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svp [43]
3 years ago
10

A survey was taken of students in math classes to find out how many hours per day students spend on social media. The survey res

ults for the first-, second-, and third-period classes are as follows:
First period: 2, 4, 3, 1, 0, 2, 1, 3, 1, 4, 9, 2, 4, 3, 0

Second period: 3, 2, 3, 1, 3, 4, 2, 4, 3, 1, 0, 2, 3, 1, 2

Third period: 4, 5, 3, 4, 2, 3, 4, 1, 8, 2, 3, 1, 0, 2, 1, 3

Which is the best measure of center for third period and why?

A: Interquartile range, because there is 1 outlier that affects the center

B: Standard deviation, because there are no outliers that affect the center

C: Mean, because there are no outliers that affect the center

D: Median, because there is 1 outlier that affects the center
Mathematics
2 answers:
MissTica3 years ago
6 0

Answer: D : Median, because there is 1 outlier that affects the center

Step-by-step explanation:

Mean and median terms are use to measure the central tendency in a data set.

Mean is the best measure of center if there is no outlier present in the data otherwise median is the best measure of center if there is outlier present because outliers effects the means value of a data but not the median value.

In the third period, there is an outlier present as 8 that affects the center, then the best measure of center for third period must be Median.

Kamila [148]3 years ago
6 0

Answer:

I reckon,the correct option is option C,Mean, because there are no outliers that affect the center.

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a: Answer is cuboid

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Shakira went bowling with her friends. She paid $3 to rent shoes and then $4.75 for each game of bowling. If she spent a total o
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4 0
3 years ago
X²+y²-6x+14y-1=0<br> and please show your work so I can learn
Eva8 [605]

Hello there,

I hope you and your family are staying safe and healthy during this winter season.

x^2 + y^2 -6x+14y-1=0

We need to use the Quadratic Formula*

x =\frac{-b+\sqrt{b^2}-4ac }{2a} , \frac{-b-\sqrt{b^2} -4ac }{2a}

Thus, given the problem:

a = 1, b=-6, c=y^2+14y-1

So now we just need to plug them in the Quadratic Formula*

x=\frac{6+2\sqrt{(-6)^2-4(y^2+14y-1)} }{2} , x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}

As you can see, it is a mess right now. Therefore, we need to simplify it

x=\frac{6+2\sqrt{10-y^2-14y} }{2}, x = \frac{6-2\sqrt{10-y^2-14y} }{2}

Now that's get us to the final solution:

x=3+\sqrt{10-y^2-14y}, x=3-\sqrt{10-y^2-14y}

It is my pleasure to help students like you! If you have additional questions, please let me know.

Take care!

~Garebear

3 0
3 years ago
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