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Bad White [126]
3 years ago
11

Excluding any outliers, what is the mean of the data set {21, 14, 11, 26, 33, 61, 29}?

Mathematics
2 answers:
Mumz [18]3 years ago
5 0

Answer: The mean is 22.

Step-by-step explanation:

If you reorder the terms, (11, 14, 21, 26, 29, 33, 61) you can see that 61 is an outlier. Exclude the value.

Now, add the leftover values together (11, 14, 21, 26, 29, 33) which equals 134.

Finally divide 134/6 (Amount of values) and this equals 22.3(repeating).

Round down to 22.

Hope this helps!

sashaice [31]3 years ago
3 0

Answer: 22

Step-by-step explanation:

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Answer:

cos^2\theta + sin^2\theta = 1

Step-by-step explanation:

Given

(\frac{b}{r})^2  + (\frac{a}{r})^2

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

(\frac{b}{r})^2  + (\frac{a}{r})^2  = (\frac{r}{r})^2

Going by the Pythagoras theorem, we can assume the following.

  • a = Opposite
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So, we have:

Sin\theta = \frac{a}{r}

Cos\theta = \frac{b}{r}

Having said that:

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(\frac{b}{r})^2  + (\frac{a}{r})^2  = 1

Substitute values for sin and cos

(\frac{b}{r})^2  + (\frac{a}{r})^2  = 1 becomes

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3 0
3 years ago
Find the Y-coordinate of point P that lies 1/3 along segment RS, where R (-7, -2) and S (2, 4).
QveST [7]

Solution:

Given that the point P lies 1/3 along the segment RS as shown below:

To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have

\begin{gathered} RP:PS \\ \Rightarrow\frac{1}{3}:\frac{2}{3} \\ thus,\text{ we have} \\ 1:2 \end{gathered}

Using the section formula expressed as

[\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}]

In this case,

\begin{gathered} m=1 \\ n=2 \end{gathered}

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\begin{gathered} x_1=-7 \\ y_1=-2 \\ x_2=2 \\ y_2=4 \end{gathered}

Thus, by substitution, we have

\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}

Hence, the y-coordinate of the point P is

0

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1 year ago
Help!
Anastaziya [24]
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