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I am Lyosha [343]
2 years ago
11

Subtract 5 from 1 third of 23. please help me solve this

Mathematics
2 answers:
Travka [436]2 years ago
6 0

Answer:

5

Step-by-step explanation:

noname [10]2 years ago
6 0
5-(1/3 *23) so its 2.6666666667
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The weights of cockroaches living in a typical college dormitory are approximately distributed with a mean of 80 grams and a sta
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Answer:

The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.

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 = 80, \sigma = 4

The percentage of cockroaches weighing between 77 grams and 83 grams

This is the pvalue of Z when X = 83 subtracted by the pvalue of Z when X = 83. So

X = 83

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

Z = \frac{83 - 80}{4}

Z = 0.75

Z = 0.75 has a pvalue of 0.7734

X = 77

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

Z = \frac{77 - 80}{4}

Z = -0.75

Z = -0.75 has a pvalue of 0.2266

0.7734 - 0.2266 = 0.5468

Rounded to the nearest whole number, 55%

The percentage of cockroaches weighing between 77 grams and 83 grams is about 55%.

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