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maxonik [38]
3 years ago
11

What is an adaptive test?

Mathematics
2 answers:
svetlana [45]3 years ago
5 0

Answer:

Step-by-step explanation:

A form of computer-based test that adapts to the examinee's ability level.

aksik [14]3 years ago
4 0

Answer:

A test that adjusts the difficulty of the questions based on the student's answer to previous questions

Explanation:

APE

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The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

So

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

Z = \frac{0.62 - 0.5}{0.05}

Z = 2.4

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

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What substitution should be used to rewrite 4x^4-21x^2+20=0 as a quadratic equation
Andreyy89

let y = x^2

4y^2 -21y + 20 = 0

solve for y

then substitute back in to get x

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Find the radius of a circle with the given circumference.<br> 12<br> in
vampirchik [111]

Step-by-step explanation:

<em>Given </em>

<em>circumference </em><em>=</em><em> </em><em>1</em><em>2</em><em> </em><em>inch</em>

<em>π</em><em> </em><em>=</em><em> </em><em>3</em><em>.</em><em>1</em><em>4</em>

<em>radius </em><em>(</em><em>r) </em><em> </em><em>=</em><em> </em><em>?</em>

<em>We </em><em>know </em>

<em>circumference </em><em>=</em><em> </em><em>2</em><em>π</em><em>r </em>

<em>1</em><em>2</em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em>*</em><em> </em><em>3</em><em>.</em><em>1</em><em>4</em><em> </em><em>*</em><em> </em><em>r</em>

<em>r </em><em>=</em><em> </em><em>1</em><em>2</em><em> </em><em>/</em><em> </em><em>6</em><em>.</em><em>2</em><em>8</em>

<em>Therefore </em><em> </em><em>r </em><em>=</em><em> </em><em>1</em><em>.</em><em>9</em><em>1</em><em> </em><em>inch</em>

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Well using distance formula we get the length of ab to be 3, knowing a'b'=6 we know the dilator must be 2
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