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kotegsom [21]
3 years ago
9

X(x-5)+3(x-5) please I really need help with this

Mathematics
1 answer:
Maru [420]3 years ago
5 0
Is this the answer you were looking for? If not I’m sorry

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Is 1.75 larger than 6.2
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No 6.2 could be put as 6.20 and 1.75

6 is bigger than 1

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Kyle has a total of 16 coins in his pocket. All the coins are nickels or quarters, and they have a total value of $2.20. How man
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Answer:

9 nickels

Step-by-step explanation:

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3 years ago
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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
Erin and her 7 friend want to share 6 small pizzas equally how much pizza will each of the 8 children get
enot [183]
Each child can get 3/4 of a pizza if the pizzas are divided into fourths
3 0
3 years ago
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assume 1 u.s dollar=7.80 h.k dollars if a supplier quotes us a price of 23.40 h.k each,what is our cost in u.s dollars?
frez [133]
3 US dollars each

23.40 divided by 7.8 = 3
8 0
3 years ago
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