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Lerok [7]
4 years ago
11

Use the data to find the interquartile range. [Note: Type your answers as numbers. Do not round.] Movie Ticket Fees at a Local T

heater $10.25 $9.75 $8.75 $12.75 $11.00 $6.50 . Range: $ ___a0 Interquartile Range: $ ___a1
Mathematics
1 answer:
Tanya [424]4 years ago
7 0

Answer:

The Inter quartile range is $2.25.

Step-by-step explanation:

The Inter quartile range (<em>IQR</em>) is a quantity which tells amount of data contained between the first quartile (<em>Q</em>₁) and third quartile (<em>Q</em>₃).

The IQR can be computed using the formula:

IQR=Q_{3}-Q_{1}

The first quartile (<em>Q</em>₁) is well defined as the mid-value between the minimum value and the median of the data set. Or the mid-value of the first half of the data set is the first quartile.

The third quartile (<em>Q</em>₃) is the mid-value amid the median and the maximum value of the data set. Or the mid-value of the second half of the data set is the Third quartile.

The data set in arranged order is:

S = {$6.50, $8.75, $9.75, $10.25, $11.00, $12.75}

The first half of the data set is:

S₁ = {$6.50, $8.75, $9.75}

The mid-value is $8.75.

The first quartile is, <em>Q</em>₁ = $8.75.

The second half of the data set is:

S₁ = {$10.25, $11.00, $12.75}

The mid-value is $11.00.

The first quartile is, <em>Q</em>₃ = $11.00.

Compute the Inter quartile range as follows:

IQR=Q_{3}-Q_{1}

        =11.00-8.75\\=2.25

Thus, the Inter quartile range is $2.25.

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2. Andy bought a television that was on sale
Marina CMI [18]

Answer:

Step-by-step explanation:

You will set up a portion tha put 35% over a 100 and 1270.75 on top of the other portion and cross multiply and divide .

6 0
3 years ago
A circle has a diameter of 25cm. How far from the centre of this circle is a chord 16cm long? Draw a picture before you start.
yan [13]

Answer:

The cord is 9.6 cm away from the centre of the circle.

Step-by-step explanation:

<u>See The Image First</u>

The image below shows a picture of everything that’s given, labeled, and drawn in the circle.

<u>Description of the Picture:</u>

You can use the Pythagorean to find the what the distance (the length <em>b</em>) the cord is from the center. You can use the radius as the hypotenuse. The radius is half of the diameter, and the diameter is 25. So the radius is 12.5, which means that 12.5 is the hypotenuse of the triangle (also known as side <em>c</em>). The whole length of the cord is 16cm long, and we can use half of the cord, which is 8cm, for the triangle. So we know two sides already: the hypotenuse (the radius) and the length <em>a</em> of the triangle (half of the length of the cord).

<u>To Solve (variables </u><em><u>a</u></em><u>, </u><em><u>b</u></em><u>, and</u><em><u> c</u></em><u>, are in italics so you don’t get confused):</u>

⇒ Use the formula of the Pythagorean Theorem (a^{2}+b^{2}=c^{2}), where <em>b </em>is the length of the distance from the cord to the centre of the circle (what we need to find out), <em>a </em>is the length of the side of the triangle (half of the cord size – 8cm), and <em>c </em>is the hypotenuse (the radius – 12.5cm). So plug the given values of <em>b</em> and <em>c</em> into the formula:

a^{2}+8^{2}=12.5^{2}

⇒ Solve for <em>a </em>by simplify:

a^{2}+64=156.25

⇒ Subtract 64 from both sides:

a^{2}=92.25

⇒ Find the square root of both sides to get<em> a</em> by itself:

\sqrt{a^{2}}=\sqrt{92.25}

⇒ Simplify to get:

a=\sqrt{92.25}  which convert’s to a=9.60468635...  round the answer to the nearest tenth is a=9.6

<u />

<u>Answer:</u> The cord is 9.6 cm away from the centre of the circle.

I hope you understand and that this helps with your question! :)

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Answer:

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Step-by-step explanation:

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I think it’s yes good luck!!!
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