to define the other 3 numbers, they can all three of them be equal to the biggest number, there can be 2 equal to the biggest number and 1 to the smallest one, there can be 1 equal to the biggest number and 2 to the smallest one, and they can all three of them be equal to the smallest number.

The total amount of possibilities for this case is

$4 * {n \choose 2}$

Case (5): All numbers are the same

This is easy, he have as many possibilities as numbers the set has. In other words, n

Conclussion

By summing over all 5 cases, the total amount of possibilities to form 5-tuples of integers from 1 through n is

$n + 4 {n \choose 2} + 6 {n \choose 3} + 4 {n \choose 4} + {n \choose 5}$

I hope that works for you!

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Answer:

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Step-by-step explanation:

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In mathematical notation, as shown in the figure below (for a standard normal distribution), the empirical rule is described as

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where the symbol $\Phi$ (the uppercase greek alphabet phi) is the cumulative density function of the normal distribution, $\mu$ is the mean and $\sigma$ is the standard deviation of the normal distribution defined as $N(\mu, \ \sigma)$ .

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Therefore, the price of 99.7% of college textbooks falls inclusively between $77 and $149.

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