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amid [387]
3 years ago
15

Place these numbers on the number line. 1/4, 2/3, 3/4, 6/8

Mathematics
1 answer:
Lelu [443]3 years ago
4 0
The best way to put them into a number line is to make sure that they've all got the same denominator, which in this case, will be 24, as this is the lowest common multiple of 3,4 and 8. You will then need to multiply these fractions accordingly. The first one (1/4) will need to be multiplied by 6 to get a denominator of 24, so this will then give you a result of 6/24. The second one (2/3), will need to be multiplied by 8, which will then give you a result of 16/24. The third one (3/4)will also need to be multiplied by 6, giving you 18/24. The final one (6/8) will need to be multiplied by 3, giving you 18/24. You will then need to put these in order, which will be- 1/4, 2/3, 3/4, and 6/8. The alternative method is to turn these into decimals- 1/4 is equivalent to 25%, which is equal to 0.25. 2/3 is equivalent to 66.66%, which is equivalent to 0.66. 3/4 is equivalent to 75%, which is equal to 0.75. 6/8 is also equivalent to 75%, which is equal to 0.75, you can then easily line them up this way. Hope this has been able to help you
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Answer:

15.87% probability that a randomly selected individual will be between 185 and 190 pounds

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 180, \sigma = 8

What is the probability that a randomly selected individual will be between 185 and 190 pounds?

This probability is the pvalue of Z when X = 190 subtracted by the pvalue of Z when X = 185. So

X = 190

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

Z = \frac{190 - 180}{8}

Z = 1.25

Z = 1.25 has a pvalue of 0.8944

X = 185

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

Z = \frac{185 - 180}{8}

Z = 0.63

Z = 0.63 has a pvalue of 0.7357

0.8944 - 0.7357 = 0.1587

15.87% probability that a randomly selected individual will be between 185 and 190 pounds

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