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3 years ago

The number of points that Shira scored each basketball game so far this season is shown on the dot plot.

Mathematics

2 answers:

Colt1911 [192]3 years ago

5 0

The data is skewed to the left and shows that she never scored fewer than 2 points or more than 12 points in a game.

valentina_108 [34]3 years ago

3 0

Answer:

C. The data is skewed to the left and shows that she never scored fewer than 2 points or more than 12 points in a game.

Step-by-step explanation:

In the given image that is representing the number of points that Shira scored, each basketball game so far this season we can observe that the data is skewed to the left.

Lets consider the definition of skewness of a data set:

A left-skewed distribution has a long left tail. Left-skewed distributions are also known as negatively-skewed distributions. That’s because there is a long tail in the negative direction on the number line. So in this case peak remains on the right side.

A right-skewed distribution has a long right tail. Right-skewed distributions are also called positive-skew distributions because there is a long tail in the positive direction on the number line. So in this case peak remains on the left side.

In our case it can be see that the data has peak on to the right side and hence it is left skewed. Also it can be seen from the dot plot that Shira has never scored fewer than 2 points and the maximum she has scored in a game is 12 points.

Therefore, the data is skewed to the left and shows that she never scored fewer than 2 points or more than 12 points in a game

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2 years ago

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

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3 years ago

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masya89 [10]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers