We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

$d=\sqrt{\frac{3h}{2} }$

Where, we have

d- the distance they can see in thousands

h- their eye-level height in feet

For Kaylib

$d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)$

For Addison h=85(1/3)

$d=\sqrt{\frac{3\times 85\frac{1}{3} }{2} }\\d\sqrt{\frac{256}{2} } \\d=\sqrt{128} \\d=8\sqrt{2} .....(2)$

Subtracting both distances we get

$8\sqrt{2}-6\sqrt{2} =2\sqrt{2}$

Therefore, the addison see to the horizon at 2 root 2mi.

To learn more about the eye level visit:

brainly.com/question/1392973

5 0

2 years ago

Barron's reported that the average number of weeks an individual is unemployed is 17.5 weeks. Assume that for the population of

Olin [163]

Answer:

$P(16.5$

And using the normal standard table or excel we find the probability:

$P(-1.768< Z< 1.768) = P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the avergae number of weeks an individual is unemployed of a population, and for this case we know the distribution for X is given by:

$X \sim N(17.5,4)$