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VikaD [51]
3 years ago
9

If 50% of an item is $34.00, what is the original price? ​

Mathematics
2 answers:
masya89 [10]3 years ago
4 0

Answer:

68.00

Step-by-step explanation:

Arlecino [84]3 years ago
3 0

Answer:

Step-by-step explanation: take it times two to get the answer! 34 times 2= 68. So the answer is 68.00 dollars!

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Dakota runs a half-marathon every 6 weeks, and madison runs a half-marathon every 15 weeks. if they start now, in how many weeks
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Dakota runs every 6 weeks......so he runs at 6 weeks, 12 weeks, 18 weeks, 24 weeks, 30 weeks

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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
Galina-37 [17]

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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