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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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Developing Proof: Fill in the reasons for this algebraic proof. Given: 5x + 1 = 21 Prove: x = 41) 5x + 1 = 21 Give the reasons f
jenyasd209 [6]

Answer:

21 = 21

Step-by-step explanation:

5x + 1 = 21

-1           -1

5x = 20

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5        5

x = 4

5x + 1 = 21           First substitute x for 4

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3 years ago
The scale factor from rectangle ABCD to rectangle WXYZ is 4. What is the length of side WX?
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Read 2 more answers
. The time required for a technician to machine a specific component is normally distributed with a mean of 2 hours and a standa
erma4kov [3.2K]

Answer:

a) There is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

b) There is a 0.42% probability that the technician will require at least 2.75 hours to complete one component

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.

In this problem, we have to be careful. The mean is in hours, while the standard deviation is in minutes. I am going to work with both in hours, as the problem states. 17 minutes is 0.283 hours, so:

\mu = 2, \sigma = 0.283

(a.) What is the probability that the technician can machine one component in 1.5 hours or less?

This probability is the pvalue of the Zscore when X = 1.5. So:

Z = \frac{1.5 - 2}{0.283}

Z = \frac{-0.5}{0.283}

Z = -1.77

Z = -1.77 has a pvalue of 0.0384.

This means that there is a 3.84% probability that the technician can machine one component in 1.5 hours or less.

(b.) What is the probability that the technician will require at least 2.75 hours to complete one component?

The pvalue of the score of X = 2.75 is the probability that the technican will require less than 2.75 hours to complete one component. The probability that he will require at least 2.75 hours to complete one component is 1 subtracted by this pvalue. So:

Z = \frac{2.75 - 2}{0.283}

Z = \frac{0.75}{0.283}

Z = 2.65

Z = 2.65 has a pvalue of 0.99598.

This means that the probability that the technican will require at least 2.75 hours to complete one component is 1 - 0.99598 = 0.0042 = 0.42%.

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