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nasty-shy [4]
3 years ago
12

5.818, 5 9/25, 53/10, 5.81 from least to greatest. THanKs :)

Mathematics
2 answers:
Igoryamba3 years ago
7 0

Answer: \frac{53}{10}, 5\ \frac{9}{25},5.81,5.818

Step-by-step explanation:

You can rewrite the fraction as a decimal number. Divide the numerator by the denominator. Then:

\frac{53}{10} =5.3

You can also rewrite the mixed number as a decimal number. Divide the numerator by the denominator and add this to the whole number. Then:

5\ \frac{9}{25}=5+0.36=5.36

Now, you can order the number from least to greatest:

5.3,5.36,5.81,5.818

Or:

\frac{53}{10}, 5\ \frac{9}{25},5.81,5.818

Nastasia [14]3 years ago
4 0

Answer:

53/10, 5 9/25 ; 5.81 ; 5.818

Step-by-step explanation:

To find the solution we need to make some calculations:  

53/10 = 5.3

5 9/25 = 5.36

So we have that the solution is:

53/10, 5 9/25 ; 5.81 ; 5.818

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The mean points obtained in an aptitude examination is 159 points with a standard deviation of 13 points. What is the probabilit
Korolek [52]

Answer:

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are 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.

Central limit theorem:

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 s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 159, \sigma = 13, n = 60, s = \frac{13}{\sqrt{60}} = 1.68

What is the probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled?

This is the pvalue of Z when X = 159+1 = 160 subtracted by the pvalue of Z when X = 159-1 = 158. So

X = 160

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

By the Central Limit Theorem

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

Z = \frac{160 - 159}{1.68}

Z = 0.6

Z = 0.6 has a pvalue of 0.7257

X = 150

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

Z = \frac{158 - 159}{1.68}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.7257 - 0.2743 = 0.4514

0.4514 = 45.14% probability that the mean of the sample would differ from the population mean by less than 1 point if 60 exams are sampled

7 0
3 years ago
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katen-ka-za [31]

Answer:

0 - 10x + 1.5

Step-by-step explanation:

-2(5x - 0.75)

Expand the bracket

-10x + 1.5

0 - 10x + 1.5 is the equivalent expression

3 0
3 years ago
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Answer:

12 with a remainder of 5

Step-by-step explanation:

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2 years ago
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Complete The Square:<br> (Z^2)-19z=66
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To solve the equation using complete square method we proceed:
z^2-19z=66
but
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factoring the LHS we get:
1/4(2z-19)^2=625/4

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8 0
3 years ago
Find equivalent ways to rewrite the expression 6a^8÷3a^4​
solong [7]

Answer:

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Step-by-step explanation:

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