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frez [133]
2 years ago
8

The histogram to the right represents the weights​ (in pounds) of members of a certain​ high-school math team. How many team mem

bers are included in the​ histogram? A histogram has a horizontal axis labeled from less than 115 to 235 plus in intervals of 20 and a vertical axis labeled from 0 to 6 in intervals of 1. Vertical bars, 20 units wide and centered on the horizontal axis values, have heights as follows: 115, 1; 135, 4; 155, 3; 175, 1; 195, 5; 215, 2; 235, 3. 115 135 155 175 195 215 235 0 2 4 6 The histogram represents nothing math team members.
Mathematics
1 answer:
Luden [163]2 years ago
5 0

Answer:

19

Step-by-step explanation:

The vertical axis of a histogram indicates the total amount of people that meet the weights indicated in the x-axis in the analysed population.  From data we can construct the following table:

x-axis   y-axis  

115        1

135       4

155       3  

175        1  

195       5  

215       2  

235      3

So there are: 1+4+3+1+5+2+3 = 19 team members included in the​ histogram.

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Step explanation:

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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
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Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

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Hey!


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Took the test <3

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