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astraxan [27]
2 years ago
13

There are 7 red cards and 13 blue cards in a deck of cards. If Chayse randomly selects a card from the deck, what is the probabi

lity that it is a red card?
Mathematics
1 answer:
True [87]2 years ago
8 0

Answer:

35%

Step-by-step explanation:

There is a 7 in 20 chance of the card being red

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X to the power of two plus five x plus 4
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Answer: (x + 1) ( x + 4)

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Compare the triangles and determine whether they can be proven congruent, if possible, by SSS, SAS, ASA, AAS, or HL. If the tria
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I can't see the picture.

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Jerry buys a case of bar soap. The case holds 12 packages. Each package contains four bars of soap. If each bar of soap is four
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The answer is D. 12 pounds

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2 years ago
Anya’s team won 4 games out of every 9 games this season. If the team played 45 games, how many games did they win?
tiny-mole [99]

Answer:20

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8 0
2 years ago
g If a random sample of 18 deliveries was selected instead and the standard deviation of these delivery times was found to be 9.
lana [24]

Answer:

We can use the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we use this formula we got:

z=\frac{21-20}{\frac{9.31}{\sqrt{18}}}= 0.456

z=\frac{25-20}{\frac{9.31}{\sqrt{18}}}= 2.279

And using a calculator, excel or the normal standard table and we have that:

P(0.456

Step-by-step explanation:

We assume this previous info:  It is known that the amounts of time required for room-service delivery at a certain Marriott Hotel are Normally distributed with the average delivery time of 20 minutes.

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 delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(20,9.31)  

Where \mu=20 and \sigma=9.31

Since the distribution for X is normal then the distribution for the sample mean \bar X is given by:

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

And we can use the z score formula given by:

z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

If we use this formula we got:

z=\frac{21-20}{\frac{9.31}{\sqrt{18}}}= 0.456

z=\frac{25-20}{\frac{9.31}{\sqrt{18}}}= 2.279

And using a calculator, excel or the normal standard table and we have that:

P(0.456

6 0
3 years ago
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