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Studentka2010 [4]
3 years ago
7

In the above figure, m∠P = 125° and m∠Q = 5x. If angles P and Q are supplementary angles.

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
4 0

9514 1404 393

Answer:

  the correct choice is marked: x = 11; Q = 55°

Step-by-step explanation:

If P and Q are supplementary, then ...

  Q = 180° -P

  Q = 180° -125°

  Q = 55°

Then we have ...

  5x = 55°

  x = 11°

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The second number is going to be 3 times as big as the first ine

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the path of a kicked soccer ball can be modeled by the equation y=3x^2. State whether the equation represents a linear or a nonl
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Answer:

  nonlinear

Step-by-step explanation:

The degree of x in the equation is not 0 or 1, so the function is nonlinear.

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2 years ago
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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