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yKpoI14uk [10]
3 years ago
8

A class of 32 students is organised in 33 teams every team consists of 3 students and there are no identical teams . show that t

here are two teams with exactly one common student
Mathematics
1 answer:
Tresset [83]3 years ago
6 0

Answer:

Step-by-step explanation:

Let's start by making up as many teams as we can with the 32 student. Given that each team is different, we can make 10 teams of 3 each. (we still have 23 more teams to make).

The last two people make a team of only 2. No matter which student from the 30 other students is picked, the team of two and the one the student is coming from will have one student in common. Though there are more borrowings that take place (many more), the results remain as stated. At least 2 teams will have 1 person in common.

The method is called the pigeon hole method.

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Just need someone to check my answer
mixas84 [53]
<span>The correct answer is 216x</span>⁶<span>y</span>⁵<span>.

Explanation:
The first thing we do is raise the last monomial to the third power.

(4xy)(2x</span>²<span>y)(3xy)</span>³
<span>=(4xy)(2x</span>²<span>y)(3</span>³<span>x</span>³<span>y</span>³<span>)
=4xy(2x</span>²<span>y)(27x</span>³<span>y</span>³<span>).

Now we can multiply the first two monomials. When we multiply powers with the same base, we add the exponents:
8x</span>³<span>y</span>²<span>(27x</span>³<span>y</span>³<span>).

We multiply these last two monomials, again adding the exponents:
216x</span>⁶<span>y</span>⁵<span>.</span>
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Line I and h intersect at what point
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please show where the lines are graphed in a picture

Step-by-step explanation:

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3 years ago
Marty earns $24 per hour. Marissa earns a base weekly salary of $100 plus $8 per hour. How long will Marty and Marissa have to w
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A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a variance of
iris [78.8K]

Answer:

The probability that the mean battery life would be greater than 533.2 minutes (in a sample of 75 batteries) is \\ P(z>0.48) = P(x>533.2) = 0.3156

Step-by-step explanation:

The main thing we have to take into account in this question is that we are about to find the probability of a <em>sample mean</em>. The distribution for <em>sample means</em> follows a <em>normal distribution</em> with mean \\ \mu and standard deviation \\ \frac{\sigma}{\sqrt{n}}. Mathematically

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

For values of the sample \\ n \ge 30, no matter the distribution the data come from.

And the variable <em>z</em> follows a <em>standard normal distribution</em>, and, as we can remember, this distribution has a mean = 0 and a standard deviation = 1. Mathematically

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} [1]

That is

\\ z \sim N(0, 1)

We have a variance of 3364. That is, a <em>standard deviation</em> of

\\ \sigma^2 = 3364; \sigma = \sqrt{3364} = 58

The population mean is

\\ \mu = 530

The sample size is \\ n = 75

The sample mean is \\ \overline{x} = 533.2

With all this information, we can solve the question

The probability that the mean battery life would be greater than 533.2 minutes

Using equation [1]

\\ z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ z = \frac{533.2 - 530}{\frac{58}{\sqrt{75}}}

\\ z = \frac{3.2}{\frac{58}{8.66025}}

\\ z = \frac{3.2}{6.69726}

\\ z = 0.47780

With this value of z we can consult a <em>cumulative standard normal table</em> (or use some statistic program) to find the cumulative probability for <em>z</em> (and remember that this variable follows a standard normal distribution).

Most standard normal tables have values for z for only two decimals, so we can round the previous value for z as z = 0.48.

Then

\\ P(z

However, in the question we are asked for \\ P(z>0.48) = P(x>533.2). As well as all normal distributions, the standard normal distribution is symmetrical around the mean, and we have

\\ P(z>0.48) = 1 - P(z

Thus

\\ P(z>0.48) = 1 - 0.68439

\\ P(z>0.48) = 0.31561

Rounding to four decimal places, we have

\\ P(z>0.48) = 0.3156

So, the probability that the mean battery life would be greater than 533.2 minutes is (in a sample of 75 batteries) \\ P(z>0.48) = P(x>533.2) = 0.3156.

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alexgriva [62]
X+1
I hope that right
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