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tigry1 [53]
3 years ago
8

The number of patients treated at Dr. McCormick’s dentist office each day was recorded for nine days. These are the data: 13, 16

, 16, 10, 6, 12, 4, 12, 16. Find the mean, median, mode, and range of the data. If necessary, round to the nearest tenth.
13.7, 12, 16, 20

11.7, 12, 16, 12

12, 13.7, 16, 12

11.7, 16, 12, 20
Mathematics
1 answer:
arlik [135]3 years ago
4 0
Hi!

Let's start with the mean.

The mean is all the numbers added together, divided by how many numbers there are. 

13+16+16+10+6+12+4+12+16=105
there are 9 numbers so..
105/9 = 11.7

The mean is 11.7 

Find the median. 

The median is the number in the middle of the data set when the numbers are in numerical order.

4, 6, 10, 12, 12, 13, 16, 16, 16

The median is 12. 

Find the mode.

The mode is the number that is repeated the most.

4, 6, 10, 12, 12, 13, 16, 16, 16

The mode is 16

Find the range.

The range is the difference of the smallest number from the biggest number.

16 - 4 = 12

The range is 12

The answer is 11.7, 12, 16, 12 

Hope this helps! :)
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Joey and Armando live on the same st as the park. The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando'
AleksandrR [38]
Let's start by assuming Armando's house is between Joey's and the park. 

Let x be the distance Joey walked to Armando's house.

<span>The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando's home. Then Joey and Armando walk 3/5 mile to the park. 

</span>\dfrac{9}{10} = x + \dfrac{3}{5}

x = \dfrac{9}{10} - \dfrac{3}{5} = \dfrac{9}{10} -\dfrac{6}{10} = \dfrac{3}{10}

That's probably the answer they're looking for.  But what if the park is between Joey and Armando's houses or Joey is between the park and Armando?  (The latter isn't really possible with the given distances.)

Let a, b, c be the distances between three collinear points like we have here.  Our equation is really a few equations in one, something like

\pm a \pm b = \pm c

Let's get rid of the plus/minuses. Squaring,

a^2 + b^2\pm 2ab = c^2

\pm 2ab = c^2-a^2-b^2

4a^2b^2 = (c^2-a^2-b^2)^2

For us, that's a quadratic equation for c^2

4(9/10)^2(3/5)^2= (c^2-(9/10)^2 - (3/5)^2)^2

I'll skip right to the solutions,

c^2=\dfrac{9}{100} \textrm{ or } c^2=\dfrac{9}{4}


c=\dfrac{3}{10} \textrm{ or } c=\dfrac{3}{2}

We could have gotten the 3/2 just by adding 9/10+3/5 but this was more fun.

6 0
3 years ago
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