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const2013 [10]
3 years ago
14

How do I work out 5% of 360?

Mathematics
2 answers:
Anna11 [10]3 years ago
6 0
5% of 360

5% = 5/100 = 0.5

0.05 × 360

= 18.
fiasKO [112]3 years ago
6 0
360\cdot\frac{5}{100}=3.6\cdot5=18
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Answer:

3.844 x 10^5

3.844E-5

3.844 × 10-5 kilometers

Step-by-step explanation:

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The weight of an object above the surface of the Earth varies inversely with the square of the
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By using the similarity method:

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Can somebody please help me and tell me how did you get the answer please thank you!!!
Aneli [31]

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10

Step-by-step explanation:

(6x-5)+(12x+5)=180

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What're the least common multiple of 3,5, and 7
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3 years ago
2.According to www.city-data, the mean price for a detached house in Franklin County, OH in 2009 was $192,723. Suppose we know t
igor_vitrenko [27]

Answer:

0.7123 = 71.23% probability that a random sample of 75 detached houses in Franklin County had a mean price greater than $190,000 in 2009.

Step-by-step explanation:

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

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean price for a detached house in Franklin County, OH in 2009 was $192,723. Suppose we know that the standard deviation was $42,000.

This means that \mu = 192723, \sigma = 42000

Sample of 75:

This means that n = 75, s = \frac{42000}{\sqrt{75}}

What is the probability that a random sample of 75 detached houses in Franklin County had a mean price greater than $190,000 in 2009?

1 subtracted by the p-value of Z when X = 190000. So

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

By the Central Limit Theorem

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

Z = \frac{190000 - 192723}{\frac{42000}{\sqrt{75}}}

Z = -0.56

Z = -0.56 has a p-value of 0.2877

1 - 0.2877 = 0.7123

0.7123 = 71.23% probability that a random sample of 75 detached houses in Franklin County had a mean price greater than $190,000 in 2009.

5 0
3 years ago
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