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vlada-n [284]
3 years ago
8

Shop only had 2 types of stamps left using

Mathematics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

  {3¢, 28¢} or {4¢, 19¢} or {7¢, 10¢}

Step-by-step explanation:

54 = 1×54 = 2×27 = 3×18 = 6×9

Possible values of the stamps are 1 more than the values of a pair of factors. Of course, a 2¢ and 55¢ stamp will not permit paying 54¢ in postage, so that combination won't work. However, other pairs that will work are ...

  • 3¢ and 28¢
  • 4¢ and 19¢
  • 7¢ and 10¢

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Parker is testing 2 different types of ant killer in his yard. He purchases Brand A and Brand B from the home improvement store.
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Corey has 510 marbles he fills one jar with 165 marbles how many are not in the jar
Luba_88 [7]

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A survey found that women's heights are normally distributed with mean 62.7 in, and standard deviation 2.8 in. The survey also f
Viefleur [7K]

Using the normal distribution, we have that:

The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation of men's heights are given as follows:

\mu = 69.3, \sigma = 3.9.

The proportion of men who meet the height requirement is is the <u>p-value of Z when X = 62 subtracted by the p-value of Z when X = 55</u>, hence:

X = 62:

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

Z = \frac{62 - 69.3}{3.9}

Z = -1.87

Z = -1.87 has a p-value of 0.0307.

X = 55:

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

Z = \frac{55 - 69.3}{3.9}

Z = -3.67

Z = -3.67 has a p-value of 0.0001.

0.0307 - 0.0001 = 0.0306 = 3.06%.

The percentage of men who meet the height requirement is 3.06%. This suggests that the majority of employees at the park are females.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

8 0
2 years ago
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