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alekssr [168]
3 years ago
6

A blackjack is a 10-point card and an ace for a value of 21. Use your answers to parts (a), (b), and (c) to determine the probab

ility that a player is dealt blackjack. (Hint: Part (d) is not a hypergeometric problem. Develop your own logical relationship as to how the hypergeometric probabilities from parts (a), (b), and (c) can be combined to answer this question.)
Mathematics
1 answer:
kondor19780726 [428]3 years ago
7 0

Answer:

Results:

0.1432

0.0045

0.0905

0.0483

Step-by-step explanation:

Step a:

P(A or 10, A or 10) = 20/52 * 19/51 = 5/13 / 19/51 = 95/663 = 0.1432

Step b:

P(A A) = 4/52 * 3/51 = 1/13 * 1/17 = 1/221 = 0.0045

Step c:

P(10 10) = 16/52 * 15/51 = 4/13 * 5/17 = 20/221 = 0.0905

Step d:

P(A 10 or 10 A) = 2 * 4/52 * 16/51 = 2/13 * 16/51 = 32/663 = 0.0483

As well we get the probability by subtracting a,  b and c:  

P(blackjack): 0.1432 - 0.0905 - 0.0045 = 0.0482  

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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 139 millimeters,
FinnZ [79.3K]

Answer:

0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters

Step-by-step explanation:

To solve this question, we need to understan 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 random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 139, \sigma = 6, n = 40, s = \frac{6}{\sqrt{40}} = 0.9487

Either the sample mean differs by 0.8 mm or less from the population mean, or it differs by more. The sum of these probabilities is decimal 1.

Probability it differs by less than 0.8mm

pvalue of Z when X = 139 + 0.8 = 139.8mm subtracted by the pvalue of Z when X = 139 - 0.8 = 138.2 mm

X = 139.8

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

By the Central Limit Theorem

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

Z = \frac{139.8 - 139}{0.9487}

Z = 0.84

Z = 0.84 has a pvalue of 0.7995

X = 138.2

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

Z = \frac{138.2 - 139}{0.9487}

Z = -0.84

Z = -0.84 has a pvalue of 0.2005

0.7995 - 0.2005 = 0.5990

What is the probability that the sample mean would differ from the population mean by more than 0.8 millimeters?

p + 0.5990 = 0.4010

0.4010 = 40.10% probability that the sample mean would differ from the population mean by more than 0.8 millimeters

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Simplify 2x+6=4x+-2 reorder the terms 6+2x=4x+-2
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Square oabc is drawn on a centimetre grid
Vikentia [17]

Answer:

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Step-by-step explanation:

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2 years ago
The following equation is being multiplied by the LCD. Complete the multiplication to eliminate the denominators
frosja888 [35]

Answer:

(x+2)(x-2) -1(3x)=(x-3)(x-2)

Step-by-step explanation:

x+2/3x - 1/x-2 = x-3/3x

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LCD is (3x)(x-2). so we multiply the whole equation by LCD

(3x)(x-2)(\frac{x+2}{3x}-\frac{1}{x-2}=\frac{x-3}{3x})

We multiply each term by LCD

(3x)(x-2)(\frac{x+2}{3x})-(3x)(x-2)(\frac{1}{x-2})=(3x)(x-2)(\frac{x-3}{3x})

(x+2)(x-2) -1(3x)=(x-3)(x-2)

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