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Eddi Din [679]
3 years ago
9

|x|, if x=10; .3; 0; −2.7; −9

Mathematics
1 answer:
alekssr [168]3 years ago
5 0

Answer:

For \left | x \right | we will have positive values only.

\left | -2.7 |\right =2.7

\left | -9 |\right =9

Rest are positive so will give positive values out of the modulus function.

Step-by-step explanation:

The absolute value of any number is its positive value only.

As for example (-2.7) and (2.7) both are having equal distance from 0 in a number line, one on the left other on the right side of the number line but are equidistant from 0.

So absolute value of a number is its distance on number line from reference point 0.

One by one we will work with all of our question.

\left | 10 |\right =10

\left | 0.3 |\right =0.3

\left | 0 |\right =0

\left | -2.7 |\right =2.7

\left | -9 |\right =9

So we have to take the positive number out of the modulus function that is also called absolute value.

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A model rocket is launched with an initial upward velocity of 60 m/s. The rocket's height h (in meters) after t seconds is given
Vedmedyk [2.9K]

Answer:

0.47 and 11.53

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h = 60t − 5t²

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5t² − 60t + 27 = 0

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. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

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}

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.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

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

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
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