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allsm [11]
2 years ago
13

19. What is the probability that the student only plays football?

Mathematics
1 answer:
Nikolay [14]2 years ago
8 0

Using the probability concept, we have that:

  • The probability that the student only plays football is: (c) 13 /33.
  • The probability that the student plays baseball, but not football is: (d) 8/33.

<h3>What is a probability?</h3>

A probability is given by the <u>number of desired outcomes divided by the number of total outcomes</u>.

The total number of students is given by:

26 + 3 + 9 + 10 + 6 + 5 + 7 = 66

Of those, 26 play only football, hence the probability is:

p = 26/66 = 13/33

Which means that option c is correct for question 19.

Of those same 66 students, 9 + 7 = 16 play baseball but not football, hence the probability is:

p = 16/66 = 8/33

Which means that option d is correct for question 20.

More can be learned about probabilities at brainly.com/question/14398287

#SPJ1

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<span>2 + 8n ≤ 50, so n ≤ 6 </span>
8 0
4 years ago
If a watch has a 0.12 defect rate. what is the probability that the watch has fewer than 3 defects in a run of 75 chimes?
Maslowich

Answer:

(\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}

Step-by-step explanation:

If a watch has fewer than three defects, then either 1.) It has no defects, 2.) it has exactly 1 defect, or 3.) it has exactly 2 defects.

1.) The probability that the watch has no defects is (\frac{88}{100})^{75}, because for every chime there is a probability of \frac{88}{100} that there is no defect

2.) The probability that the watch has exactly 1 defect is (\frac{88}{100})^{74}(\frac{12}{100}) times the number of ways you can choose 1 of 75 of the chimes to be defective, which is \binom{75}{1}, so the probability that the watch has exactly 1 defect is \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}).

3.) For the same reason as 2.), the probability that the watch has exactly two defects is \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}

Since 1.), 2.), and 3.) are mutually exclusive events, the probability of their union is simply the probability of each of them added together, which is (\frac{88}{100})^{75} + \binom{75}{1} (\frac{88}{100})^{74}(\frac{12}{100}) + \binom{75}{2} (\frac{88}{100})^{73}(\frac{12}{100})^{2}

8 0
3 years ago
For a field trip 29 students rode in cars and the rest filled eight buses. how many students were in each bus if 237 students we
Inessa [10]
So if 29 students rode in cars, we can subtract 29 from 237 because we don't need to know about the people in the cars, which means that 208 students rode on buses. If there were 8 buses, the we need to divide 208 by 8 which makes 26 students per bus.

I hope this helps!
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Answer:2

Step-by-step explanation:

Slope=rise/run=(2-0)/(1-0)=2/1=2

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3 years ago
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Step-by-step explanation:

can you retake the picture so i can see it more clearly. please and thank you

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