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3 years ago

6

Solve the inequality 17u≥−8, and write the solution in interval notation.

1 answer:

xxTIMURxx [149]3 years ago

8 0

Answer:

u= -8/17 [-8, ∞)

Step-by-step explanation:

solve it like regular equation

17u≥-8

divide 17 on both side

then you get u≥-8/17

so, (u) is any value greater than or equal to -8/17

less than equal means solid dot on number line i think its called also mean ([) meaning it includes this value too. It is a multiplication so it can be any number above -8 or -8 itself. Thus, the interval notation will be [-8, ∞). Infinite is not a fix number so we use this (). hope this helps good luck

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3 years ago

The army reports that the distribution of waist sizes among female soldiers is approximately normal, with a mean of 28.4 inches

____ [38]

Using the normal distribution, it is found that:

a) Her waist is at the 2.74th percentile.

b) 99.86% of female soldiers requires custom uniform pants.

<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 are given, respectively, by:

$\mu = 28.4, \sigma = 1.2$ .

Item a:

The percentile is the <u>p-value of Z when X = 26.1,</u> hence:

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

$Z = \frac{26.1 - 28.4}{1.2}$

Z = -1.92

Z = -1.92 has a p-value 0.0274 = 2.74th percentile.

Item b:

The proportion is the <u>p-value of Z when X = 32 subtracted by the p-value of Z when X = 24</u>, hence:

X = 32:

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

$Z = \frac{32 - 28.4}{1.2}$

Z = 3.

Z = 3 has a p-value 0.9987.

X = 24:

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

$Z = \frac{24 - 28.4}{1.2}$

Z = -3.67.

Z = -3.67 has a p-value 0.0001.

0.9987 - 0.0001 = 0.9986 = 99.86% of female soldiers requires custom uniform pants.

More can be learned about the normal distribution at brainly.com/question/25800303

#SPJ1

4 0

2 years ago