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9966 [12]
3 years ago
5

You are part of a school and have to report the results of standardized test scores. Which measure of central tendency would you

use? Assume that you don't know how your students will perform; so consider the general effects of outliers on the three measures of central tendency. Research and summarize the general effects that outliers have on each of the measures of central tendency. Then, using some real values, provide an example of how the three measures of central tendency vary due to outlier effects.
Mathematics
1 answer:
USPshnik [31]3 years ago
4 0
You should know this
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If v = 35 , a = 4 , t = 5 and v = u + a t , evaluate u .
Llana [10]

v = u + at

=> 35 = u + (4×5)

=> 35 = u + 20

=> 35 - 20 = u

=> 15 = u

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3 years ago
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What is 2/3 of $300?
wel
Well one third is 100
 so two thirds must be 200
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4 years ago
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Ella has a points card for a movie theater.
MAXImum [283]

Answer:

175<=12.5x+40

Step-by-step explanation:

She starts with 40 points already so you add in those 40 then since she gets 12.5 points per visit that is your x, so you set that as greater than or equal to 175

7 0
3 years ago
There is 5/6 of an apple pie left from dinner. Tomorrow, victor plans to eat 1/6 of the pie that was left. How much of the whole
Tatiana [17]
If there is 5/6 left and he plans to eat 1/6 you do 5/6-1/6=4/6 of the pie
8 0
3 years ago
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Identify the x-intercepts of the function below f(x)=x^2+12x+24
damaskus [11]

<u>ANSWER:  </u>

x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

<u>SOLUTION:</u>

Given, f(x)=x^{2}+12 x+24 -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

Here, for (1) a = 1, b= 12 and c = 24

X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}

\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}

Hence the x-intercepts of  \mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})

8 0
3 years ago
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