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3 years ago

10

9, 2, 1, 5, 9, 7, 1, 2, 2, 5, 6 Question What is the median for the set of data

2 answers:

fenix001 [56]3 years ago

6 0

Count:11Sum:49Mean:4.4545454545455Median:5Mode:2

scZoUnD [109]3 years ago

4 0

To find the median average for a set of numbers, you first line them up by size.

So, the set will become 1, 1, 2, 2, 2, 5, 5, 6, 9, 9.

Now find the middle one. The problem is, there is no middle number. So, we take the two middle numbers of the set then find the mean average.

The two middle numbers are 2 and 5.

Add them up.

2 + 5 = 7

Then divide by 2/

7 / 2 = 3.5

So, 3.5 is the median average for this set.

You might be interested in

Work out 10.3 % of 853.93 km Give your answer rounded to 2 DP.

Vilka [71]

Answer: 87.95

Step-by-step explanation: 10.3% of 853.93

10.3%= 0.103

So you would do 0.103 times 853.93=87.95479 round that to 2 Dp. You get 87.95

7 0

3 years ago

masya89 [10]

So you subtract $22.89 from $54.73 to get a total of $31.84 and that’s how much the customer paid.

6 0

3 years ago

13. A glass can hold 3 cups of water. A bowl can

miv72 [106K]

Answer: 6

Explanation:
3 x 2 = 6

8 0

3 years ago

2/3 minus 16/32 what’s the answer

antiseptic1488 [7]

Answer:

1/6

Step-by-step explanation:

LCD = 96

2(32) - 16(3)/ 3(32) = 64 - 48/96

= 16/96

Simplify:

16/ 16 = 1

96/16 = 6

2/3 - 16/32 = 1/6

Sorry if my work might seem confusing. If it is, then try searching "fraction calculator." It really helps :)

6 0

3 years ago

maw [93]

The sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

<h3>How to find a sector area, and arc length?</h3>

For a sector that has a central angle of θ, and a radius of r;

The sector area, and the arc length are:

$A = \frac{\theta}{360} * \pi r^2$ --- sector area

$L = \frac{\theta}{360} * 2\pi r$ ---- arc length

<h3>How to find the given sector area, and arc length?</h3>

Here, the given parameters are:

Central angle, θ = 160

Radius, r = 5 inches

The sector area is

$A = \frac{\theta}{360} * \pi r^2$

So, we have:

$A = \frac{160}{360} * \frac{22}{7} * 5^2$

Evaluate

A = 34.92

The arc length is:

$L = \frac{\theta}{360} * 2\pi r$

So, we have:

$L = \frac{160}{360} * 2 * \frac{22}{7} * 5$

L = 13.97

Hence, the sector area and the arc length are 34.92 square inches and 13.97 inches, respectively

Read more about sector area and arc length at:

brainly.com/question/2005046

#SPJ1

8 0

2 years ago