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serious [3.7K]
3 years ago
11

Pseudorandom numbers exhibit a ________ in order to be considered truly random.

Mathematics
1 answer:
stellarik [79]3 years ago
5 0
I believe it is B, uniform distribution
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Which of the following is the MOST accurate definition of standard deviation
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The MOST accurate definition of standard deviation is the mean absolute deviation of the sum of the squared deviation from the average. Option 4

<h3>Definition of standard deviation</h3>

Standard deviation can be defined as a statistic tool that measures the dispersion of a dataset in relation to its mean and is calculated as the square root of the available variance of the set.

It is calculated as the square root of the given variance.

Thus, the MOST accurate definition of standard deviation is the mean absolute deviation of the sum of the squared deviation from the average.

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Find the output for each input of y = -2x.
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Two fractions have a common denominators of 8. What could the two fractions be?
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A researcher wishes to be 95% confident that her estimate of the true proportion of individuals who travel overseas is within 3%
andre [41]

Answer:  a) 683   b) 1067

Step-by-step explanation:

The confidence interval for population proportion is given by :-

p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}

a) Given : Significance level :\alpha=1-0.95=0.05

Critical value : z_{\alpha/2}}=\pm1.96

Margin of error : E=0.03

Formula to calculate the sample size needed for interval estimate of population proportion :-

n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.2(0.8)(\dfrac{1.96}{0.03})^2=682.951111111\approx683

Hence, the required sample size would be 683 .

b) If no estimate of the sample proportion is available then the formula to calculate sample size will be :-

n=0.25(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.25(\dfrac{1.96}{0.03})^2=1067.11111111\approx1067

Hence, the required sample size would be 1067 .

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3 years ago
A number k is multiplied by 3/4. the product is 12. What is the value of k?
elena55 [62]

Answer:

3/4 × 12/1

3×12/4×1

36/4

9

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3 years ago
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