How to solve a mixed fraction?

1 answer:

5 0

To solve a mixed fraction you will have to multiply the mixed number by the denominator then add the numerator for example:
7 3/4 you are gonna go 7x4=28 then you are gonna take that 28 and add 3 to it so it would be 31. I hope that helped some c:

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Orla and Eduardo each looked at a strand of their hair under a microscope and measured the diameter. Orla's strand was 0.005cm i

bezimeni [28]

Answer: The diameter of Eduardo's hair is greater than Orla's hair by 0.007 cm.

Step-by-step explanation:

Given: The diameter of Orla's hair strand = 0.005 cm

The diameter of Eduardo's hair strand = 0.012 cm

Clearly , $0.012>0.005$

The difference between the diameters= $0.012-0.005=0.007\ cm$

Thus, the diameter of Eduardo's hair is greater than Orla's hair by 0.007 cm.

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4 years ago

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An HP laser printer is advertised to print text documents at a speed of 18 ppm (pages per minute). The manufacturer tells you th

aliya0001 [1]

Answer:

0.227 = 22.7% probability that the mean printing speed of the sample is greater than 18.12 ppm.

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .