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Lisa [10]
3 years ago
15

When should you use the median and interquartile range as your measure of center and measure of variability to compare populatio

ns shown on a box plot? Check all that apply.
when there are outliers in the data sets
when there are no big gaps in the middle of the data sets
when the plots representing the data are symmetrical
when the plots representing the data are nonsymmetrical
when there are no outliers in the data set
(yes it IS multiple choice)
Mathematics
2 answers:
Over [174]3 years ago
6 0
Because I don’t like math sorry
Rus_ich [418]3 years ago
5 0

Answer:loooo

Kk

Step-by-step explanation:ooojjjjjjj

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<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2
Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

but the integrand is even, so this is really just

\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx

Substitute x = 1/2 arccot(u/2), which transforms the integral to

\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du

There are lots of ways to compute this. What I did was to consider the complex contour integral

\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}

and it follows that

\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}

7 0
1 year ago
In how many different ways can seven children line up in one row to have their picture taken?
wolverine [178]

49 should be the correct answer.

-TheOneandOnly003


8 0
3 years ago
Read 2 more answers
Karen gave an equal amount of m and m's to herself and four friends. Each person receives m and m's equivalent to the largest on
Mariana [72]

Answer:

Karen had 45 m and m's candy.

Step-by-step explanation:

Let the number of m and m's candy be 'x'.

Now given:

Karen gave an equal amount of m and m's to herself and four friends.

So we can say that;

Number of people m and m's candy distributed equally = 5

Also Given:

Each person receives m and m's equivalent to the largest one digit number.

Now we know that;

Largest one digit number is 9.

So we can say that;

Each person receives m and m's = 9

We need to find number of m and m's Karen have.

Solution:

So we can say that;

Total number of m and m's Karen have is equal to Number of people m and m's candy distributed equally multiplied by number of m and m's can each person receives.

framing in equation form we get;

Total number of m and m's Karen had = 9\times5 = 45

Hence Karen had 45 m and m's candy.

8 0
2 years ago
A 28-year-old man pays $181 for a one-year life insurance policy with coverage of $150,000. If the probability that he will live
Marina CMI [18]
Expected value E(x) = 0.9994(150000 - 181) = 0.9994 x 149819 = $149,729.11
3 0
3 years ago
Divide. Give special attention to the order of division. a. 72 ÷ 9 ÷ 2 b. (18 ÷ 6) ÷ 3 c. 45 ÷ 5 ÷ 3 d. 144 ÷ (12 ÷ 2)
Shkiper50 [21]
A. 72÷9÷2 = 8÷2 = 4
b. (18 ÷ 6) ÷ 3 = 3 ÷ 3 = 1
c. 45 ÷ 5 ÷ 3 = 9 ÷ 3 = 3
d. 144 ÷ (12 ÷ 2) = 144 ÷6 = 24
4 0
3 years ago
Read 2 more answers
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