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Kisachek [45]
3 years ago
13

Juanita wants participate in 20 km bicycle race. She is considering using 1 of the 2 training programs below: Program A: Ride 10

km on the first day and increase the distance by an additional 500 m each day Program B: Ride 7,500 m on the first day and increase the distance by an additional 0.75 km each day. If Juanita chooses Program A, on which day she ride on the first 15 km for the first time?
Mathematics
2 answers:
Gemiola [76]3 years ago
8 0
Day 11
this is nothnig im just saying this cause it has to be 20 chharaters long
allsm [11]3 years ago
7 0

She rides 10 kilometer on the first day and 11k on the 3d day I think...........

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The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of m
UkoKoshka [18]

Answer:

Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

Step-by-step explanation:

Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of \bar X Round your answers to two decimal places.

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 delivery times of a population, and for this case we know the distribution for X is given by:

X \sim N(14.7,3.7)  

Where \mu=14.7 and \sigma=3.7

Since the distribution of X is normal then we know that the distribution for the sample mean \bar X is given by:

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

And we have;

\mu_{\bar X}= 14.70

\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59

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Answer:

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Step-by-step explanation:

Dimensions of the large ∆:

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Dimensions of the small ∆:

base (b) = 2cm, height (h) = 6cm

Area = 0.5*b*h = 0.5*2*6 = 6 cm^2

Difference between the area of the large and the small ∆ = 13.5 - 6 = 7.5 cm²

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Step-by-step explanation:

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well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

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\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

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