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KATRIN_1 [288]
3 years ago
11

Each baseball team has nine players and each football team has eleven players. Six schools have both baseball and football teams

. Four schools have only a baseball team. How many players are there for all ten schools?
Mathematics
2 answers:
jolli1 [7]3 years ago
6 0
156 players in all ten schools
svetoff [14.1K]3 years ago
5 0
<span>Each baseball team has nine players and each football team has eleven players. Six schools have both baseball and football teams. Four schools have only a baseball team. How many players are there for all ten schools. The answer is 56 players. The six schools have 54 plus 66 players (baseball and football respectively ie 9x6 and 11x6) and the four schools have 9x4 =36 players so 54 + 66 +36= 156 players total.</span>
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Answer:

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29 x 6 = 174

Step-by-step explanation:

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Step-by-step explanation:


17 + 2 = 19 km is 1/2 of the total length, that is,


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Pls help ;-;
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C

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A closed circle means inclusive, and -2 is included in values that would make the equation true. Then you just have to test another number greater than and less than -2 to see which way the arrow should point.

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4 years ago
Consider the probabilities of people taking pregnancy tests. Assume that the true probability of pregnancy for all people who ta
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Using conditional probability, it is found that there is a 0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

<h3>What is Conditional Probability?</h3>

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • P(A \cap B) is the probability of both A and B happening.
  • P(A) is the probability of A happening.

In this problem, the events are:

  • Event A: Positive test result.
  • Event B: Pregnant.

The probability of a positive test result is composed by:

  • 99% of 10%(truly pregnant).
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Hence:

P(A) = 0.99(0.1) + 0.02(0.9) = 0.117

The probability of both a positive test result and pregnancy is:

P(A \cap B) = 0.99(0.1)

Hence, the conditional probability is:

P(B|A) = \frac{0.99(0.1)}{0.117} = 0.8462

0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

You can learn more about conditional probability at brainly.com/question/14398287

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