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shutvik [7]
2 years ago
6

Drake needs to drill holes in a wooden plank. The holes will have a diameter of 0.5 millimeters, with an error allowance of 0.05

millimeters. Which equation can be used to find the minimum and maximum acceptable diameters?
Mathematics
1 answer:
Tanzania [10]2 years ago
6 0

Answer:

 |x − 0.5| = 0.05

Step-by-step explanation:

Here x represents the diameter,

Since, an error of 0.05 millimeters is allowance in diameter,

Thus, the maximum acceptable diameter = x + 0.05,

And, the minimum acceptable diameter = x - 0.05

According to the question,

The holes will have a diameter of 0.5 millimeters,

Thus, x - 0.05 < 0.5  and 0.5 < x + 0.05

⇒ x - 0.5 < 0.05 and x - 0.5 > - 0.05  

⇒ x - 0.5 < 0.05 and -(x -0.5) < 0.05

⇒ |x-0.5| < 0.05

⇒ Option D is correct.

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Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years (Consumer
grandymaker [24]

Answer:

There is a 0.18% probability that a randomly selected TV will have a replacement time less than 5.0 years.

To provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, the length of the warranty is 10.76 years.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. This means that \mu = 8.2, \sigma = 1.1.

Find the probability that a randomly selected TV will have a replacement time less than 5.0 years.

This is the pvalue of Z when X = 5.

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

Z = \frac{5 - 8.2}{1.1}

Z = -2.91

Z = -2.91 has a pvalue of 0.00181

This means that there is a 0.18% probability that a randomly selected TV will have a replacement time less than 5.0 years.

If you want to provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, what is the time length of the warranty?

This is the value of X when Z has a pvalue of 0.99. This is Z = 2.33.

So

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

2.33 = \frac{X - 8.2}{1.1}

X - 8.2 = 2.33*1.1

X = 10.76

To provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, the length of the warranty is 10.76 years.

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