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Airida [17]
3 years ago
11

A card game has 50 cards. After dealing 7 cards to each player, Tupi has 15 cards left over. Solve the equation 50 7p = 15 to fi

nd the number of players.
Mathematics
2 answers:
Charra [1.4K]3 years ago
4 0
5 players
50-15=35
35/7=5
mafiozo [28]3 years ago
3 0

Answer:

50 - 7p = 15....subtract 50 from both sides

-7p = 15 - 50

-7p = - 35...divide both sides by -7

p = -35/-7

p = 5 <=== there are 5 players

Step-by-step explanation:

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Read 2 more answers
It is known that IQ scores form a normal distribution with a mean of 100 and a standard deviation of 15. If a researcher obtains
sdas [7]

Answer:

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the IQ scores of a population, and for this case we know the distribution for X is given by:

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

6 0
3 years ago
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