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cestrela7 [59]
2 years ago
11

Marcus asked 10 people at a juggling festival what age they were when they started to juggle

Mathematics
1 answer:
Scrat [10]2 years ago
6 0

Question

Marcus asked 10 people at a juggling festival what age they were when they started to juggle. Which interval contains the median age?

Answer:

See Explanation

Step-by-step explanation:

Given

n = 10

Required

The median interval

The question is incomplete, as the required data is not given.

To solve this question, I will use the following assumed dataset.

Age:17\ 17\ 18\ 20\ 21\ 22\ 22\ 24\ 28\ 28

First, calculate the median position.

Median = \frac{n+1}{2}\ th

Median = \frac{10+1}{2}\ th

Median = 5.5\ th

This implies that the median is the mean of the 5th and 6th data

So, we have the interval to be.

Median = [5th, 6th]

Median=[21,22]

<em>Generally, the median of 10 data set is located at interval 5 to 6</em>

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RSB [31]

Hey there!

The median is 8.5

The mean is 7

The range is 11

Hope this helps!

God bless ❤️

xXxGolferGirlxXx

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3 years ago
Solve for x.
Ivahew [28]
ANSWER

x=\frac{2-\sqrt{10}} {3}

or

x=\frac{\sqrt{10}+2} {3}

We have

3x^2-4x-2=0

Since we cannot factor easily, we complete the square.

Adding 2 to both sides give,

3x^2-4x=2

Dividing through by 3 gives

x^2-\frac{4}{3}x= \frac{2}{3}

Adding (-\frac{2}{3})^2 to both sides gives

x^2-\frac{4}{3}x+(-\frac{2}{3})^2= \frac{2}{3}+(-\frac{2}{3})^2

The expression on the Left Hand side is a perfect square.

(x-\frac{2}{3})^2= \frac{2}{3}+\frac{4}{9}

\Rightarrow (x-\frac{2}{3})^2= \frac{10}{9}

\Rightarrow (x-\frac{2}{3})=\pm \sqrt{\frac{10}{9}}

\Rightarrow (x)=\frac{2}{3} \pm {\frac{\sqrt{10}}{3}

Splitting the plus or minus sign gives

x=\frac{2- \sqrt{10}} {3}

or

x=\frac{\sqrt{10}+2} {3}
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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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