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lutik1710 [3]
2 years ago
9

The coach recorded the scores of nine different basketball players in their last game. ● The median number of points scored by t

hese nine players was 13. ● The mean number of points scored by these nine players was 12. Part A: List 9 values that could represent this data set. Part B: A 10th basketball player on this team scored 12 points in the last game. If her score is added to this data set, how will the median and mean change? Explain your reasoning.
Mathematics
1 answer:
IRINA_888 [86]2 years ago
4 0

The values that could represent this data set include 5, 6, 9, 10, 12, 10, 4, 6, and 19.

<h3>How to calculate the mean?</h3>

It should be noted that a mean simply means the average of a set of numbers.

In this case, the values that could represent this data set include 5, 6, 9, 10, 12, 10, 4, 6, and 19. Therefore, the mean will be:

= (5 + 6 + 9 + 10 + 12 + 4 + 6 + 19)/9

= 81/9

= 9

When a 10th basketball player on this team scored 12 points in the last game, the new mean will be:

= (81 + 12)/10

= 93/10

= 9.3

Therefore, there's an increase in mean.

Learn more about mean on:

brainly.com/question/1136789

#SPJ1

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\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\&#10;% function transformations for trigonometric functions&#10;\begin{array}{rllll}&#10;% left side templates&#10;f(x)=&{{  A}}sin({{  B}}x+{{  C}})+{{  D}}&#10;\\\\&#10;f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}\\\\&#10;f(x)=&{{  A}}tan({{  B}}x+{{  C}})+{{  D}}&#10;&#10;\end{array}\qquad

\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks}\\&#10;\quad \textit{horizontally by amplitude } |{{  A}}|\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\end{array}

\bf \begin{array}{llll}&#10;&#10;&#10;\bullet \textit{vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{function period or frequency}\\&#10;\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\&#10;\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)&#10;\end{array}

now, with that template in mind, let's see

\bf \begin{array}{llccll}&#10;cos(&4x&+0)&+2\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&B&C&D&#10;\end{array} &#10;\\\\\\&#10;period\qquad \cfrac{2\pi }{B}\implies \cfrac{2\pi }{4}\implies \cfrac{\pi }{2}&#10;\\\\\\&#10;\textit{horizontal/phase shift}\qquad \cfrac{C}{B}\implies \cfrac{0}{4}\implies 0\impliedby none&#10;\\\\\\&#10;\textit{vertical shift}\qquad D=+2\impliedby \textit{2 units up}

now the range, how far up and down it goes on the y-axis

well, for the graph of cos(x), the range is, goes up to 1, down to -1, is all,

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but with the midline at 2, goes up to 3 and down to 1, so the range is 3 ⩽ y ⩽ 1
4 0
4 years ago
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