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Akimi4 [234]
3 years ago
15

Franco has 7 quarters in his pocket. Of these, 3 depict the state of Delaware, 2 depict Georgia, 1 depicts Connecticut and 1 dep

icts Pennsylvania. Franco removes 1 quarter from his pocket and then removes a second quarter without replacing the first. What is the probability that both will be Delaware quarters?
Mathematics
2 answers:
tiny-mole [99]3 years ago
6 0

2/6

because after you take one quarter out you have a total of 6 left & you have 3 delaware ones and you already have 1 delaware one out of your pocket it makes it 2/6

ICE Princess25 [194]3 years ago
5 0

Answer:

1/7 is the correct answer

Step-by-step explanation:

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The width of a casing for a door is normally distributed with a mean of 24 in and a standard deviation of 0.14 in. The width of
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Answer:

0.2296 = 22.96% probability that the width of the casing exceeds the width of the door by more than 0.25 in.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Subtraction of normal variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

The width of a casing for a door is normally distributed with a mean of 24 in and a standard deviation of 0.14 in.

This means that \mu_{C} = 24, \sigma_{C} = 0.14

The width of a door is normally distributed with a mean of 23.87 in and a standard deviation of 0.08 in.

This means that \mu_{D} = 23.87, \sigma_{D} = 0.08.

Find the probability that the width of the casing exceeds the width of the door by more than 0.25 in?

This is P(C - D > 0.25).

Distribution C - D:

The mean is:

\mu = \mu_{C} - \mu_{D} = 24 - 23.87 = 0.13

The standard deviation is:

\sigma = \sqrt{\sigma_{C}^2+\sigma_{D}^2} = \sqrt{0.14^2+0.08^2} = 0.1612

Probability:

This probability is 1 subtracted by the pvalue of Z when X = 0.25. So

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

Z = \frac{0.25 - 0.13}{0.1612}

Z = 0.74

Z = 0.74 has a pvalue of 0.7704

1 - 0.7704 = 0.2296

0.2296 = 22.96% probability that the width of the casing exceeds the width of the door by more than 0.25 in.

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