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Marina CMI [18]
3 years ago
10

What is the interquartile range 53, 62, 67, 71, 83, 94, 102, 105

Mathematics
2 answers:
Leona [35]3 years ago
3 0
I believe it's 36.75

Hope this helped!
zalisa [80]3 years ago
3 0
83 is the answer because I like asked my teAcher rn and she said that's the answer
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What is 451 rounded to the nearest hundred?
USPshnik [31]

Answer:

500

Step-by-step explanation:

If the tens digit is 50 or above then you round up to the next hundred but if it is 49 or below you round down to the nearest hundred

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3 years ago
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Original price of a phone: $129.99<br> Tax: 2%
Solnce55 [7]

Answer:

$132.59 is the answer

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2 years ago
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What is the distance between point A(2,9) and B(-2,6)
Katyanochek1 [597]

Answer:

The distance among those points is 5 units

Step-by-step explanation:

See the picture please

6 0
3 years ago
For a school play, Rico bought 3 adult tickets and 5 child tickets for $40. Sasha bought 1 adult ticket and 5 child tickets for
Serhud [2]

The cost of one adult ticket is $7.5 and one child ticket is $3.5

Julia will pay $48 for 5 adult tickets and 3 child tickets.

Step-by-step explanation:

Let,

Cost of one adult ticket = x

Cost of one child ticket = y

According to given statement;

3x+5y=40    Eqn 1

x+5y=25      Eqn 2

Subtracting Eqn 2 from Eqn 1

(3x+5y)-(x+5y)=40-25\\3x+5y-x-5y=15\\2x=15

Dividing both sides by 2

\frac{2x}{2}=\frac{15}{2}\\x=7.5

Putting x=2.5 in Eqn 1

3(7.5)+5y=40\\22.5+5y=40\\5y=40-22.5\\5y=17.5

Dividing both sides by 5

\frac{5y}{5}=\frac{17.5}{5}\\y=3.5

The cost of one adult ticket is $7.5 and one child ticket is $3.5

5 adult tickets = 5*7.5 = $37.5

3 child tickets = 3*3.5 = $10.5

Total = 37.5+10.5 =$48

Julia will pay $48 for 5 adult tickets and 3 child tickets.

Keywords: linear equation, subtraction

Learn more about subtraction at:

  • brainly.com/question/3071107
  • brainly.com/question/3126500

#LearnwithBrainly

8 0
3 years ago
Ments
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
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