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Nuetrik [128]
3 years ago
15

Solve the following equation for a a-32=47​

Mathematics
2 answers:
bekas [8.4K]3 years ago
7 0

Answer:

5

Step-by-step explanation:

because use the calculator pls ......................

Verizon [17]3 years ago
6 0

Answer:

value of a = 79

Step-by-step explanation:

by solving it we get,

=》a - 32 = 47

=》a = 47 + 32

=》a = 79

hence, a = 79

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How many solutions can be found for the equation 5x + 3(x − 1) = 10x − 2x − 3?
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Using the Box plot you drew for the following data set, describe how the data is dispersed: 22, 35, 18, 30, 37, 20, 40, 18, 38,
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3 years ago
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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