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AURORKA [14]
3 years ago
11

Find the number of possible 5 card hands that contain the cards specified the cards are taken from a standard 52 card deck

Mathematics
1 answer:
Zarrin [17]3 years ago
5 0

Answer:

Step-by-step explanation:

Find the number of possible 5-card hands that contain the cards specified. The cards are taken from a standard 52-card deck.

5 red cards.

Ans: 26C5 = 26!/[21!*5!] = 65780

--------------------------------------------

4 spades and 1 card that is not a spade.

Ans: [13C4*39C1] = 27885

--------------------------------------------

3 face cards (kings, queens, or jacks) and 2 cards that are not face cards.

Ans: 12C3*40C2 = 171,600

============================

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The answer I believe is b

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3 years ago
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lions [1.4K]

Answer:

2. Side length =9

Step-by-step explanation:

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An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

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3 years ago
-5x+y=6<br> -3x+6y=-12<br> Solving systems of equations by substitution
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then if you subtract 36 from both sides you would get -3x+30x=-48 then combine like terms to get 27x=-48 then divide both sides by 27 to get x=\frac{-48}{27} or x≈1.8

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Step-by-step explanation:

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