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earnstyle [38]
2 years ago
8

What is the positive solution of x2 – 36 = 5x?

Mathematics
1 answer:
goldenfox [79]2 years ago
3 0

Answer:

9

Step-by-step explanation:

x² − 36 = 5x

x² − 5x − 36 = 0

(x − 9) (x + 4) = 0

x = 9 or -4

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Students in a class were asked to name their favorite sport they like to watch on television. How many more time students want t
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First, we need to add up all the percentages to make sure we have 100%.

25.5% + 0.03% = 25.53%

This means that 74.47% of the students chose something other than basketball or soccer.

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25.5% of 2553 is 651 students and

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Basketball was chosen 81 times over again compared to soccer.

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Basketball was chosen 643 times more than soccer.
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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
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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
1 year ago
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