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ohaa [14]
2 years ago
10

This question is too hard

Mathematics
2 answers:
Bond [772]2 years ago
7 0

Answer:270

Step-by-step explanation:180/2=90

90*3=270

Hope this helps :)

aliina [53]2 years ago
6 0

Answer:

270

Step-by-step explanation:

You can do 180/2 and that gives you 90. Since we know 90:1, we can do 90x3. And that gives 270

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13/15 > 4/5 is a true stament.

Step-by-step explanation:

The left side is 0.86 repeating turned into a decimal. which means 0.86 is greater than the right which the right is 0.8 turned into a decimal. which in this case this means that the given stament is always true.

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The United States Coast Guard assumes the mean weight of passengers in commercial boats is 185 pounds. The previous value was lo
Valentin [98]

Answer:

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

Step-by-step explanation:

To solve this problem, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 185, \sigma = 26.7, n = 48, s = \frac{26.7}{\sqrt{48}} = 3.85

The weights of a random sample of 48 commercial boat passengers were recorded. The sample mean was determined to be 177.6 pounds. Find the probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

The probability of an extreme value below the mean.

This is the pvalue of Z when X = 177.6.

So

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

By the Central Limit Theorem

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

Z = \frac{177.6 - 185}{3.85}

Z = -1.92

Z = -1.92 has a pvalue of 0.0274.

So there is a 2.74% of having a sample mean as extreme than that and lower than the mean.

The probability of an extrema value above the mean.

Measures above the mean have a positive z score.

So this probability is 1 subtracted by the pvalue of Z = 1.92

Z = 1.92 has a pvalue of 0.9726.

So there is a 1-0.9726 = 0.0274 = 2.74% of having a sample mean as extreme than that and above than the mean.

Total:

2*0.0274 = 0.0548 = 0.055

There is a 5.5% probability that a random sample of passengers will have a mean weight that is as extreme or more extreme (either above or below the mean) than was observed in this sample.

4 0
3 years ago
a farmer has one square mile of land. suppose the farmer buys another 1/2 square mile of land and divides all his land into squa
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Total area of land = 1 + 1/2 = 3/2 square miles

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