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3 years ago

Ana bikes for 5.5 hours. She covers a distance of 57.75 mi. Assuming she bikes at a constant speed, how fast was Ana biking?

Mathematics

2 answers:

Anastaziya [24]3 years ago

5 0

Answer:

D. 10.5 mi/h

Step-by-step explanation:

I took the test online in K12 and when I reviewed it said that 10.5 is the correct answer.

Doss [256]3 years ago

3 0

The answer is D. because if you do 57.75 divided by 5.5 you get 10.5 mi/h

hope this helps :)

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If an airplane descends 15 feet every 1.45 seconds, how far would the airplane travel in three and a half minutes? The equation

NemiM [27]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Here we go ~

The air plane descends 15 feels every 1.45 seconds, so it's unit rate is :

$\qquad \sf \dashrightarrow \: \dfrac{15}{1.45} \: \: ft / sec$

Now we have to calculate the descent in 3.5 minutes ~

$\qquad \sf \dashrightarrow \: \dfrac{15}{1.45} \: \: \times 3.5 \times 60$

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So, the airplane descends 2172.41 feels in three and a half minutes ~

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1 year ago

General form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)

jonny [76]

hello :<span>
<span>an equation of the circle Center at the A(a,b) and ridus : r is :
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3 years ago

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Sidana [21]

Answer:

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Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

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$\mu_{\bar X} = \mu$

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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".

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$X \sim N(\mu,\sigma)$

Where $\mu$ the mean and $\sigma$ the deviation

Since the distribution for X is normal then we can conclude that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = \mu$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}$

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