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2 years ago

What are the last two digits of 7^1867?

1 answer:

3 0

Considering that the powers of 7 follow a pattern, it is found that the last two digits of $7^{1867}$ are 43.

<h3>What is the powers of 7 pattern?</h3>

The last two digits of a power of 7 will always follow the following pattern: {07, 49, 43, 01}, which means that, for $7^n$ , we have to look at the remainder of the division by 4:

If the remainder is of 1, the last two digits are 07.

If the remainder is of 2, the last two digits are 49.

If the remainder is of 3, the last two digits are 43.

If the remainder is of 0, the last two digits are 01.

In this problem, we have that n = 1867, and the remainder of the division of 1867 by 4 is of 3, hence the last two digits of $7^{1867}$ are 43.

More can be learned about the powers of 7 pattern at brainly.com/question/10598663

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In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. In addition, there

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Answer:

15.87% probability that a random sample of 16 people will exceed the weight limit

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 normally distributed samples are solved using 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 sums, the theorem can be applied, with mean $n*\mu$ and standard deviation $s = \sqrt{n}*\sigma$

In this problem, we have that:

$n = 16, \mu = 16*150 = 2400, s = \sqrt{16}*27 = 108$

If a random sample of 16 persons from the campus is to be taken, what is the chance that a random sample of 16 people will exceed the weight limit

This is 1 subtracted by the pvalue of Z when X = 2508. So

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

By the Central Limit Theorem

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