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SOVA2 [1]
3 years ago
15

Amanda surveyed 13 students in her class about their heights in inches. Her data are listed below: 58, 59, 55, 52, 57, 59, 62, 5

8, 55, 56, 59, 65, 53
What is the IQR of this data set?
What is the Median (Q2) of this data set?
Mathematics
2 answers:
Anton [14]3 years ago
6 0
Iqr is 4, median is 58 hope that answers your question.
Aleonysh [2.5K]3 years ago
3 0

Answer:

IQR is 4, median is 58.

Step-by-step explanation:

Enter these values in a list, hit stat > calc > 1-var stats. That gives you the 5 number summary. To get IQR, subtract Q3 from Q1.

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According to the graph, what is the value of the constant in the equation<br> below?
Anna007 [38]

Answer:

75 or A

Step-by-step explanation:

Just answered it myself! ^^

4 0
3 years ago
Prove the formula for (d/dx)(cos−1(x)) by the same method as for (d/dx)(sin−1(x)). Let y = cos−1(x). Then cos(y) = and 0 ≤ y ≤ π
gtnhenbr [62]

Answer:

\frac{d(cos^{-1}x )}{dx} = \frac{-1}{\sqrt{1-x^2} }

Step-by-step explanation:

Given the differential (d/dx)(cos−1(x)), to find the equivalent formula we will differentiate the inverse function using chain rule as shown below;

let;

y = cos^{-1} x \\\\taking \ cos\ of\ both\ sides\\\\cosy = cos(cos^{-1} x)\\\\cosy = x\\\\x = cosy\\\\\frac{dx}{dy} = -siny\\

\frac{dy}{dx} = \frac{-1}{sin y}  \\\\from\ trigonometry\ identity,\ sin^{2} x+cos^{2}x = 1\\sinx = \sqrt{1-cos^{2} x}

Therefore;

\frac{dy}{dx} = \frac{-1}{\sqrt{1-cos^{2}y } }

Since x = cos y from the above substitute;

\frac{dy}{dx} = \frac{-1}{\sqrt{1-x^{2}} }

Hence, \frac{d(cos^{-1}x )}{dx} = \frac{-1}{\sqrt{1-x^2} } gives the required proof

5 0
3 years ago
2^3 • 3 evaluate the expression plz help me
Scilla [17]

Answer:

24

Step-by-step explanation:

2^3 * 3 = 8 * 3 = 24

Hope this helped!

3 0
3 years ago
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Solve the linear equation 3(x + 3) = 4(x-6)
vova2212 [387]
Solution for your linear equation is 33
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Question 13 of 20 :
Mekhanik [1.2K]

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Step-by-step explanation:

None of these can be divided by anything except themselves and 1.

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