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svlad2 [7]
3 years ago
10

Below are the heights (in inches) of students in a third-grade class. Find the median height. 39, 37, 48, 49, 40, 42, 48, 53, 47

, 42, 49, 51, 52, 45, 47, 48
Mathematics
1 answer:
Nutka1998 [239]3 years ago
5 0

Given:

The heights (in inches) of students in a third-grade class are:

39, 37, 48, 49, 40, 42, 48, 53, 47, 42, 49, 51, 52, 45, 47, 48

To find:

The median height.

Solution:

The given data set is:

39, 37, 48, 49, 40, 42, 48, 53, 47, 42, 49, 51, 52, 45, 47, 48

Arrange the data set in ascending order.

37, 39, 40, 42, 42, 45, 47, 47, 48, 48, 48, 49, 49, 51, 52, 53

Here, the number of observations is 16. So, the median of the given data set is:

Median=\dfrac{\dfrac{n}{2}\text{th term}+\left(\dfrac{n}{2}+1\right)\text{th term}}{2}

Median=\dfrac{\dfrac{16}{2}\text{th term}+\left(\dfrac{16}{2}+1\right)\text{th term}}{2}

Median=\dfrac{8\text{th term}+9\text{th term}}{2}

Median=\dfrac{47+48}{2}

Median=\dfrac{95}{2}

Median=47.5

Therefore, the median height of the students is 47.5 inches.

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sattari [20]

We're looking for a solution of the form

y=\displaystyle\sum_{n\ge0}a_nx^n

with derivatives

y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n

y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n

Substituting these into the ODE gives

\displaystyle\sum_{n\ge0}\left(\bigg(2(n+2)(n+1)a_{n+2}-4a_n\bigg)x^{n+2}+2(n+1)a_{n+1}x^{n+1}+(n+2)(n+1)a_{n+2}x^n\right)=0

Shifting indices to get each term in the summand to start at the same power of x and pulling the first few terms of the resulting shifted series as needed gives

2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0

Then the coefficients in the series solution are given according to the recurrence

\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of y(0)=y'(0)=1, a degree-8 approximation of

y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).

5 0
3 years ago
Help pls 40 points with review 3
Marianna [84]

Answer:

Rectangle A is not a scale drawing.

Rectangle B is a scale drawing.

Step-by-step explanation:

A scale drawing involves ratios.

Rectangle A:

10cm is double the original 5cm giving us a 1 : 2 ratio.

However, 25cm is <em>not</em> double the original 20cm. This means that this rectangle is not to scale of the given rectangle.

Rectangle B:

10cm is double the original 5cm giving us a 1 : 2 ratio once again.

This time however, 40cm is double the original 20cm. This means that this rectangle is a scale drawing of the given rectangle.

~Hope this Helps!~

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3 years ago
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antoniya [11.8K]
D=20/3 or in decimal form d=6.666, it keeps going
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If central right angle A intercepts BC, which measures 113, then right angle A measures
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Arc's get their angle measurement from the central angle they're in.

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Igoryamba

Answer:

A is not a function

Step-by-step explanation:

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A is not a function

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