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olya-2409 [2.1K]
3 years ago
9

I need some DETAILED examples of Dividing Integers! Urgent, pls help ASAP!

Mathematics
1 answer:
vfiekz [6]3 years ago
8 0

Here are some examples:

Integers are positive and negative numbers, but dividing integers are dividing positive and negative numbers:

Remember these rules:

  1. Negative divided by a negative equals a positive.
  2. Negative divided by a positive equals a negative.
  3. Positive divided by a positive equals a positive .

Examples:

  • -4 \div 2 = -2
  • -4 \div -4 = 1
  • 4 \div 2 = 2

Hope this helps!

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Need Help Badly<br>What is the solution
Kamila [148]
Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
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Solve the problem. Use the Central Limit Theorem.The annual precipitation amounts in a certain mountain range are normally distr
bazaltina [42]

Answer:

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

Step-by-step explanation:

To solve this question, we use 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.

Mean of 109.0 inches, and a standard deviation of 12 inches.

This means that \mu = 109, \sigma = 12

Sample of 25.

This means that n = 25, s = \frac{12}{\sqrt{25}} = 2.4

What is the probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches?

This is the p-value of Z when X = 112. So

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

By the Central Limit Theorem

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

Z = \frac{112 - 109}{2.4}

Z = 1.25

Z = 1.25 has a p-value of 0.8944.

0.8944 = 89.44% probability that the mean annual precipitation during 25 randomly picked years will be less than 112 inches.

7 0
3 years ago
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