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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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Answer:

The probability that A selects the first red ball is 0.5833.

Step-by-step explanation:

Given : An urn contains 3 red and 7 black balls. Players A and B take turns (A goes first) withdrawing balls from the urn consecutively.

To find : What is the probability that A selects the first red ball?

Solution :

A wins if the first red ball is drawn 1st,3rd,5th or 7th.

A red ball drawn first, there are E(1)= ^9C_2 places in which the other 2 red balls can be placed.

A red ball drawn third, there are E(3)= ^7C_2 places in which the other 2 red balls can be placed.

A red ball drawn fifth, there are E(5)= ^5C_2 places in which the other 2 red balls can be placed.

A red ball drawn seventh, there are E(7)= ^3C_2 places in which the other 2 red balls can be placed.

The total number of total event is S= ^{10}C_3

The probability that A selects the first red ball is

P(A \text{wins})=\frac{(^9C_2)+(^7C_2)+(^5C_2)+(^3C_2)}{^{10}C_3}

P(A \text{wins})=\frac{36+21+10+3}{120}

P(A \text{wins})=\frac{70}{120}

P(A \text{wins})=0.5833

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