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SVEN [57.7K]
4 years ago
12

The amount of money that high school students spend on fast food each month is usually between $50 and $200. However, there are

a few students who do not eat fast food at all. What measure of spread would be most appropriate to measure the amount of money that high school students spend on fast food per month?
A)Mean

B)Interquartile range

C)Range

D)Standard deviation
Mathematics
1 answer:
sashaice [31]4 years ago
4 0

Answer:

B) Interquartile range

Step-by-step explanation:

Since there are some students that do not eat fast food at all, their amounts spent monthly would be $0.  These are much different than the rest of the data, and thus are outliers.

Outliers can greatly affect the mean.  This tells us the mean may not be the best measure.

Since the mean is affected by outliers, the standard deviation is not the best measure either.

Adding data points of 0 will also affect the range.  This leaves the interquartile range as the better measure.

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You have to divide 189.25 hours by 23 days.
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3 years ago
What is the answer in the picture above
Rudiy27

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3 years ago
Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

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3 years ago
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Step-by-step explanation:

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Step-by-step explanation:

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