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AURORKA [14]
3 years ago
9

Steve wanted to build a small racetrack for his remote-control car in his back yard. He used makeshift equipment to make a perfe

ct circle for his racetrack. The circle had a radius of 32 feet. If a car takes 3 complete turns around the racetrack, approximately, how many feet did the car travel?
F) 200.96 feet
G) 602.88 feet
H) 301.44 feet
J) 614.43 feet
Mathematics
1 answer:
Bad White [126]3 years ago
3 0

Answer:

Steve wanted to build a small racetrack for his remote-control car in his back yard. He used makeshift equipment to make a perfect circle for his racetrack. The circle had a radius of 32 feet. If a car takes 3 complete turns around the racetrack, approximately, how many feet did the car travel?

F) 200.96 feet

G) 602.88 feet

H) 301.44 feet

J) 614.43 feet

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A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

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

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

In this case we have:  \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

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

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Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

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For X =5.8 ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒4^t^h statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

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