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4 years ago

How many years are there in one bimillennium

1 answer:

5 0

A bi-millennium means two (bi) times thousand of years (millenium), this means that a bi millennium has 2000 years in it: two thousand years.

The origin of this word is latin: because of this it also has the plural of "bimillennia"

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What is the calculation of <img src="https://tex.z-dn.net/?f=2%205%20%5C6%20-%20%284%201%20%5C3%20-%203%20%5C2%29" id="TexFo

shepuryov [24]

25 ( 39 -9)

6 (6)

25  × 30

6 6

125

20 5  [ans]

8 0

2 years ago

The sum of 3 times a number and 6 is equal to 2 times the number

Mrrafil [7]

Answer:

3x + 6 = 2x

Solving for x gives x = -6.

Step-by-step explanation:

Let the number be x, then:

3x + 6 = 2x

3x - 2x = -6

x = -6.

7 0

3 years ago

Help plz

Vlad [161]

I think the answer is C hope it helps

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3 years ago

Juries should have the same racial distribution as the surrounding communities. According to the U.S. Census Bureau, 18% of resi

Ymorist [56]

Answer:

0.997 = 99.7% probability that the resulting sample proportion to be between 0.066 and 0.294

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

18% of residents in Minneapolis, Minnesota, are African Americans. Suppose a local court will randomly sample 100 state residents and will then observe the proportion in the sample who are African American.

This means that $p = 0.18, n = 100$