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DanielleElmas [232]
3 years ago
5

The relationship between the length and weight of certain sea turtles can be approximated by the equation L = 0,55 W,

Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
5 0

Answer:

4.55 pounds

Step-by-step explanation:

From the above question , we are told that:

The relationship between the length and weight of certain sea turtles can be approximated by the equation

L = 0.55 W,

where

L is the length in feet

W is the weight in pounds. If a sea turtle is 2.5 feet long, which is closest to its weight?

L = 2.5 feet

Hence,

2.5 = 0.55W

W = 2.5/0.55

W = 4.5454545455 pounds

Approximately, weight of the sea turtle in pounds = 4.55 pounds

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The heights of a certain type of tree are approximately normally distributed with a mean height y = 5 ft and a standard
Arisa [49]

Answer: Fourth Option:

<em>"A tree with a height of 6.2 ft is 3 standard deviations above the mean"</em>

Step-by-step explanation:

It is said that an X value is found Z standard deviations from the mean mu if:

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In this case we have that:

\mu=5\ ft

\sigma=0.4\ ft

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

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

Z=\frac{5.4-5}{0.4}=1

This means that: <em>A tree with a height of 5.4 ft is 1 standard deviation </em><em>above </em><em>the mean</em>

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

Z=\frac{4.6-5}{0.4}=-1

This means that: <em>A tree with a height of 4.6 ft is 1 standard deviation </em><em>below </em><em>the mean</em>

Second Option: False

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

Z=\frac{5.8-5}{0.4}=2

This means that: <em>A tree with a height of 5.8 ft is </em><em>2 </em><em>standard deviation </em><em>above</em><em> the mean</em>

Third Option: False

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

Z=\frac{6.2-5}{0.4}=3

This means that: <em>A tree with a height of 6.2 ft is 3 standard deviations </em><em>above</em><em> the mean.</em>

Fourth Option: True

3 0
3 years ago
Read 2 more answers
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