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Marina CMI [18]
3 years ago
8

What is the are of a triangle if the B=12 and the H=4?

Mathematics
2 answers:
Nitella [24]3 years ago
7 0
A=B*H/1/2

So you would multiply the Base times Height 
so you should do this 
12x4=48
1/2 of 48 is 24
so the area is 24
Anuta_ua [19.1K]3 years ago
6 0
24 units because 12*4=48 and 48/2=24
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Step-by-step explanation:

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Myra has a 17 foot roll of crepe paper to make 8 streamers to decorate for a party. How long will each streamer be if she cuts t
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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
Can someone please help me with these last 3 questions? <br> Offering 30 points (10 per question)
suter [353]
10. 4:5....added = 9
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     5/9 (1800) = 9000/9 = 1000 (green)
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11. 7/4 = 21/r
      cross multiply
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      7r = 84
       r = 84/7
       r = 12....Rod has 12 stickers

12. 14/4 = 3.5 eggs per cup
      for 16 cups...16 * 3.5 = 56 eggs
3 0
3 years ago
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