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Ainat [17]
2 years ago
7

Solve the inequality: |n/10| ≤ 4

Mathematics
1 answer:
Shalnov [3]2 years ago
8 0

Given the Absolute Value Inequality:

|\frac{n}{10}|\leq4

You can solve it as follows:

1. Split the inequality into two inequalities:

- Inequality 1:

\frac{n}{10}\leq4

- Inequality 2:

\frac{n}{10}\ge-4

2. Solve for "n" on each inequality:

- Inequality 1:

\begin{gathered} n\leq4\cdot10 \\ n\leq40 \end{gathered}

- Inequality 2:

\begin{gathered} n\ge(-4)(10) \\ n\ge-40 \end{gathered}

3. You can write the solution as follows:

-40\leq n\leq40

Hence, the answer is:

-40\leq n\leq40
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The scale factor in the dilation of a mathematical object determines how much larger or smaller the image will be (compared to the original object). When the absolute value of the scale factor is greater than one, an expansion occurs. When the absolute value of the scale factor is less than one, a compression occurs.

Step-by-step explanation:

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3 years ago
The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100
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Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of 660, hence \mu = 660.
  • The standard deviation is of 90, hence \sigma = 90.
  • A sample of 100 is taken, hence n = 100, s = \frac{90}{\sqrt{100}} = 9.

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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

Z = \frac{670 - 660}{9}

Z = 1.11

Z = 1.11 has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

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