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densk [106]
3 years ago
15

A - Frequency

Mathematics
2 answers:
Marina CMI [18]3 years ago
8 0

Answer:

Step-by-step explanation:

When we collect a large data we may find a single entry repeated.  In these cases we prepare frequency distribution with x = the item in one column and f = the no of times it repeats i.e. frequency in other column.

Similarly for class intervals also, we write as frequency to the right side of interval column which gives no of items which fall within the class.

This process ensures compact presenting of data.

Hence we have

a)The number of observations that fall in a class

answer:   Frequency

b) The relative frequency of a class multiplied by 100

answer:   Percentage.  Because when we express probability as a percentage we get total 100

c) The ratio of the frequency of a class to the total number of observations

answer:  Relative frequency

(Relative frequency also known as probability is frequency/total entries)

lina2011 [118]3 years ago
7 0
The Answer to your question is Relative Frequency!
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In the tenth month, 550 copies will be sold in the tenth month.

In a year, 780 copies will be sold.

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Please answer this thanks<br><br>(●´⌓`●)(●´⌓`●)(●´⌓`●)​
Karo-lina-s [1.5K]

Answer:

1) A

2) A

3) D

4) B

5) D

Step-by-step explanation:

<u>__________________________________________________________</u>

<u>FACTS TO KNOW BEFORE SOLVING :-</u>

If any 2 parallel lines are cut by a transversal line , then :-

  • its corresponding angles on same side [a pair of an interior angle and it's corresponding exterior angle (but not its adjacent exterior angle)] are equal.
  • its alternate interior angles are equal.
  • its alternate exterior angles are equal.

<u>__________________________________________________________</u>

Q1)

According to the figure on the right side of the question paper ,

  • ∠2 & ∠8 are interior angles on the same side
  • ∠2 = ∠7 (∵ Alternate interior angles are equal)

So,

∠7 + ∠8 = 180°

⇒ ∠2 + ∠8 = 180°    (∵ ∠2 = ∠7)

Hence , we can conclude that interior angles of same side are supplementary angles. So the correct option is A.

Q2)

According to the figure on the question paper ,

  • ∠5 & ∠3 are exterior angles on the same side
  • ∠1 = ∠5 (∵ Corresponding angles are equal)

So ,

∠1 + ∠3 = 180°

⇒ ∠5 + ∠3 = 180°  (∵ ∠1 = ∠5)

Hence , we can conclude that exterior angles on the same side are supplementary angles. So , the correct option is A.

Q3)

According to the figure , (∠2 , ∠7) & (∠1 , ∠8) are alternate interior angles. But as (∠1 , ∠8) is there as an option , so the correct option is D.

Q4)

According to the figure , m∠1 = 129°.

Also , ∠1 & ∠7 are interior angles on the same side.

⇒ They are supplementary angles.

⇒ ∠1 + ∠7 = 180°

⇒ ∠7 = 180° - 129° = 51°

So , the correct option is B.

Q5)

According to the figure , m∠2 = 3x - 10 and m∠6 = 2x + 20

Also , ∠2 = ∠6 (∵ Corresponding angles are equal)

⇒ 3x - 10 = 2x + 20

⇒ 3x - 2x = 20 + 10

⇒ x = 30

So , ∠2 = 3×30 - 10 = 80° = ∠6 (∵ Corresponding angles are equal)

Hence , the correct option is D.

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Six times the sum of a number and twelve is forty. Which equation represents the statement above? 6N + 12N = 40 6N + 12 = 40 6(N
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By de Moivre's theorem,

1 - i = \sqrt2\,e^{-i\pi/4} \implies (1-i)^2 = 2\,e^{-i\pi/2}

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where k\in\{0,1,2,3\}. The fourth roots of (1-i)^2 are then

k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}

k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}

k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}

k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}

or more simply

\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}

We can go on to put these in rectangular form. Recall

\cos^2(x) = \dfrac{1 + \cos(2x)}2

\sin^2(x) = \dfrac{1 - \cos(2x)}2

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\cos\left(-\dfrac\pi8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac{1 + \cos\left(\frac\pi4\right)}2} = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}

\sin\left(-\dfrac\pi8\right) = -\sin\left(\dfrac\pi8\right) = -\sqrt{\dfrac{1 - \cos\left(\frac\pi4\right)}2} = -\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}

\cos\left(\dfrac{3\pi}8\right) = \sin\left(\dfrac\pi8\right) = \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}

\sin\left(\dfrac{3\pi}8\right) = \cos\left(\dfrac\pi8\right) = \sqrt{\dfrac12 + \dfrac1{2\sqrt2}}

and the roots are equivalently

\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}

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Learn more in brainly.com/question/20162367

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