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3 years ago

What is the mean,median,mode(s) an range if the numbers are 15,30,17,15,24,31,15

Mathematics

2 answers:

Phoenix [80]3 years ago

7 0

The mean is 21. The median is 15. The mode is 15. The range is 16.

Leto [7]3 years ago

7 0

Hello There!

Mean: Average of numbers in a set of data.

STEP 1: Add the numbers together. In this case, we would add

15 + 30 + 17 + 15 + 24 + 31 + 15 = 147

STEP 2: Divide the sum by the number of digits you added. We would divide

147 "sum of numbers" by 7 "number of digits" and we would get an answer of

21.

The mean for the set of data is 21

Median: The middle number when all the numbers are listed in order.

To find the median, we would put the lowest number on the left and the biggest number on the right and continue to go from least top greatest.

-Least To Greatest-

15-15-15-17-24-30-31

In our set of numbers, we cross out one number from the left and the right each time until we come across one number in the middle. The number that is in the middle in 17.

The median for this set of data is 17

Mode: The number that occurs most often in a set of numbers.

15-15-15-17-24-30-31

Since the number 15 occurs three times, the mode for the set of data is 15.

The mode for this set of data is 15

Range: The difference between the highest and the lowest number.

15-15-15-17-24-30-31

The highest number in the set of numbers is 31 and the lowest number in this set is 15.

31 - 15 = 16

The range for this set of data is 16

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3 years ago

What is the slope of a line that passes through (–12,15) and (0,4)?

Romashka [77]

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5 0

3 years ago

Read 2 more answers

Find the radius of the circle with equation x²+y²+8x+8y+28=0

galina1969 [7]

<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

Radius: 2 units.

To solve the problem, using the given formula of a circle, we need to find its standard equation form which is equal to:

$(x-h)^{2}+(y-k)^{2}=r^{2}$

Where:

"h" and "k" are the coordinates of the center of the circle and "r" is its radius.

So, we need to complete the square for both variable "x" and "y".

The given equation is:

$x^{2}+y^{2}+8x+8y+28=0$

So, solving we have:

$x^{2}+y^{2}+8x+8y=-28$

$(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4$

$(x^2-(-4))+(y^2-(-4))=4$

Now, we have that:

$h=-4\\k=-4\\r=\sqrt{4}=2$

So,

Center: (-4,-4)

Radius: 2 units.

Have a nice day!

Note: I have attached a picture for better understanding.

3 0

3 years ago

Please help and show work

melisa1 [442]

Answer:

first one is simnple cross multiply and divide

12*7=4x

84=4x

x=21

2nd same process but more steps

7x=5(x+8)

7x=5x+40

7x-5x=40

2x=40

x=20

Step-by-step explanation:

4 0

3 years ago

Twice a number plus four equals 6

maks197457 [2]

Answer:

2(x) +4 = 6

2(x) +4 -4 = 6 - 4

2x = 2

2x/2 = 2/2

x = 1

Step-by-step explanation:

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3 years ago