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Karo-lina-s [1.5K]
3 years ago
9

Here is the histogram of a data distribution. All class widths are 1. What is the median of the distribution?

Mathematics
2 answers:
lidiya [134]3 years ago
8 0
First of all, we have to know what median is.

Median- the middle number.

So we have to organize the numbers in roofer from least to greatest, so we have to count how many 2,3,4,5,6s there are and so fourth.

I’ll have them listed for you.

2 2 2 2 3 4 5 6 7 7 7 8 8 8 8 9 9 9 9 10 10 10 10 10 10.

Next step is to find the middle number (s). If it has an even amount of numbers, we have to take the two middle numbers and divided by two.
If there are an odd amount of numbers, we just pick the middle number.
[i.e: 2 3 4; the middle number here is 3]
[i.e: 2 3 4 5; the middle number is 3+4=7 divided by 2 which equals 3.5]

Okay so let’s count how many numbers there are.
There are 25 numbers in total, so therefore there are 2 middle number (s).

So let’s divided 25/2 to get the two middle numbers, which is 12.5, therefore numbers 12-13 are middle numbers.

So let’s count what number is 12 and 13. When you count the numbers in order, you can tell that 8 and 8 are numbers of 8.

The final step is to just calculate by adding both numbers and dividing by two because we are trying to find the middle number(average number).

8 + 8 = 16
16/2= 8

The median for this question is 8.

Can you give me the brainliest? Or give me a heart? Thank you.

Please tell me if you are stuck in a part, if you need to contact me, please feel free to contact me on Brainly.com or on Instagram.
@jessmyybest
fredd [130]3 years ago
4 0

Answer:

8

Step-by-step explanation:

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Simplify.<br> y = (x + 1)2 -
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NO LINKS!!! Find the arc measure and arc length of AB. Then find the area of the sector ABQ.​
Norma-Jean [14]

Answer:

<u>Arc Measure</u>:  equal to the measure of its corresponding central angle.

<u>Formulas</u>

\textsf{Arc length}=2 \pi r\left(\dfrac{\theta}{360^{\circ}}\right)

\textsf{Area of a sector of a circle}=\left(\dfrac{\theta}{360^{\circ}}\right) \pi r^2

\textsf{(where r is the radius and the angle }\theta \textsf{ is measured in degrees)}

<h3><u>Question 39</u></h3>

Given:

  • r = 7 in
  • \theta = 90°

Substitute the given values into the formulas:

Arc AB = 90°

\textsf{Arc length of AB}=2 \pi (7) \left(\dfrac{90^{\circ}}{360^{\circ}}\right)=3.5 \pi=11.00\:\sf in\:(2\:d.p.)

\textsf{Area of the sector AQB}=\left(\dfrac{90^{\circ}}{360^{\circ}}\right) \pi (7)^2=\dfrac{49}{4} \pi=38.48\:\sf in^2\:(2\:d.p.)

<h3><u>Question 40</u></h3>

Given:

  • r = 6 ft
  • \theta = 120°

Substitute the given values into the formulas:

Arc AB = 120°

\textsf{Arc length of AB}=2 \pi (6) \left(\dfrac{120^{\circ}}{360^{\circ}}\right)=4\pi=12.57\:\sf ft\:(2\:d.p.)

\textsf{Area of the sector AQB}=\left(\dfrac{120^{\circ}}{360^{\circ}}\right) \pi (6)^2=12 \pi=37.70\:\sf ft^2\:(2\:d.p.)

<h3><u>Question 41</u></h3>

Given:

  • r = 12 cm
  • \theta = 45°

Substitute the given values into the formulas:

Arc AB = 45°

\textsf{Arc length of AB}=2 \pi (12) \left(\dfrac{45^{\circ}}{360^{\circ}}\right)=3 \pi=9.42\:\sf cm\:(2\:d.p.)

\textsf{Area of the sector AQB}=\left(\dfrac{45^{\circ}}{360^{\circ}}\right) \pi (12)^2=18 \pi=56.55\:\sf cm^2\:(2\:d.p.)

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SOMEONE PLS HELP ME WITH THIS QUESTION!!! I WILL MARK BRAINLIEST
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Answer:

The answer is 18% which is F

Step-by-step explanation:

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